\filename{rfex.tex}
\versiondate{9.6.03}
\def\qnA{\needed{1A} Let $R$ be the set of continuous functions from
$[0,\infty[$ to
$\Bbb R$, with ordinary addition of functions and with a multiplication
$*$ given by
\centerline{$(f*g)(t) = \int_0^t f(u)g(t-u)du\enskip\forall\enskip
t\ge 0,\,f,\,g\in R$.}
\noindent Show that $R$ is a commutative ring. Show that $R$ has no
multiplicative identity. ({\it Hint.} Consider $(f*g)(0)$.)
}
\def\qnB{\needed{1A, 1B, 1C} (a) Define {\it ring.}
(b) Which of the following are rings? In each case (other than (v) and (viii))
state explicitly the operations you think it natural to use for $+$ and
$\times$. For each object which is not a ring name a condition in your
definition of \lq ring' which it does not satisfy.
\quad(i) upper triangular $3\times 3$ real matrices;
\quad(ii) orthogonal $3\times 3$ real matrices;
\quad(iii) $\Bbb N$;
\quad(iv) $\Bbb Z_{12}$;
\quad(v) $\Bbb R^3$ with ordinary vector addition and
cross-product multiplication (i.e., $(a_1,a_2,a_3)\times(b_1,b_2,b_3)
=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1)$):
\quad(vi) symmetric $3\times 3$ real matrices;
\quad(vii) the set $\Cal P\Bbb N$ of all subsets of $\Bbb N$;
\quad(viii) $\Bbb Z$ with ordinary addition and the mutiplication $*$, where $m*n=m + n - mn$.
}
\def\qnC{\needed{1A, 1D} (a) Define {\it ring,} {\it subring.}
(b) Let $(G,+)$ be an abelian group. Show that the set of group
homomorphisms from $G$ to itself is a ring under suitable operations (to
be described explicitly).
(c) Let $V$ be a vector space over a field $F$. Show that the set of
linear maps from $V$ to itself is a subring of the ring of homomorphisms
of the additive group $(V,+)$ to itself.
}
\def\qnD{\needed{1B, 1H, 1K} Define {\it ideal.} (a) Describe (with proof) the
construction of the
quotient ring $R/I$ where $R$ is a ring and $I$ is an ideal of $R$.
(b) If $R$ is a ring and $a\in R$, under what circumstances is $aR=\{ar:r\in R\}$ an ideal of $R$?
(c) If $R$ is a ring and $M_n(R)$ is the ring of $n\times n$ matrices with coefficients in $R$, show that any ideal of $M_n(R)$ is of the form $M_n(I)$ where $I$ is an ideal of $R$.
(d) Let $R$ be the ring of $2\times 2$ matrices with integer coefficients.
Let $I$ be the set of matrices in $R$ all of whose coefficients are
even. How many elements does $R/I$ have?
Is $R/I$ commutative? (Justify your answer.)
}
\def\qnE{\needed{1D, 1F, 1H, 1K, 1L} (a) Define {\it homomorphism} (between
rings), {\it subring, ideal,
isomorphism} (of rings).
(b) Describe (without proof) the construction of the quotient ring
$R/I$ where $R$ is a ring and $I$ is an ideal of $R$.
(c) Show that if $R$ and $S$ are rings and $\phi:R\to S$ is a
homomorphism then $I=\{r:\phi(r)=0\}$ is an ideal of $R$ and
$\phi[R]=\{\phi(r):r\in R\}$ is a subring of $S$.
(d) Show that $R/I$ is isomorphic to $\phi[R]$. ({\it Hint.} For
$s\in\phi[R]$ set $\theta(s)=\{r:\phi(r)=s\}$. Show that
$\theta(s)=r+I$ whenever $\phi(r)=s$. Show that $\theta:\phi[R]\to
R/I$ is an isomorphism.)
}
\def\qnF{\needed{1H, 1K, 2A} Define {\it ideal,} {\it ring of polynomials} over
a given ring.
Describe (without proof) the quotient ring $R/I$ where $R$ is a ring and
$I$ is an ideal of $R$.
Let $R$ be the ring of polynomials over the field $\Bbb R$. Let
$I\subseteq R$ be the ideal of multiples of the polynomial $x^2$. Show
that the quotient $R/I$ can be identified with the vector space $\Bbb
R^2$ for a suitable multiplication, to be described.
}
\def\qnG{\needed{1D, 2C} (a) Let $R$ be any ring. Let $e$ be a suitable
symbol and set
$R_1=\{ne+a:n\in\Bbb Z,\,a\in R\}$. Show that with appropriate
definitions of $+$ and $\times$ (to be described) $R_1$ is a ring with
{\bf 1} in which $R$ is embedded as a subring.
(b) Show that (i) if $R$ is commutative, so is $R_1$ (ii) if $R$ is an integral domain, so is $R_1$.
(c) Give an example in which $R$ is commutative and without zero divisors, but $R_1$ is not an integral domain.
}
\def\qnH{\needed{1A, 1H, 1K, 1L, 2A, 2C} (a) Define {\it ring,} {\it ideal,}
{\it integral domain.}
(b) Show that $R=C([0,1])$, the set of continuous
real-valued functions on $[0,1]$, is a ring with $\tbf{1}$. (Explicitly describe the
operations $+$, $\times$ you use and give clear statements of any
theorems in analysis on which you rely.) Is it an integral domain?
Justify your answer.
(c) Define $e\in R$ by setting $e(t)=t$ for $t\in[0,1]$. Let $I$ be
the ideal $eR$. Is $R/I$ an integral domain? Justify your answer.
(d) Show that $J=\{f:f\in R,\,f(0)=0\}$ is an ideal of $R$. Show that
$R/J$ is isomorphic to a familiar field.
[Standard theorems concerning quotient rings may be used without proof
but should be clearly stated.]
}
\def\qnI{\needed{2A, 2C, 2E} (a) What is meant by saying that a ring $R$ has a
\lq{\bf 1}' or
\lq{unity}'? Show that a ring can have at most one {\bf 1}. In a
ring with a {\bf 1}, what is an {\it invertible element} or \lq{unit}'? Show that an invertible element can have at most one inverse,
and that the invertible elements of a ring with {\bf 1} form a group.
(b)(i) Define {\it integral domain,} {\it field.}
\quad(ii) Show that a finite integral domain is a field.
\quad(iii) Give an example of an integral domain which is not a field,
and explain the point at which the argument you gave in (ii) fails when
applied to your example.
}
\def\qnAJ{\needed{1A, 2A, 2C, 2G} Define {\it ring,} {\it integral domain,}
{\it field.}
Explain, with proof, how the field of fractions of an integral domain is
constructed.
Describe the field of fractions of the ring $\{m+ni:m,\,n\in\Bbb Z\}$ of
Gaussian integers.
}
\def\qnAA{
11.\needed{1C, 2A, 3E} Let $R$ be a ring. Explain what is meant by
the ring
$R[x]_{\infty}$
of
formal power series with coefficients in $R$, and prove that it is a
ring.
If $F$ is a field, show that every element of $F[x]_{\infty}$ is of the
form $x^ne$ where $e$ is an invertible element of $F[x]_{\infty}$.
Hence
show that $F[x]_{\infty}$ is a Euclidean domain.
}
\def\qnAB{\needed{1B, 1C, 1D, 1H, 2C, 3E, 3F, 3G} Define {\it integral domain,}
{\it Euclidean domain,} {\it ideal}.
Show that if $R$ is a Euclidean domain and $I$ is an ideal of $R$ then
$I$ is of the form $aR$ for some $a\in R$.
Which of the following are integral or Euclidean domains?
\quad(i) matrices of the form
$\left(\matrix
a&b\\
0&a\endmatrix\right)$,
where $a$, $b\in\Bbb R$;
\quad(ii) $\Bbb Z$;
\quad(iii) the ring $\Bbb R[x]$ of polynomials over $\Bbb R$;
\quad(iv) the ring $\Bbb Z[x]$ of polynomials over $\Bbb Z$;
\quad(v) the ring $\Bbb R[x,y]$ of polynomials in two variables
over $\Bbb R$;
\quad(vi) the subring $2\Bbb Z$ of $\Bbb Z$.
\noindent Justify your answers.
}
\def\qnAC{\needed{1A, 1B, 1H, 3H} (a) Define {\it ring,} {\it ideal,} {\it
principal ideal.}
(b) Let $X$ be a set. Show that the set $\Cal PX$ of subsets of $X$ is
a ring under suitable operations (to be described explicitly). Does
$\Cal PX$ have a multiplicative identity?
(c) Show that every ideal of $\Cal PX$ is principal.
}
\def\qnAD{\needed{1C, 2A, 3H} (a) Explain (with an outline of the proof) how
the polynomial ring
$R[x]$ is constructed from a ring $R$.
(b) Define {\it principal ideal domain.} Show that if $R[x]$ is a
principal ideal domain then $R$ is a field.
}
\def\qnAE{\needed{1B, 1C, 3B, 3H, 3I} (a) Define {\it principal ideal domain.}
Show that if $R$ is a
principal ideal domain
and $a_1,\ldots,a_k\in R$ then there are $r_1,\ldots,r_k\in R$ such that
$r_1a_1+\ldots+r_ka_k$ is a factor of every $a_j$.
(b) Let {\bf A} be an $n\times n$ complex matrix, where $n\ge 2$. Show
that there is a
non-zero irreducible polynomial $p\in\Bbb C[x]$ such that
$p(\tbf{A})=\tbf{0}$. Expressing $p$ as
$(x-\lambda_1)^{m_1}\ldots
(x-\lambda_k)^{m_k}$, where each $\lambda_j\in\Bbb C$ and each $m_j\ge
1$, show that $\Bbb R^n=U_1\oplus\ldots\oplus U_k$, where
$U_j=\{\tbf{u}:\tbf{u}\in\Bbb R^n,\,
(\tbf{A}-\lambda_j\tbf{I})^{m_j}(\tbf{u})=\tbf{0}\}$.
Show that $\tbf{A}\tbf{u}\in U_j$ whenever $j\le k$, $\tbf{u}\in U_j$.
}
\def\qnAF{\needed{1K, 2A, 3B, 3E, 3F, 3H, 3I, 3L} Let $R$ be a principal ideal
domain.
(a) Show that for any $a$, $b\in R$ there are $r$, $s\in R$ such that
$ar+bs$ is a factor of both $a$ and $b$.
(b) Show that if $p\in R$ is irreducible then the quotient ring $R/pR$
is a field.
(c) Set
$R=\{m+n\omega:m,\,n\in\Bbb Z\}$
where
$\omega=-{1\over 2}+i{{\sqrt 3}\over 2}
\in\Bbb C$. Give a formula for a Euclidean valuation on $R$ (you
need not show that it is a valuation) and show that $\zeta=2+\omega$ is
a
prime in $R$. Describe the field $R/\zeta R$.
}
\def\qnAG{\needed{1D, 2A, 3B, 3E, 3F, 3L} Set $\omega=-{1\over 2}+i{{\sqrt
3}\over 2}
\in\Bbb C$, $R=\{m+n\omega:m,\,n\in\Bbb Z\}\subseteq\Bbb C$.
(a) Show that $R$ is a subring of $\Bbb C$.
(b) Show that $R$ is a Euclidean domain.
(c) Show that $2$ and $2+\omega$ are irreducible in $R$.
(d) Describe the field $R/2R$, giving its addition and multiplication
tables.
}
\def\qnAH{\needed{2A, 2C, 3B, 4A, 4B, 4C} (a) Define {\it unique factorisation
domain.} Show that an
integral domain $R$ is a unique factorisation domain iff (i)
irreducibles in $R$ are prime (ii) every
non-zero non-invertible element of $R$ is expressible as a product of
irreducibles.
(b)(i) Give an example (with proof) of an integral domain $R$ in which
every
non-zero non-invertible element of $R$ is expressible as a product of
irreducibles, but $R$ is not a unique factorisation domain;
\quad(ii) give an example (with proof) of an integral domain in which
there
are no irreducible elements (so that every irreducible is prime) but $R$
is not a field (so is not a unique factorisation domain).
}
\def\qnAI{\needed{2A, 2C, 2G, 3H, 4D} (a) Define {\it integral domain,} {\it
principal ideal domain.}
Describe
(without proof) the construction of the field of fractions of an
integral domain.
(b) State the Unique Factorisation Theorem for principal ideal domains.
(c) Let $R$ be a principal ideal domain, $F$ its field of fractions.
Suppose
$x\in F$ is of the form $q/p_1^{k_1}\ldots p_r^{k_r}$ where $q\in R$ and $p_1,\ldots,p_r$ are distinct irreducible
elements of $R$. Show that $x$ is expressible as
$$\sum_{j=1}^r{{q_j}\over{p_j^{k_j}}},$$
\noindent where $q_1,\ldots,q_r\in R$.
(d) Illustrate (c) by expressing ${{17}\over{60}}$ in the form ${l\over
3}+{m\over 4}+{n\over 5}$ where $l$, $m$, $n\in\Bbb Z$.
}
\def\qnBJ{\needed{2G, 3E, 3F, 3I, 4D} Let $R$ be a Euclidean domain with
Euclidean valuation $\nu$, and $F$ its field of fractions. Show that
any element of $F$ is uniquely expressible in the form
$$a+\sum_{j=1}^n{{q_j}\over{p_j^{k_j}}}$$
\noindent where $a\in R$, $q_1,\ldots,q_n\in R\setminus\{0\}$,
$p_1,\ldots,p_n\in R$ are irreducible elements of $R$,
$k_1,\ldots,k_n\ge 1$ are integers and $\nu(q_j)<\nu(p_j)$ for every
$j$. (It is understood that $n$ may be $0$.) Illustrate your proof
by tracing it step-by-step through the representation of
$${{x^4-x^3+3x^2-x+1}\over{x^5+x^3}}$$
\noindent as partial fractions in the field of rational functions with integer
coefficients.
}
\def\qnBA{\needed{3B, 3E, 3F, 3H, 4D} Define {\it Euclidean domain,} {\it
principal
ideal domain.} State
the Unique Factorisation
Theorem
for principal ideal domains.
Let $R$ be the ring
of Gaussian integers, viz. $\{m+ni:m,\,n\in\Bbb Z\}$. Show that $R$ is
a Euclidean domain.
Show that an element $m+ni$ of $R$ is irreducible in $R$ iff {\it either}
$m^2+n^2$ is a prime in $\Bbb Z$ {\it or} one of $|m|$, $|n|$ is
zero and the other is a prime in $\Bbb Z$ which is not the sum of two
squares in $\Bbb Z$.
}
\def\qnBB{\needed{1D, 3B, 3C, 3E, 3F, 3G, 3J, 4D} Define {\it Euclidean
domain,}
{\it irreducible} element of an integral domain, {\it prime} element of
an integral domain, {\it principal ideal domain.}
Show that primes are always irreducible, and that in a principal ideal
domain
irreducible elements are prime.
State the Unique Factorisation Theorem for principal ideal domains.
Let $R$ be the set
$\{m+n\omega:m,\,n\in\Bbb Z\}\subseteq\Bbb C$
where
$\omega=-{1\over 2}+i{{\sqrt 3}\over 2}$.
Show that $R$ is a subring of $\Bbb C$ and is a Euclidean domain.
Express $6$ as a product of prime factors in $R$.
}
\def\qnBC{\needed{2A, 2C, 2F, 3B, 3F} (a) Let $R$ be an integral domain. Show that if $a^2=b^2$ in $R$ then $a=\pm b$.
(b) Let $F$ be a finite field. Show that the product of the
non-zero elements of $F$ is $-1$. ({\it Hint.} Collect them, as far as possible, into inverse pairs.)
(c) Let $p\in\Bbb N$ be a prime. Show that $(p-1)!\cong -1$ mod $p$.
(d) Let $p\in\Bbb N$ be a prime of the form $4m+1$ where $m\in\Bbb N$; set $q=(p-1)/2$. Show that $(q!)^2\cong -1$ mod $p$.
(e) Let $R$ be the ring $\{m+ni:m,\,n\in\Bbb Z\}$ of Gaussian integers. Is $R$ is a unique factorisation domain? Justify your answer.
(f) Let $p\in\Bbb N$ be a prime of the form $4m+1$ where $m\in\Bbb N$. Show that $p$ is not a prime in $R$. ({\it Hint.} Find $n$, $k$ such that $(n-i)(n+i)=kp$.) Hence show that $p$ is expressible as the $a^2+b^2$ where $a$, $b\in\Bbb Z$. ({\it Hint.} Show that $p$ has a
non-trivial factor $r\in R$ and consider $|r|^2$.)
}
\def\qnBD{\needed{3B, 3E, 4E} Define {\it Euclidean domain,} {\it irreducible}
element. Show that if
$R$ is a Euclidean domain then any two elements $a$, $b\in R$ have a
greatest common divisor (in a sense which you should explain).
Show that the ring
$R=\{m+ni:m,\,n\in\Bbb Z\}$
of Gaussian integers is a Euclidean domain.
Find the greatest common divisor in $R$ of $3+i$ and $5$. Show that
$3$ and $7$ are irreducible in $R$.
}
\def\qnBE{\needed{2C, 3B, 3C, 3K, 4A, 4B, 4E, 4F} (a) Define the terms {\it
integral
domain,} {\it irreducible
element,} {\it unique factorisation domain,} {\it prime element,} {\it
greatest common divisor.}
(b) Show that in an integral domain all primes are irreducible and that
in a unique factorisation domain all irreducibles are prime.
(c) Give an example (with proof) of an integral domain which is not a
unique factorisation domain.
(d) Show that in a unique factorisation domain $R$
\quad(i) any non-zero $a_1,\ldots,a_k\in R$ have a greatest common
divisor in $R$;
\quad(ii) if $d$ is a greatest common divisor of $a_1,\ldots,a_k$ and
$b\in R\setminus\{0\}$ then $bd$ is a greatest common divisor of
$ba_1,\ldots,ba_k$.
}
\medskip
\def\qnBF{\needed{4H} (a) Let $R$ be a unique factorisation domain and $F$ its
field of fractions. Show that if a polynomial with coefficients in $R$
is irreducible in $R[x]$ then it is irreducible in $F[x]$.
(b) For each of the following polynomials either find a factorisation in
$\Bbb Q[x]$ or show that it is irreducible in $\Bbb Q[x]$:
\quad(i) $x^2+{1\over 2}x+1$;
\quad(ii) $x^3+{3\over 4}x+{1\over 2}$;
\quad(iii) $x^4+x^3+x^2+x+1$.
\noindent Are any of these polynomials irreducible in $\Bbb C[x]$?
Justify your answer.
}
\medskip
\def\qnBG{\needed{1C, 2A, 2G, 3B, 4A, 4G, 4H} Define {\it unique factorisation
domain.}
Let $R$ be a unique
factorisation domain and $F$ its field of fractions.
(a) Recall that a polynomial is {\it primitive} if its coefficients have
no
non-trivial common factor. Show that if $p$, $q$ are primitive
polynomials in the polynomial ring $R[x]$ then their product $pq$ is
also primitive.
(b) Show that $p\in R[x]$ is irreducible in $R[x]$ iff it is primitive
and irreducible in $F[x]$.
}
\medskip
\def\qnBH{\needed{1C, 2C, 2F, 3E, 3F, 3G, 3H, 4A, 4D, 4I} Define {\it integral
domain,}
{\it Euclidean domain,} {\it principal
ideal domain,} {\it unique factorisation domain.} For each of the
following rings state exactly which of the four definitions it
satisfies; briefly justify your answers.
\quad(i) $\Bbb R$;
\quad(ii) $\Bbb Q[x]$;
\quad(iii) $\Bbb C[x,y]$;
\quad(iv) $\Bbb Z$;
\quad(v) $\Bbb Z[x]$;
\quad(vi) $\Bbb Z_{12}$;
\quad(vii) $\Bbb Z_{13}$.
}
\def\qnBI{\needed{2A, 2B, 3F, 5A} A field $F$ is {\it of characteristic 2} if $1+1=0$.
(a) Let $F$ be a field which is not of characteristic $2$ and $K\subseteq F$ a subfield of $F$. Suppose that every element of $K$ has a square root in $F$. Show that every polynomial over $K$ of degree 2 is reducible in $F[x]$.
(b) Show that if $F$ is a finite field of characteristic 2 then every element of $F$ has a square root, but there is a quadratic of the form $x^2+ax+1$ which is irreducible in $F[x]$.
}
\def\qnCJ{\needed{1A, 1H, 2A, 2C, 2E, 5E} Define {\it ring,} {\it ideal,} {\it
integral domain.}
Prove that a finite integral domain is a field.
Describe a field with four elements, giving its addition and
multiplication tables.
}
\def\qnCA{\needed{1C, 2A, 2F, 3B, 5E} (a) Define {\it irreducible element} in a
ring. List the
irreducible quadratics and cubics in the ring $\Bbb Z_2[x]$ of
polynomials over the
field $\Bbb Z_2$.
(b) Is $x^4+x^2+1$ irreducible in $\Bbb Z_2[x]$? Is $x^4+x+1$
irreducible in $\Bbb Z_2[x]$? Justify your answers.
}
\def\qnCB{\needed{1C, 2A, 2F, 3B, 3L, 5F} Give the addition and multiplication
tables of the field $\Bbb Z_3$.
Show that $p=x^2+a$ is irreducible in the ring $R=\Bbb Z_3[x]$
for just one value of $a\in\Bbb Z_3$; what is it?
Give a list of the elements of the field $F=R/pR$. Show that
the group $F\setminus\{0\}$ is cyclic and find a generator for it.
}
\def\qnCC{\needed{1B, 2A, 2F, 5E, 5F} Let $F$ be a finite field. Show that there
is a
quadratic with
coefficients in $F$ which is irreducible in $F[x]$.
({\it Hint.} If $1+1\ne 0$, show that not every element of $F$ is a
square. If $1+1=0$, show that not every element of $F$ is of the form
$x(x+1)$.)
Find such quadratics for the fields $\Bbb Z_2$, $\Bbb Z_3$ and $\Bbb
Z_5$.
}
\def\qnCD{\needed{2A, 5A, 5H} Define {\it subfield.}
Suppose that $F$ is a field and $K\subseteq F$ is a subfield. Let
$c\in F$ be such that $c$ is a root of a quadratic with coefficients in
$K$. Show that $L=\{a+bc:a,\,b\in K\}$ is a subfield of $F$. Show
that if $x^3+px+q$ is a cubic with coefficients in $K$ which has no root
in $K$ then it has no root in $L$.
Hence show that there is no
straight-edge-and-compasses construction of an angle of $10^{\circ}$.
}
\def\ansA{(a) Defining $(f+g)(t)=f(t)+g(t)$ for $f$, $g\in R$, $t\ge 0$
we have $f+g\in R$ for all $f$, $g\in R$ (because the sum of continuous functions is continuous).
So this gives us a binary operation $+$ on $R$.
It is associative
and commutative because
addition on $\Bbb R$ is associative and commutative.
It has an identity $\tbf{0}$, the function with constant value $0$,
because
this function is continuous (so belongs to $R$) and $0$ is the identity
of
$(\Bbb R,+)$. Defining $(-f)(t)=-f(t)$ for $f\in R$, $t\ge 0$
we have $-f\in R$ for every $f\in R$ (because a scalar multiple of a
continuous
function is continuous) and $(-f)+f=\tbf{0}=f+(-f)$ for every $f\in R$.
Thus
$(R,+)$ is an abelian group.
\medskip
(b) The given definition of $f*g$ gives a continuous function (indeed, a
differentiable one, by the fundamental theorem of calculus) for all $f$,
$g\in R$, so is a binary operation.
To determine $(f*g)*h$, where $f$, $g$ and $h\in R$, consider
$$\eqalign{((f*g)*h)(t)&=\int_0^t(f*g)(u)h(t-u)du
\cr
&=\int_0^t\bigl(\int_0^vf(v)g(u-v)h(t-u)dv\bigr)du\cr
&=\int_0^t\bigl(\int_v^tf(v)g(u-v)h(t-u)du\bigr)dv
\cr
&=\int_0^t\bigl(\int_0^{t-v}f(v)g(w)h(t-v-w)dw\bigr)dv\cr
&=\int_0^tf(v)(g*h)(t-v)dv
\cr
&=(f*(g*h))(t).\cr}$$
\noindent (In the formulae above, we use the substitution $w=u-v$.)
Thus $*$ is associative.
\medskip
(c) To determine $f*(g+h)$ we have
$$\eqalign{(f*(g+h))(t)&=\int_0^tf(u)(g+h)(t-u)du\cr
&=\int_0^tf(u)g(t-u)du+\int_0^tf(u)h(t-u)du\cr
&=(f*g)(t)+(f*h)(t).\cr}$$
\noindent (The point here is that integration is additive.)
Thus $f*(g+h)=f*g + f*h$ for all $f$, $g$, $h\in R$. Similarly
$(f+g)*h=f*h + g*h$.
\medskip
(d) Multiplication is commutative because
$$\eqalign{(f*g)(t)&=\int_0^tf(u)g(t-u)du
=\int_0^tf(t-w)g(w)dw\cr
&=\int_0^tg(w)f(t-w)dw
=(g*f)(t).\cr}$$
\medskip
(e) Finally, there is a function $f\in R$ such that $f(0)=1$, but
for any $g\in R$ we have $(f*g)(0)=\int_0^0f(u)g(-u)du=0$, so that
$f*g\ne f$; thus no $g\in R$ can be a multiplicative identity.
}
\def\ansB{(a) A {\it ring} is a set $R$ together with two binary operations
$+$, $\cdot$ such that
\quad(i)($\alpha$) $x+y\in R$ for all $x$, $y\in R$;
\qquad ($\beta$) $(x+y)+z=x+(y+z)$ for all $x$, $y$, $z\in R$;
\qquad ($\gamma$) there is an additive identity $0\in R$ such that $x+0=
0+x=x$ for all $x\in R$;
\qquad ($\delta$) for every $x\in R$ there is an additive inverse $-x\in
R$
such
that $(-x)+x=x+(-x)=0$;
\qquad $(\epsilon)$ $x+y=y+x$ for all $x$, $y\in R$;
\quad (ii)($\alpha$) $xy\in R$ for all $x$, $y\in R$;
\qquad($\beta$) $(xy)z=x(yz)$ for all $x$, $y$, $z\in R$;
\quad (iii)($\alpha$) $x(y+z)=xy+xz$ for all $x$, $y$, $z\in R$;
\qquad($\beta)$ $(x+y)z=xz+yz$ for all $x$, $y$, $z\in R$.
\medskip
(b)(i) Taking ordinary matrix addition and multiplication, this is a
ring.
\quad(ii) Taking ordinary matrix addition and multiplication, this is
not a
ring; it fails (a-i-$\alpha$).
\quad(iii) Taking ordinary addition and multiplication, this is not a
ring;
it fails (a-i-$\gamma$).
\quad(iv) Taking addition and multiplication mod 12, this is a ring.
\quad(v) Not a ring; it fails (a-ii-$\beta$).
\quad(vi) Taking ordinary matrix addition and multiplication, this is
not a ring; it fails (a-ii-$\alpha$).
\quad(vii) Taking symmetric difference for addition and intersection
for
multiplication, this is a ring.
\quad(viii) Not a ring; it fails (a-iii-$\alpha$).
}
\def\ansCa{(a) A {\it ring} is a set $R$ together with two binary operations
$+$, $\cdot$ such that
\quad(i)($\alpha$) $x+y\in R$ for all $x$, $y\in R$;
\qquad ($\beta$) $(x+y)+z=x+(y+z)$ for all $x$, $y$, $z\in R$;
\qquad ($\gamma$) there is an additive identity $0\in R$ such that $x+0=
0+x=x$ for all $x\in R$;
\qquad ($\delta$) for every $x\in R$ there is an additive inverse $-x\in
R$ such that $(-x)+x=x+(-x)=0$;
\qquad $(\epsilon)$ $x+y=y+x$ for all $x$, $y\in R$;
\quad (ii)($\alpha$) $xy\in R$ for all $x$, $y\in R$;
\qquad($\beta$) $(xy)z=x(yz)$ for all $x$, $y$, $z\in R$;
\quad (iii)($\alpha$) $x(y+z)=xy+xz$ for all $x$, $y$, $z\in R$;
\qquad($\beta)$ $(x+y)z=xz+yz$ for all $x$, $y$, $z\in R$.
\noindent A {\it subring} of $R$ is a set $S\subseteq R$ such that
\qquad $x+y\in S$ for all $x$, $y\in S$;
\qquad $0\in S$;
\qquad $-x\in S$ for all $x\in S$;
\qquad $xy\in S$ for all $x$, $y\in S$.}%end of ansCa
\def\ansCb{(b) Let $R$ be the set of group homomorphisms from $G$ to itself.
\quad(i) For $\phi$, $\psi\in R$ define $\phi+\psi:G\to G$ by setting
$(\phi+\psi)(x)=\phi(x)+\psi(x)$ for all $x\in G$.
\qquad($\alpha$) If $\phi$, $\psi\in R$, $\phi+\psi\in R$.
\Prf\ For any $x$, $y\in G$ we have
$$\align(\phi+\psi)(x+y)&=\phi(x+y)+\psi(x+y)=(\phi(x)+\phi(y))+
(\psi(x)+\psi(y))\\
\intertext{because $\phi$ and $\psi$ are homomorphisms}
&=(\phi(x)+\psi(x))+(\phi(y)+\psi(y))
\\
\intertext{because $G$ is abelian}
&=(\phi+\psi)(x)+(\phi+\psi)(y).
\Qed\endalign$$
\qquad($\beta$) If $\psi$, $\phi$ and $\chi\in R$, then for any $x\in G$
we have
$$\align((\psi+\phi)+\chi)(x)&=(\phi+\psi)(x)+\chi(x)
=(\phi(x)+\psi(x))+\chi(x)\\
&=\phi(x)+(\psi(x)+\chi(x))=
(\phi+(\psi+\chi))(x),
\endalign$$
\noindent so $(\phi+\psi)+\chi=\phi+(\psi+\chi)$.
\qquad($\gamma$) Set $\tbf{0}(x)=0$, the identity of $G$, for every
$x\in G$.
Then $\tbf{0}(x+y)=0=\tbf{0}(x)+\tbf{0}(y)$ for all $x$, $y\in G$ so
$\tbf{0}\in R$. If now $\phi\in R$ we have
\centerline{$(\phi+\tbf{0})(x)=\phi(x)+\tbf{0}(x)=\phi(x)+0=\phi(x)=
0+\phi(x)=\tbf{0}(x)+\phi(x)=(\tbf{0}+\phi)(x)$}
\noindent for all $x\in G$, so $\phi+\tbf{0}=\phi=\tbf{0}+\phi$.
\qquad($\delta$) If $\phi\in R$, define $-\phi:G\to G$ by setting $(-
\phi)(x)
=-\phi(x)$ for all $x\in G$. Then for any $x$, $y\in G$
$$\align(-\phi)(x+y)&=-(\phi(x+y))=-(\phi(x)+\phi(y))
=-\phi(y)+(-\phi(x))\\
&=-\phi(x)+(-\phi(y))=(-\phi)(x)+(-\phi(y)),
\endalign$$
\noindent so $-\phi\in R$. Next, for any $x\in G$,
$$\align (\phi+(-\phi))(x)&=\phi(x)+(-\phi)(x)=\phi(x)-\phi(x)=
0=\tbf{0}(x)\\
&=-\phi(x)+\phi(x)=(-\phi+\phi)(x),
\endalign$$
\noindent so $\phi+(-\phi)=\tbf{0}=-\phi+\phi$.
\qquad($\epsilon$) If $\phi$, $\psi\in R$ and $x\in G$, then
\centerline{$(\phi+\psi)(x)=\phi(x)+\psi(x)=\psi(x)+\phi(x)
=(\psi+\phi)(x)$,}
\noindent so $\phi+\psi=\psi+\phi$.
\quad(ii) For $\phi$, $\psi\in R$ define $\phi\psi:G\to G$ by setting
$(\phi\psi)(x)=\phi(\psi(x))$ for all $x\in G$.
\qquad($\alpha$) $\phi\psi\in R$ for all $\phi$, $\psi\in R$ because the
composition of two group homomorphisms is a homomorphism.
\qquad($\beta$) $(\phi\psi)\chi=\phi(\psi\chi)$ for all $\phi$, $\psi$,
$\chi
\in R$ because composition of functions is associative.
\quad(iii)($\alpha$) If $\phi$, $\psi$, $\chi\in R$ and $x\in G$ then
$$\align(\phi(\psi+\chi))(x)&=\phi((\psi+\chi)(x))=
\phi(\psi(x)+\chi(x))=\phi(\psi(x))+\phi(\chi(x))
\\
&=(\phi\psi)(x)+(\phi\chi)(x)
=(\phi\psi+\phi\chi)(x);
\endalign$$
\noindent as $x$ is arbitrary, $\phi(\psi+\chi)=\phi\psi+\phi\chi$.
\qquad($\beta$)
If $\phi$, $\psi$, $\chi\in R$ and $x\in G$ then
\centerline{$((\phi+\psi)\chi)(x)=(\phi+\psi)(\chi(x))
=\phi(\chi(x))+\psi(\chi(x))
=(\phi\chi)(x)+(\psi\chi)(x)
=(\phi\chi+\psi\chi)(x)$;}
\noindent as $x$ is arbitrary, $(\phi+\psi)\chi=\phi\chi+\psi\chi$.
}%end of ansCB
\def\ansCc{(c) A function $T:V\to V$ is linear iff it is a homomorphism of the
additive group and moreover
\centerline{$T(\alpha v)=\alpha Tv$}
\noindent for all $v\in V$ and all $\alpha\in F$, the field of scalars.
\quad (i) If $S$, $T$ are linear then
\centerline{$(S+T)(\alpha v)=S(\alpha v)+T(\alpha v)
=\alpha Sv+\alpha Tv
=\alpha(Sv+Tv)
=\alpha(S+T)(v)
$}
\noindent for all $v\in V$, $\alpha\in F$. So $S+T$ is linear.
\quad(ii) Defining $\tbf{0}$ as above we have
\centerline{$\tbf{0}(\alpha v)=0=\alpha\tbf{0}(v)$}
\noindent for all $\alpha$, $v$, so $\tbf{0}$ is linear.
\quad(iii) If $T:V\to V$ is linear, then defining $-T$ as above we have
\centerline{$(-T)(\alpha v)=-T(\alpha v)=-\alpha Tv = \alpha(-Tv)
=\alpha(-T)v$}
\noindent for all $\alpha$, $v$, so $-T$ is linear.
\quad(iv) If $S$ and $T$ are linear then
\centerline{$(ST)(\alpha v)=S(T(\alpha v))=S(\alpha T(v))
=\alpha S(T(v)))=\alpha(ST)(v)$}
\noindent for all $\alpha$, $v$, so $ST$ is linear.
\medskip
}
\def\ansD{If $R$ is a ring, an {\it ideal} of $R$ is a set $I\subseteq R$
such that
\qquad $x+y\in I$ for all $x$, $y\in I$;
\qquad $0\in I$;
\qquad $-x\in I$ for all $x\in I$;
\qquad $ax$, $xa\in I$ for all $x\in I$, $a\in R$.
\medskip
(a)(i) Consider the relation $\equiv$ on $R$ defined by saying
\centerline{$x\equiv y$ if $x-y\in I$.}
\noindent Then $\equiv$ is an equivalence relation on $R$. \Prf\
($\alpha$) $x-x=0\in I$ so $x\equiv x$ for every $x\in I$.
($\beta$) If $x\equiv y$ then $y-x=-(x-y)\in I$ so $y\equiv x$.
($\gamma$) If $x\equiv y$ and $y\equiv z$ then $x-z=(x-y)+(y-z)\in I$
and
$x\equiv z$. \Qed
\medskip
\qquad(ii)
If $x\equiv x'$ and $y\equiv y'$ then $x+y\equiv x'+y'$ and $xy\equiv
x'y'$.
\Prf\ ($\alpha$) $(x+y)-(x'+y')=(x-x')+(y-y')\in I$. ($\beta$) $xy-x'y'
=(x-x')(y-y')+x'(y-y')+(x-x')y'\in I$.
\Qed
\qquad(iii)
Because $\equiv$ is an equivalence relation it splits $R$ into
equivalence
classes. Write $a^{\bullet}$ for the equivalence class of $a\in R$, so
that $a^{\bullet}=b^{\bullet}$ iff $a\equiv b$. Write $R/I$ for the
set
$\{a^{\bullet}:a\in R\}$ of equivalence classes.
Now part (ii) tells us that we can define addition and multiplication on
$R/I$ by the formulae
\centerline{$a^{\bullet}+b^{\bullet}=(a+b)^{\bullet},\enskip,
a^{\bullet}b^{\bullet}=(ab)^{\bullet}$.}
\qquad(iv) Thus we have two binary operations on $R/I$ and $R/I$ is
closed
under both. If $a$, $b$, $c\in R$ then
\qquad $(a^{\bullet}+b^{\bullet})+c^{\bullet}
=(a+b)^{\bullet}+c^{\bullet}
=((a+b)+c)^{\bullet}
=(a+(b+c))^{\bullet}
=a^{\bullet}+(b+c)^{\bullet}
=a^{\bullet}+(b^{\bullet}+c^{\bullet})$.
\qquad $a^{\bullet}+0^{\bullet}
=(a+0)^{\bullet}
=a^{\bullet}
=(0+a)^{\bullet}
=0^{\bullet}+a^{\bullet}$.
\qquad $a^{\bullet}+(-a)^{\bullet}
=(a+(-a))^{\bullet}
=0^{\bullet}
=(-a+a)^{\bullet}
=(-a)^{\bullet}+a^{\bullet}$.
\qquad $a^{\bullet}+b^{\bullet}
=(a+b)^{\bullet}
=(b+a)^{\bullet}
=b^{\bullet}+a^{\bullet}$.
\qquad $(a^{\bullet}b^{\bullet})c^{\bullet}
=(ab)^{\bullet}c^{\bullet}
=((ab)c)^{\bullet}
=(a(bc))^{\bullet}
=a^{\bullet}(bc)^{\bullet}
=a^{\bullet}(b^{\bullet}c^{\bullet})$.
\qquad $(a^{\bullet}+b^{\bullet})c^{\bullet}
=(a+b)^{\bullet}c^{\bullet}
=((a+b)c)^{\bullet}
=(ac+bc))^{\bullet}
=(ac)^{\bullet}+(bc)^{\bullet}
=(a^{\bullet}c^{\bullet})+(b^{\bullet}c^{\bullet})$.
\qquad $a^{\bullet}(b^{\bullet}+c^{\bullet})
=a^{\bullet}(b+c)^{\bullet}
=(a(b+c))^{\bullet}
=(ab+ac)^{\bullet}
=(ab)^{\bullet}+(ac)^{\bullet}
=(a^{\bullet}b^{\bullet})+(a^{\bullet}c^{\bullet})$.
\noindent Thus $R/I$ is a ring, with zero $0^{\bullet}$ and negatives
$-(a^{\bullet})=(-a)^{\bullet}$.
\medskip
(b)
$aR$ is always a subring of $R$; it is an ideal iff $bx\in aR$ whenever
$x\in aR$, $b\in R$; a sufficient condition for this is that $ab=ba$
for
every $b\in R$.
\medskip
(c){\bf Apology:} Question should read `If $R$ is a ring with $\tbf{1}
\ldots$'.
For matrices {\bf A} in $M_n(R)$, let their coefficients be $a_{ij}$.
If $J$ is an ideal of $M_n(R)$,
set $I=\{a_{11}:\tbf{A}\in J\}\subseteq R$. If
$\tbf{A}+\tbf{B}=\tbf{C}$
then $c_{11}=a_{11}+b_{11}$, so $I$ is closed under addition.
Considering
the zero matrix {\bf 0}, we see that $0\in I$. If $\tbf{A}\in J$ then
$-\tbf{A}\in J$; so if $a\in I$ then $-a\in I$. If $a\in I$ and $b\in
R$,
take some $\tbf{A}\in J$ such that $a=a_{11}$, and let $\tbf{B}\in
M_n(R)$
be the matrix with top left coefficient $b$, all others zero; then
$\tbf{AB}$, $\tbf{BA}$ have top left coefficients $ab$, $ba$
respecively, and
belong to $J$, so $ab$ and $ba$ belong to $I$.
Thus $I$ is an ideal of $R$. For $i$, $j\le n$ let $\tbf{E}_{ij}$
be the matrix with $(i,j)$ coefficient $1$, all others $0$.
Now observe that if $\tbf{A}\in J$ and $i$,
$j\le n$, then
$\tbf{E}_{1i}\tbf{A}\tbf{E}_{j1}$
belongs to $J$ and has top left
coefficient $a_{ij}$. Thus $a_{ij}\in I$. As $\tbf{A}$, $i$ and $j$
are arbitrary, $J\subseteq M_n(I)$.
Finally, if $\tbf{A}\in M_n(I)$, then for each $(i,j)$ choose
$\tbf{A}_{ij}\in
J$ such that the top left coefficient of $\tbf{A}_{ij}$ is $a_{ij}$.
Now
$\tbf{E}_{i1}\tbf{A}_{ij}\tbf{E}_{1j}$
belongs to $J$ and has $(i,j)$ coefficient
$a_{ij}$, all others $0$; so $\tbf{A}
=\sum_{i,j\le n}\tbf{E}_{i1}\tbf{A}_{ij}\tbf{E}_{1j}
\in J$. Thus $J$ is precisely
$M_n(I)$.
\medskip
(d) Every member of $R$ can be uniquely expressed as the sum of a member
of $I$ and a zero-one matrix. So there are $2^4=16$ equivalence
classes.
Because $\tbf{E}_{11}\tbf{E}_{12}=\tbf{E}_{12}\notin I$,
and $\tbf{E}_{12}\tbf{E}_{11}
=\tbf{0}$,
$\tbf{E}_{11}^{\bullet}\tbf{E}_{12}^{\bullet}
\ne\tbf{E}_{12}^{\bullet}\tbf{E}_{11}^{\bullet}$, so $R/I$ is not
commutative.
}
\def\ansE{(a) If $R$ and $S$ are rings, a function $\phi:R\to S$ is a
{\it homomorphism} if $\phi(x+y)=\phi(x)+\phi(y)$,
$\phi(xy)=\phi(x)\phi(y)$
for all $x$, $y\in R$.
If $R$ is a ring, a {\it subring} of $R$ is a set $S\subseteq R$ such
that
$0\in S$ and $x+y$, $-x$, $xy\in S$ for all $x$, $y\in S$.
An {\it ideal} of $R$ is a set $I\subseteq R$ which is a subring and
moreover
is such that $ax$, $xa\in I$ for all $x\in I$, $a\in R$.
If $R$ and $S$ are rings, a function $\phi:R\to S$ is an {\it
isomorphism}
if it is a bijective homomorphism.
\medskip
(b) If $R$ is a ring and $I$ an ideal of $R$, we have an equivalence
relation
$\equiv$ on $R$ given by $x\equiv y$ if $x-y\in I$. If $x\equiv x'$
and
$y\equiv y'$ then $x+y\equiv x'+y'$ and $xy\equiv x'y'$. Consequently
we may define addition and multiplication on the set $R/I$ of
equivalence
classes under $\equiv$ by setting
$x^{\bullet}+y^{\bullet}=(x+y)^{\bullet}$,
$x^{\bullet}
y^{\bullet}=(xy)^{\bullet}$ for all $x$, $y\in R$, writing $x^{\bullet}
\in R/I$ for the equivalence class of $x\in R$. Under this
multiplication and
addition $R/I$ is a ring.
\medskip
(c)(i) If $x$, $y\in I$ and $a\in R$ then
\qquad $\phi(x+y)=\phi(x)+\phi(y)=0+0=0$, so $x+y\in I$;
\qquad $\phi(0)=0$, so $0\in I$;
\qquad
$\phi(-x)=-\phi(x)=-0=0$, so $-x\in I$;
\qquad $\phi(xa)=\phi(x)\phi(a)=0\phi(a)=0$, so $xa\in I$;
\qquad $\phi(ax)=\phi(a)\phi(x)=\phi(a)0=0$, so $ax\in I$.
\noindent Thus $I$ is an ideal.
\medskip
\quad (ii) If $a$, $b\in \phi[R]$, take $x$, $y\in R$ such that
$\phi(x)=a$
and $\phi(y)=b$; then
\qquad $a+b=\phi(x+y)\in\phi[R]$;
\qquad $0=\phi(0)\in\phi[R]$;
\qquad $-a=\phi(-x)\in\phi[R]$;
\qquad $ab=\phi(xy)\in\phi[R]$.
\noindent Thus $S$ is a subring.
\medskip
(d) If $\phi(r)=s$ then $\theta(s)=\{x:\phi(r)=\phi(x)\}
=\{x:\phi(x-r)=\phi(x)+\phi(-r)=0\}
=\{x:x-r\in I\}
=\{x:x\equiv r\}=r^{\bullet}$. Thus $\theta(s)\in R/I$ for all
$s\in\phi[R]$. If $s_1$, $s_2\in\phi[R]$ take $r_1$, $r_2$ such that
$\phi(r_1)=s_1$ and $\phi(r_2)=s_2$; then $\phi(r_1r_2)=s_1s_2$ so
$\theta(s_1s_2)=(r_1r_2)^{\bullet}=r_1^{\bullet}r_2^{\bullet}
=\theta(s_1)\theta(s_2)$; similarly $\theta(s_1+s_2)=\theta(s_1)
+\theta(s_2)$.
Thus $\theta$ is a homomorphism. Also, if $\theta(s_1)=\theta(s_2)$,
then $r_1^{\bullet}=r_2^{\bullet}$ so $r_1\equiv r_2$ and $r_1-r_2\in I$
and $s_1-s_2=\phi(r_1-r_2)=0$ and $s_1=s_2$; thus $\theta$ is
injective.
Finally, every member of $R/I$ is of the
form $r^{\bullet}=\theta(\phi(r))$ so $\theta$ is surjective.
So $\theta$ is an isomorphism.
}
\def\ansF{No answer written out.}
\def\ansG{ Note first that we can define $na$, for $n\in\Bbb Z$ and $a\in R$,
as follows. First, take $0a=0$; next, set $(n+1)a=na + a$ for each
integer $n\ge 0$; this defines $na$ for all integers $n\ge 0$. It is
easy to prove by induction that $(m+n)a=ma+na$, $m(a+b)=ma+mb$, $(mn)a=
m(na)$ and $m(ab)=(ma)b=a(mb)$
for all $a$, $b\in R$ and all integers $m$, $n\ge 0$. Next, for
integers $m<0$ and $a\in R$, define $ma$ to be $-((-m)a)$, where $(-m)a$
is defined as just above and $-((-m)a)$ is the additive inverse of $(-
m)a$
in $R$. A straightforward check (considering several cases
individually)
shows that now $m(a+b)=ma+mb$, $(m+n)a=ma+na$, $(mn)a=m(na)$ and
$m(ab)=(ma)b=a(mb)$ for all
$m$, $n\in\Bbb Z$ and $a$, $b\in R$.
Define $+$, $\times$ on $R_1$ by setting
\centerline{$(me+a)+(ne+b)=(m+n)e+(a+b)$,}
\centerline{$(me+a)\times(ne+b)=mne+(mb+na+ab)$,}
\noindent using the rules above to define $ma$, $nb$. A routine check
shows that this makes $R_1$ a ring; for instance, for the associative
law of multiplication, we have
$$\align ((me+a)\times(ne+b))\times(pe+c)
&=(mne+(mb+na+ab))\times(pe+c)\\
&=mnpe+mnc+p(mb+na+ab)+(mb+na+ab)c\\
&=mnpe + npa + mpb + mnc + mbc + nac + pab + abc\\
&=mnpe + m(nc + pb + bc) + npa + a(nc+pb+bc)\\
&=(me+a)\times(npe+(nc+pb+bc))\\
&=(me+a)\times((ne+b)\times(pe+c)).
\endalign$$
To see that $R_1$ has a $\tbf{1}$, consider $\tbf{1}=1e+0$; then
$\tbf{1}\times(me+a)=1me+1a+m0+0a=me+a=m1e+m0+1a+a0=(me+a)\times
\tbf{1}$ for all $m$, $a$, so that $\tbf{1}$ is a multiplicative
identity
of $R_1$, and of course it is not the zero of $R_1$, which is $0e+0$.
The map $a\mapsto 0e+a:R\to R_1$ is an injective ring homomorphism, so
its
image is a subring of $R_1$ which is a copy of $R$.
\medskip
{\bf Apology:} It is not true that if $R$ is an integral domain so is
$R_1$.
Consider, for instance, the case $R=\Bbb Z$. Then we see that
\centerline{$(1e+0)\times(0e+(-1))=0e+0$,}
\noindent so that $R_1$ has zero divisors.
\medskip
}
\def\ansH{(a)
A {\it ring} is a set $R$ together with two binary operations
$+$, $\cdot$ such that
\quad(i)($\alpha$) $x+y\in R$ for all $x$, $y\in R$;
\qquad ($\beta$) $(x+y)+z=x+(y+z)$ for all $x$, $y$, $z\in R$;
\qquad ($\gamma$) there is an additive identity $0\in R$ such that $x+0=
0+x=x$ for all $x\in R$;
\qquad ($\delta$) for every $x\in R$ there is an additive inverse $-x\in
R$
such
that $(-x)+x=x+(-x)=0$;
\qquad $(\epsilon)$ $x+y=y+x$ for all $x$, $y\in R$;
\quad (ii)($\alpha$) $xy\in R$ for all $x$, $y\in R$;
\qquad($\beta$) $(xy)z=x(yz)$ for all $x$, $y$, $z\in R$;
\quad (iii)($\alpha$) $x(y+z)=xy+xz$ for all $x$, $y$, $z\in R$;
\qquad($\beta)$ $(x+y)z=xz+yz$ for all $x$, $y$, $z\in R$.
An {\it ideal} of a ring $R$ is a set $I\subseteq R$ such
that
\qquad $x+y\in I$ for all $x$, $y\in I$;
\qquad $0\in I$;
\qquad $-x\in I$ for all $x\in I$;
\qquad $ax$, $xa\in I$ for all $x\in I$, $a\in R$.
An {\it integral domain} is a ring $R$ such that
\qquad multiplication in $R$ is commutative;
\qquad
$R$ has a multiplicative identity $1\ne 0$;
\qquad if $ab=0$ then at least one of $a$, $b$ is $0$.
\medskip
(b)
For $f$, $g\in R$ define $f+g:[0,1]\to\Bbb R$ and $f\times g:[0,1]\to
\Bbb R$ by setting $(f+g)(t)=f(t)+g(t)$, $(f\times g)(t)=f(t)g(t)$ for
all
$t\in [0,1]$. Then $f+g$, $f\times g$ belong to $R$ because the sum
and
product of continuous functions is always continuous.
Addition in $R$ is associative
and commutative because
addition on $\Bbb R$ is associative and commutative.
It has an identity $\tbf{0}$, the function with constant value $0$,
because
this function is continuous (so belongs to $R$) and $0$ is the identity
of
$(\Bbb R,+)$. Defining $(-f)(t)=-f(t)$ for $f\in R$, $t\ge 0$
we have $-f\in R$ for every $f\in R$ (because a scalar multiple of a
continuous
function is continuous) and $(-f)+f=\tbf{0}=f+(-f)$ for every $f\in R$.
Thus
$(R,+)$ is an abelian group.
Multiplication in $R$ is associative
and commutative because
mulitplication in $\Bbb R$ is associative and commutative.
It has an identity $\tbf{1}$, the function with constant value $1$,
because
this function is continuous (so belongs to $R$) and $1$ is the identity
of
$(\Bbb R,.)$; and of course $\tbf{1}\ne\tbf{0}$ because $[0,1]$ is not
empty.
Finally multiplication in $R$ is distributive over addition because
multiplication in $\Bbb R$ is distributive over addition.
$R$ is not an integral domain because, for instance, $f\times g=\tbf{0}$
where
\centerline{$f(t) = 1-2t$ for $t\in[0,{1\over 2}]$, $0$ for $t\in
[{1\over 2},1]$,}
\centerline{$g(t) = 0$ for $t\in[0,{1\over 2}]$, $2t-1$ for $t\in
[{1\over 2},1]$.}
\medskip
(c) $R/I$ is not an integral domain because if we take $f(t)=\sqrt{t}$
for
$t\in[0,1]$, then $f\notin I$ but $f\times f=e\in I$; consequently,
writing
$f^{\bullet}\in R/I$ for the equivalence class of $f$,
we have $f^{\bullet}\ne 0^{\bullet}$, the zero of $R/I$, while
$f^{\bullet}
\times f^{\bullet}=e^{\bullet}=0^{\bullet}$.
\medskip
(d) If $f$, $g\in J$ and $h\in R$ then $(f+g)(0)=f(0)+g(0)=0$, $(-f)(0)=
-f(0)=0$, $(f\times h)(0)=(h\times f)(0)=
f(0)h(0)=0$; thus $f+g$, $-f$, $f\times h$
all belong to $J$; also $\tbf{0}(0)=0$ so $\tbf{0}\in J$.
Defining $\phi:R\to\Bbb R$ by $\phi(f)=f(0)$ we see that $\phi$ is a
homomorphism with kernel $J$ and image $\Bbb R$. By the First
Isomorphism
Theorem, $R/J$ is isomorphic to $\Bbb R$.
}
\def\ansI{(a)
A $\tbf{1}$ is a multiplicative identity not equal to $0$.
If $e$, $e'$ are both multiplicative identities in a ring $R$, then
$e=ee'=
e'$; so that $R$ can have at most one multiplicative identity.
In a ring $R$ with $\tbf{1}$, an {\it invertible element} is
an element $x\in R$ with an {\it inverse} $y$ such that $xy=yx=\tbf{1}$.
If $y$, $y'$ are both inverses of $x$, then
\centerline{$y=y1=y(xy')=(yx)y'=1y'=y'$,}
\noindent so $y=y'$.
If $x$, $y$ are both invertible, with inverses $x^{-1}$ and $y^{-1}$,
then
\centerline{$(xy)(y^{-1}x^{-1})=x(yy^{-1})x^{-1}=x1x^{-1}=xx^{-1}=1$,}
\centerline{$(y^{-1}x^{-1})xy)=y^{-1}(x^{-1}x)y=y^{-1}1y=y^{-1}y=1$,}
\noindent so $xy$ is also invertible. Thus the set $U$ of invertible
elements
is closed under multiplication. $11=1$ so $1\in U$
and is the identity of $U$. If $x\in U$ then $xx^{-1}=x^{-1}x=1$ so
$x$ is the inverse of $x^{-1}$ and $x^{-1}\in U$; of course $x^{-1}$ is
the
inverse of $x$ in $U$, so every element of $U$ has an inverse in $U$ and
$U$ is a group.
\medskip
(b)
(i) An {\it integral domain} is a ring $R$ such that
\qquad multiplication in $R$ is commutative;
\qquad
$R$ has a $\tbf{1}$;
\qquad if $ab=0$ then at least one of $a$, $b$ is $0$.
A {\it field} is a commutative ring with $\tbf{1}$ in which every
non-zero element is invertible.
\medskip
(ii) If $R$ is a finite integral domain, take any non-zero $a\in R$.
Consider the map $x\mapsto ax:R\to R$. If $ax=ax'$ then $a(x-x')=
ax-ax'=0$; because $R$ is an integral domain, $x-x'=0$, i.e., $x=x'$.
Thus the map $x\mapsto ax$ is injective. Because $R$ is finite, it
must also be surjective. In particular, there is an $x\in R$ such that
$ax=1$, the identity of $R$. Because $R$ is commutative, $xa=1$ and
$x$ is the inverse of $a$. As $a$ is arbitrary, $R$ is a field.
\medskip
(iii) $\Bbb Z$ is an integral domain but not a field; 2 has no
inverse.
If we take $R=\Bbb Z$, $a=2$ in the argument above, the map $x\mapsto
ax=2x$
is still injective, but it is not surjective; this is possible because
$\Bbb Z$ is infinite.
}
\def\ansAJa{(a) A {\it ring} is a set $R$ together with two binary operations
$+$, $\cdot$ such that
\quad(i)($\alpha$) $x+y\in R$ for all $x$, $y\in R$;
\qquad ($\beta$) $(x+y)+z=x+(y+z)$ for all $x$, $y$, $z\in R$;
\qquad ($\gamma$) there is an additive identity $0\in R$ such that $x+0=
0+x=x$ for all $x\in R$;
\qquad ($\delta$) for every $x\in R$ there is an additive inverse
$-x\in R$ such that $(-x)+x=x+(-x)=0$;
\qquad $(\epsilon)$ $x+y=y+x$ for all $x$, $y\in R$;
\quad (ii)($\alpha$) $xy\in R$ for all $x$, $y\in R$;
\qquad($\beta$) $(xy)z=x(yz)$ for all $x$, $y$, $z\in R$;
\quad (iii)($\alpha$) $x(y+z)=xy+xz$ for all $x$, $y$, $z\in R$;
\qquad($\beta)$ $(x+y)z=xz+yz$ for all $x$, $y$, $z\in R$.
\medskip
An {\it integral domain} is a ring $R$ such that
\qquad multiplication in $R$ is commutative;
\qquad
$R$ has a $\tbf{1}$ (that is, a multiplicative identity not equal to
$0$);
\qquad if $ab=0$ then at least one of $a$, $b$ is $0$.
\medskip
A {\it field} is a commutative ring with $\tbf{1}$ in which every
non-zero element is invertible.
}%end of ansAJa
\def\ansAJb{(b) Let $R$ be an integral domain.
\medskip
\quad(i) Let us observe first that we have a cancellation law: if
$ab=ac$ and $a\ne 0$ then $b=c$. \Prf\ $0=ab-ac=a(b-c)$; as $a\ne 0$,
$b-c$ must be $0$, i.e., $b=c$. \Qed
\medskip
\quad(ii) Let $X$ be the set $R\times(R\setminus\{0\})$. On $X$ define
a relation $\sim$ by saying
\centerline{$(a,b)\sim(a',b')$ if $ab'=ba'$.}
\noindent Then $\sim$ is an equivalence relation on $X$. \Prf\
($\alpha$) $ab=ba$ so $(a,b)\sim(a,b)$ for all $(a,b)\in X$. ($\beta$)
If $(a,b)\sim(a',b')$ then $b'a=ab'=ba'=a'b$ so $(a',b')\sim(a,b)$.
($\gamma$) If $(a,b)\sim(a',b')$ and $(a',b')\sim(a'',b'')$
then $b'ab''=ab'b''=ba'b''=bb'a''=b'ba''$; as $b'\ne 0$, $ab''=ba''$
and $(a,b)\sim(a'',b'')$. \Qed
\medskip
\quad(iii) Let $F$ be the set of equivalence classes in $X$ for $\sim$.
For $(a,b)\in X$ write ${a\over b}\in F$ for the equivalence class
containing $(a,b)$; thus ${a\over b}={{a'}\over{b'}}$ iff $ab'=ba'$.
\medskip
\quad(iv) If ${a\over b}={{a'}\over {b'}}$ and ${c\over
d}={{c'}\over{d'}}$ then ${{ad+bc}\over{bd}}={{a'd'+b'c'}\over{b'd'}}$.
\Prf\ Note first that as $bd\ne 0$, $b'd'\ne 0$, ${{ad+bc}\over{bd}}$
and ${{a'd'+b'c'}\over{b'd'}}$ are defined in $F$. Now examine
$$\eqalign{(ad+bc)b'd'&=adb'd'+bcb'd'=ab'dd'+bb'cd'=ba'dd'+bb'dc'
\cr
&=bda'd'+bdb'c'=bd(a'd'+b'c'). \Qed}$$
\noindent Consequently we can define an operation $+$ on $F$ by writing
\centerline{${a\over b}+{c\over d}={{ad+bc}\over{bd}}$}
\noindent whenever $(a,b)$, $(c,d)\in X$.
\medskip
\quad(v) If ${a\over b}={{a'}\over {b'}}$ and ${c\over
d}={{c'}\over{d'}}$ then ${{ac}\over{bd}}={{a'c'}\over{b'd'}}$. \Prf\
As before, ${{ac}\over{bd}}$ and ${{a'c'}\over{b'd'}}$ are defined in
$F$ because $bd$ and $b'd'$ are non-zero.
Now
\centerline{$acb'd'=ab'cd'=ba'dc'=bda'c'$,}
\noindent as required. \Qed Consequently we may define an operation
$\times$ on $F$ by writing
\centerline{${a\over b}\times{c\over d}={{ac}\over{bd}}$}
\noindent whenever $(a,b)$ and $(c,d)$ belong to $X$.
\medskip
\quad(vi) We have already seen that $+$ is a binary operation on $F$
taking values in $F$. Now
\qquad($\alpha$) ${a\over b}+({c\over d}+{e\over f})={a\over b}+
{{cf+de}\over{df}}={{adf+b(cf+de)}\over{bdf}}={{adf+bcf+bde}\over{bdf}}
={{(ad+bc)f+bde}\over{bdf}}
={{ad+bc}\over{bd}}+{e\over f}
=({a\over b}+{c\over d})+{e\over f}$, so addition on $F$ is associative.
\qquad($\beta$) Writing $0_F={0\over 1}$, we have
\centerline{${a\over b}+0_F={{a1+b0}\over{b1}}={a\over
b}={{0b+1a}\over{1b}}=0_F+{a\over b}$}
\noindent for all $(a,b)\in X$, so $0_F$ is an additive identity for
$F$.
\qquad($\gamma$) If $(a,b)\in X$, then
\centerline{${a\over b}+{{-a}\over b}={{ab+b(-a)}\over{b^2}}={{ab-
ba}\over{b^2}}={0\over{b^2}}
={{-ab+ab}\over{b^2}}={{(-a)b+ab}\over{b^2}}={{-a}\over b}+{a\over b}$,}
\noindent while ${0\over{b^2}}={{0b^2}\over{1b^2}}={0\over 1}=0_F$, so
${{-a}\over b}$ is an additive inverse of ${a\over b}$.
\qquad($\delta$) ${a\over b}+{c\over
d}={{ad+bc}\over{bd}}={{cb+da}\over{db}}
={c\over d}+{a\over b}$, so addition on $F$ is commutative.
\qquad($\epsilon$) We have already seen that $\times$ is a binary
operation on $F$. Now for any $(a,b)$, $(c,d)$, $(e,f)\in X$, ${a\over
b}\times({c\over d}\times{e\over f})
={a\over b}\times{{ce}\over{df}}={{ace}\over{bdf}}
={{ac}\over{bd}}\times{e\over f}
=({a\over b}\times{c\over d})\times{e\over f}$, so multiplication on $F$
is associative.
\qquad($\zeta$) ${a\over b}\times({c\over d}+{e\over f})
={a\over b}\times{{cf+de}\over{df}}
={{a(cf+de)}\over{bdf}}
={{acf+ade}\over{bdf}}
={{bacf+bade}\over{b^2df}}
={{acbf+bdae}\over{bdbf}}
={{ac}\over{bd}}+{{ae}\over{bf}}
=({a\over b}\times{c\over d})+({a\over b}\times{e\over f})$, and
similarly $({a\over b}+{c\over d})\times{e\over f}=({a\over
b}\times{e\over f})+({c\over d}\times{e\over f})$, so multiplication in
$F$ is distributive over addition.
\medskip
Thus $(F,+,\times)$ is a ring.
\medskip
\quad(vii) To see that $F$ is a field, we need to check the following:
\qquad($\alpha$) ${a\over b}\times{c\over
d}={{ac}\over{bd}}={{ca}\over{db}}={c\over d}\times{a\over b}$, so
$\times$ is commutative.
\qquad($\beta$) Set $1_F={1\over 1}$. Because $1.1=1\ne 0=1.0$,
$1_F\ne 0_F$. If ${a\over b}\in F$, ${a\over b}\times
1_F={{a1}\over{b1}}={a\over b}={{1a}\over{1b}}=1_F\times{a\over b}$, so
$1_F$ is a multiplicative identity for $F$.
\qquad($\gamma$) Suppose that ${a\over b}\in F\setminus\{0_F\}$. Then
$a=a1\ne b0=0$ so $(b,a)\in X$. Now ${a\over b}\times{b\over
a}={{ab}\over{ba}}={{ab1}\over{ab1}}={1\over 1}=1_F={b\over
a}\times{a\over b}$, so ${b\over a}$ is a multiplicative inverse for
${a\over b}$. As ${a\over b}$ is arbitrary, $F$ is a field.
\medskip
(c) If $R$ is the ring of Gaussian integers, then $F$ can be identified
with the set of complex numbers of the form ${{m+ni}\over{p+qi}}$, where
$m$, $n$, $p$, $q\in\Bbb Z$ and $p$, $q$ are not both $0$. Now
${{m+ni}\over{p+qi}}={{(m+ni)(p-
qi)}\over{p^2+q^2}}={{mp+nq}\over{p^2+q^2}}+{{np-mq}\over{p^2+q^2}}i$
is of the form $r+si$ where $r$, $s\in\Bbb Q$; and conversely, if $r$,
$s\in\Bbb Q$, $r+is$ can be expressed as ${{m+ni}\over{p}}$ where $m$,
$n$ and $p$ are integers. Thus $F$ is precisely the set of complex
numbers with both real and imaginary parts rational.
}
\def\ansAA{No answer written out.}
\def\ansAB{ (a) An {\it integral domain} is a ring $R$ such that
\qquad multiplication in $R$ is commutative;
\qquad
$R$ has a $\tbf{1}$, that is, a multiplicative identity $1\ne 0$;
\qquad if $ab=0$ then at least one of $a$, $b$ is $0$.
\medskip
A {\it Euclidean domain} is an integral domain $R$ with a {\it Euclidean
valuation} $\nu:R\setminus\{0\}\to\{0,1,2,\ldots\}$ such that
\qquad $\nu(a)\le\nu(ab)$ for all $a$, $b\in R\setminus\{0\}$;
\qquad for all $a\in R$, $b\in R\setminus\{0\}$ there are $q$, $r\in R$
such
that $a=bq+r$ and either $r=0$ or $\nu(r)<\nu(b)$.
\medskip
If $R$ is a ring, an {\it ideal} of $R$ is a set $I\subseteq R$
such
that
\qquad $x+y\in I$ for all $x$, $y\in I$;
\qquad $0\in I$;
\qquad $-x\in I$ for all $x\in I$;
\qquad $ax$, $xa\in I$ for all $x\in I$, $a\in R$.
\medskip
(b) If $I=\{0\}$ then $I=0R$. Otherwise, set $m=\min\{\nu(x):x\in
I\setminus\{0\}\}\ge 0$ and take $a\in I\setminus\{0\}$ such that
$\nu(a)=m$. Because $I$ is an ideal, $aR\subseteq I$. Now let $b$ be
any member of $I$. Then there are $q$, $r\in R$ such that $b=aq+r$ and
either $r=0$ or $\nu(r)<\nu(a)=m$. But $r=b-aq\in I$, so by the
definition of $m$ the second alternative is impossible, and $r=0$; thus
$b=aq\in aR$. As $b$ is arbitrary, $I\subseteq aR$ and $I=aR$.
\medskip
(c)(i) Not an integral domain;
$$\Matrix{0&1\\0&0}\Matrix{0&1\\0&0}=\Matrix{0&0\\0&0}.$$
\medskip
\quad(ii) A Euclidean domain; take $\nu(n)=|n|$ for $n\ne 0$.
\medskip
\quad(iii) A Euclidean domain; take $\nu(p)$ to be the degree of $p$
for any
non-zero polynomial $p$.
\medskip
\quad(iv) $\Bbb Z[x]$ is an integral domain ($R[x]$ is an integral
domain for any integral domain $R$) but is not a Euclidean domain,
because the ideal $\{p:$ the constant coefficient of $p$ is even$\}$ is
not of the form $p\Bbb Z[x]$ for any $p\in\Bbb Z[x]$.
\medskip
\quad(v) As in (iv), $\Bbb R[x,y]$ is an integral domain, but it is not
a Euclidean domain because $\{p: p$ is not constant$\}$ is a
non-principal ideal.
\medskip
\quad(vi) Not an integral domain, as it has no $\tbf{1}$.
}
\def\ansAC{No answer written out.}
\def\ansAD{(a) Let $R$ be any ring. Write $R[x]$ for the set of all
expressions
\centerline{$p=p_0+p_1x+p_2x^2+\ldots$}
\noindent where $p_0$, $p_1,\ldots$ belong to $R$ and all but finitely
many of them are zero. Define addition and multiplication on $R[x]$ by
writing
$$\eqalign{(p_0+p_1x+p_2x^2+\ldots)&+(q_0+q_1x+q_2x^2+\ldots)
\cr
&=(p_0+q_0)+(p_1+q_1)x+(p_2+q_2)x^2+\ldots,\cr}$$
$$\eqalign{(p_0+p_1x+p_2x^2+\ldots)&(q_0+q_1x+q_2x^2+\ldots)
\cr
&=(p_0q_0)+(p_1q_0+p_0q_1)x+(p_2q_0+p_1q_1+p_0q_2)x^2+\ldots;}$$
\noindent generally, the coefficient of $x^n$ in $p+q$ is $p_n+q_n$,
while the coefficient of $x^n$ in $pq$ is $\sum_{i=0}^np_{n-i}q_i$.
It is easy to check that $R[x]$ is an abelian group under addition, with
its zero the polynomial with all coefficients $0$, and the additive
inverse of $p=p_0+p_1x+\ldots$ the polynomial
$(-p_0)+(-p_1)x+(-p_2)x^2+\ldots$. To check the associativity of
multiplication, let $p=p_0+p_1x+\ldots$, $q=q_0+q_1x+\ldots$,
$r=r_0+r_1x+\ldots$ be three polynomials, and consider $(pq)r$ and
$p(qr)$. The coefficient of $x^n$ in $(pq)r$ is
$$\align\sum_{i=0}^n(\sum_{j=0}^{n-i}p_{n-i-j}q_j)r_i
&=\sum_{i=0}^n(\sum_{k=i}^{n}p_{n-k}q_{k-i})r_i\\
\intertext{(substituting $k=i+j$ in the internal sum)}
&=\sum_{k=0}^n\sum_{i=0}^k p_{n-k}q_{k-i}r_i,
\endalign$$
\noindent which is the coefficient of $x^n$ in $p(qr)$; so
$(pq)r=p(qr)$. The distributive law in $R[x]$ is now easy to check.
\medskip
(b)
If $R$ is an integral domain, an element $p$ of $R$ is {\it irreducible}
if it is not $0$, not invertible and whenever $p=ab$ then one of $a$,
$b$ is invertible.
If $R$ is an integral domain, an element of $R$ is {\it prime} if it is
not $0$, not invertible and whenever $p$ is a factor of a product $ab$
then it is a factor of at least one of $a$, $b$.
A {\it principal ideal domain} is an integral domain $R$ such that
every ideal in $R$ is of the form $aR$ for some $a\in R$.
Suppose now that $R[x]$ is a principal ideal domain.
\medskip
\quad(i) If $a$, $b\in R$ then consider the polynomials $\hat
a=a+0x+0x^2+\ldots$, $\hat b=b+0x+0x^2+\ldots$; we have $\hat a\hat
b=ab+0x+\ldots$ and $\hat b\hat a=ba+0x+\ldots$. But $R[x]$ is
commutative, so $ab=ba$; as $a$ and $b$ are arbitrary, $R$ is
commutative.
\medskip
\quad(ii) $R[x]$ has a $\tbf{1}$ of the form $e=e_0+e_1x+\ldots\ne
0+0x+\ldots$; so $R\ne\{0\}$. If $a\in R$, then $e\hat a=\hat a=\hat a
e$, so that, in particular, $e_0a=a=ae_0$; thus $e_0$ is a
multiplicative identity for $R$; as $R\ne\{0\}$, $e_0\ne 0$ and $R$ has
a $\tbf{1}$.
\medskip
\quad(iii) Now take any non-zero $a\in R$ and consider
\centerline{$I=\{p:p\in R[x],\,a\text{ divides }p_0\}$.}
\noindent Then $I$ is an ideal of $R[x]$ so is of the form $qR[x]$ for
some $q\in R[x]$. Now $\hat a\in I$, so is expressible as $qp$, and
$0+e_0x+0x^2+\ldots\in I$, so is expressible as $qr$. We have
$q_0p_0=a$, $q_0r_0=0$ and $q_0r_1+q_1r_0=e_0$. Because $q_0p_0=a\ne
0$, $q_0\ne 0$. Next, because $q_0r_0=0$ in $R$, $\hat q_0\hat
r_0=0_{R[x]}$; but $R[x]$ is an integral domain and $\hat q_0\ne
0_{R[x]}$, so $\hat r_0=0_{R[x]}$ and $r_0=0$. So $q_0r_1=e_0$. But
also $q=qe\in I$ so $q_0$ is divisible by $a$; say $q_0=ab$; then
$abr_1=e_0$. As $R$ is commutative, $br_1$ is a multiplicative inverse
of $a$. As $a$ is arbitrary, $R$ is a field.
}
\def\ansAE{No answer written out.}
\def\ansAF{No answer written out.}
\def\ansAG{(a) $(m+n\omega)+(m'+n'\omega)=(m+m')+(n+n')\omega$ so $R$ is closed
under addition; $0=0+0\omega\in R$; $-(m+n\omega)=(-m)+(-n)\omega$ so
$-z\in R$ for every $z\in R$ and $R$ is a subgroup of $(\Bbb C,+)$.
Finally, $\omega^2=-1-\omega$ so $(m+n\omega)(m'+n'\omega)
=(mm'-nn')+(mn'+nm'-nn')\omega$ and $R$ is closed under multiplication.
\medskip
(b) For $m+n\omega\in R\setminus\{0\}$ set
$\nu(m+n\omega)=|m+n\omega|^2=(m-{1\over 2}n)^2+{3\over 4}n^2
=m^2-mn+n^2\in\Bbb N$. Then of course $\nu(a)\le\nu(a)\nu(b)=\nu(ab)$
for all non-zero $a$, $b\in R$. If $a$, $b\in R\setminus\{0\}$, write
$z={a\over b}\in\Bbb C$. Then there is a $q\in R$ such that $|z-q|<1$,
because the members of $R$ form a triangular lattice with side $1$. So
setting $r=a-bq\in R$ we have $|r|=|b||z-q|<|b|$ and either $r=0$ or
$\nu(r)<\nu(b)$. Thus $\nu$ is a Euclidean valuation and $R$ is a
Euclidean domain.
\medskip
(c) Let us begin by examining the invertible elements of $R$. Note
first that if $u\in R$ is invertible in $R$, its inverse $u^{-1}$ in $R$
is just ${1\over u}$, taken in $\Bbb C$; so that $|u^{-1}|=1/|u|$.
Now because $|m+n\omega|^2=m^2-mn+n^2$ is an integer for all integral
$m$ and $n$, $|u|$ must be precisely $1$. To find the integers $m$,
$n$ with $m^2-mn+n^2=1$, we have $m^2-mn+n^2=(m-{1\over 2}n)^2+{3\over
4}n^2=1$, so $n$ must be $0$, $1$ or $-1$; now it is easy to see that
the only members of $R$ of modulus $1$ are $1$, $-1$, $\omega$,
$-\omega$, $1+\omega$ and $-1-\omega$, and that these are all invertible
in $R$.
\medskip
\quad(i) If $2=ab$ in $R$, we have $|a|^2|b|^2=4$, while $|a|^2$ and
$|b|^2$ are integers. If $|a|^2=1$ then $a$ is invertible; if
$|b|^2=1$ then $b$ is invertible. The only other possibility is that
$|a|^2=|b|^2=2$. \Quer\ But if $a=m+n\omega$, with $m$, $n\in\Bbb Z$
and $|a|^2=2$, we have
$4|a|^2=(2m-n)^2+3n^2=8$; because neither of $8$, $8-3$ is a perfect
square, this is not possible. \Bang\
Thus $2$ is irreducible.
\medskip
\quad(ii) If $2+\omega=ab$ we have $|a|^2|b|^2=3$ so one of $|a|^2$,
$|b|^2$ is $1$ and one of $a$, $b$ is invertible. So $2+\omega$ is
irreducible.
\medskip
(c) Examining $2R$, we see that its points form a triangular lattice; a
central triangle has vertices $0$, $2$ and $2(1+\omega)$. Every
element of $R$ is of the form $2q+r$ where $r$ lies within or on this
triangle. So every member of $R/2R$ is of the form $r^{\bullet}$ where
$r$ lies in or on the triangle, that is, is one of
\centerline{$0$, $1$, $2$, $1+\omega$, $2+\omega$, $2+2\omega$.}
\noindent Now of course $0^{\bullet}=2^{\bullet}=(2+2\omega)^{\bullet}$,
so we are reduced to at most four equivalence classes, being
$0^{\bullet}$, $1^{\bullet}$, $(1+\omega)^{\bullet}
=\bar\omega^{\bullet}$, $(2+\omega)^{\bullet}=\omega^{\bullet}$; and
these are indeed all distinct because all the representatives $0$, $1$,
$1+\omega$ and $2+\omega$ are distance less than $2$ from each other, so
their differences cannot belong to $2R$ (recalling that $R$ has no
elements of modulus strictly between $0$ and $1$).
The addition and multiplication tables of $R/2R$ are now defined by the
identities $\bar\omega^{\bullet}+\bar\omega^{\bullet}=0^{\bullet}
=\omega^{\bullet}+\omega^{\bullet}$, $\omega^{\bullet}\omega^{\bullet}
=\bar\omega^{\bullet}$, $\bar\omega^{\bullet}\bar\omega^{\bullet}
=\omega^{\bullet}$, $\omega^{\bullet}\bar\omega^{\bullet}=1$.
Written out in full they are
\medskip
\Centerline{[not available]}
}
\def\ansAHa{(a)(i)
A {\it unique factorisation domain} is an integral domain $R$ such that
($\alpha$)
every non-zero, non-invertible element of $R$ is expressible as a
product of
irreducible elements
$(\beta)$ if $p_1,\ldots,p_m$, $q_1,\ldots,q_n$ are irreducible
and $p_1p_2\ldots p_m=q_1q_2\ldots q_n$ then $m=n$ and $p_1,\ldots,p_m$
can be re-ordered in such a way as to make each $p_i$ an associate of
the corresponding $q_i$.
\medskip
\quad(ii) If $R$ is a unique factorisation domain then of course every
non-zero,
non-invertible
element is expressible as a product of irreducibles. To see that
irreducibles
are prime, let $p$ be irreducible, and suppose that $p$ is a factor of
$ab$;
that $pc=ab$. If either $a$ or $b$ is $0$, then it is a multiple of
$p$; if
either $a$ or $b$ is invertible then the other is a multiple of $p$.
If
none of these happen, then $ab\ne 0$ (because $R$ is an integral domain)
so
$c\ne 0$; also, because $p$ is ireducible, it cannot be a multiple of
$ab$,
so $c$ is not invertible.
We may therefore express $a$, $b$, $c$ as products of irreducibles,
\centerline{$a=q_1\ldots q_k$, $b=q'_1\ldots q'_m$, $c=p_1\ldots p_n$}
\noindent and get $pp_1\ldots p_n=q_1\ldots q_kq'_1\ldots q'_m$.
Because
factorisation in $R$ is unique, we must have $k+m=1+n$ and also there
must be
an invertible $e$ such that $pe$ is one of the $q_i$ or $q'_j$. But in
this
case, because every $q_i$ is a factor of $a$ and every $q'_j$ is a factor
of
$b$, $p$ is a factor of at least one of $a$, $b$. As $a$ and $b$ are
arbitrary, $p$ is prime, as required.
\medskip
\quad(iii)
Now suppose that $R$ satisfies the two conditions. To show that $R$ is
a
unique factorisation domain we need to
prove $(\beta)$ of the definition above. Do this by induction on $m$.
If
$m=1$ we have $p_1=q_1\ldots q_n$; but as $p_1$ is irreducible all but
one
of the $q_i$ must be invertible, and this means that $n=1$ and $q_1=p_1$
is an associate of $p_1$. For the inductive step to
$m+1$ we have $p_1\ldots p_mp_{m+1}=q_1\ldots q_n$. Now this means
that
$p_{m+1}$ is a factor of $q_1\ldots q_n$; but $p_{m+1}$ is irreducible
and
we are supposing that irreducibles are prime, so $p_{m+1}$ is a factor
of
$q_k$ for some $k\le n$. Since
$q_k$ is also irreducible, there must be an invertible $e$ such that
$q_k=ep_{m+1}$. Now we have
\centerline{$p_1\ldots p_m=eq_1\ldots q_{k-1}q_{k+1}\ldots q_n$}
\noindent (because $R$ is an integral domain, so we may cancel $p_{m+1}$
from both sides). But $eq_1$ (or $eq_2$, if $k=1$) is also
irreducible, so the inductive hypothesis
tells us that $m=n-1$ and that $eq_1,\ldots,q_{k-1},q_{k+1},\ldots,q_n$
can be got by shuffling $p_1,\ldots,p_m$ and multiplying them by
appropriate
invertible elements. But this means that $m=n+1$ and (since
$q_k=ep_{m+1}$)
the whole list $q_1,\ldots,q_n$ can be got by shuffling
$p_1,\ldots,p_{m+1}$
and multiplying them by invertible elements. Thus the induction
proceeds.
}%end of ansAHa
\def\ansAHb{(b)(i) Set $R=\{m+n\gamma:m,\,n\in\Bbb Z\}$ where $\gamma=i\sqrt{5}$.
Because
$\gamma^2=-5\in\Bbb Z$, $R$ is a subring of $\Bbb C$. If $r\in R$, set
$\nu(r)=|r|^2\in\Bbb Z$. Then $\nu(ab)=\nu(a)\nu(b)$ for all $a$,
$b\in R$,
and $\nu(a)\le 1$ iff $a$ is either $0$ or $\pm 1$, that is, is either
$0$ or
invertible in $R$.
It follows that every non-zero non-invertible element $a$ of $R$ is
expressible
as a product of irreducibles. \Prf\ Induce on $\nu(a)$. The
induction
starts with the vacuous case $\nu(a)=1$. For the inductive step
to $\nu(a)=n+1$, if $a$ is itself irreducible, we can stop; otherwise
it
is expressible as $bc$ where neither $b$ nor $c$ is invertible, so
$\nu(b)>1$
and $\nu(c)>1$ and $\nu(b)<\nu(a)$ and $\nu(c)<\nu(a)$. By the
inductive
hypothesis both $b$ and $c$ are expressible as the products of
irreducibles,
so $a$ also is. Thus the induction proceeds. \Qed
To see that $R$ is not a unique factorisation domain, consider
\centerline{$(1+\gamma)(1-\gamma)=6=2.3$.}
\noindent Because $\nu(1+\gamma)=6=\nu(1-\gamma)$, $\nu(2)=4$,
$\nu(3)=9$
any non-trivial factor $p$ of of any of these must have $\nu(p)=1$,
$2$ or $3$; in the first case $p$ is invertible and the other two cases
are
impossible, as neither $2$ nor $3$ is expressible as $m^2+5n^2$.
So $1+\gamma$, $1-\gamma$, $2$ and $3$ are all irreducible and
the factorisation of $6$ into irreducibles is not unique.
\medskip
\quad(ii) seems to be hard. If we are allowed Zorn's Lemma we may take
a
maximal subring $R$
of $\Bbb R$ containing $1$ but not containing ${1\over 2}$.
$R$ is a subring of $\Bbb R$ contining $1$ so is an integral domain.
It
is not a field because $2\in R$ is not invertible.
If $b\in R$ then ${1\over{1-2b}}\in R$. To see this, set $x={1\over{1-
2b}}$
and consider $R[x]=\{a_0+a_1x+\ldots+a_nx^n:a_0,\ldots,a_n\in R\}$.
\Quer\
If $x\notin R$ then $R[x]$ is a subring of $\Bbb R$ strictly bigger than
$R$;
but as $R$ is maximal, $R[x]$ must contain ${1\over 2}$, so that we have
$a_0,\ldots,a_n\in R$ such that $a_0+\ldots+a_nx^n={1\over 2}$ and
$2a_0(1-2b)^n+\ldots+2a_n=(1-2b)^n$. But multiplying out this means
that
$1=2d$ for some $d\in R$, which is impossible. \Bang
Next, if $c\in R$ and $c\ge 0$, we must have
$\sqrt{c}\in R$. \Prf\ Consider $R[\sqrt{c}]=
\{a_0+a_1\sqrt{c}:a_0,\,a_1\in R\}$. This is a subring of $\Bbb R$
including
$R$. \Quer\ If it is not $R$ it must contain ${1\over 2}$, so we have
${1\over 2}=a+b\sqrt{c}$ where $a$, $b\in R$. Now $(1-2b)\sqrt{c}=2a$;
but from the last paragraph we know that ${1\over{1-2b}}\in R$ so
$\sqrt{c}\in
R$, which is absurd, as in this case $R[\sqrt{c}]=R$. \Bang\ Thus
$R[\sqrt
{c}]=R$ and $\sqrt{c}\in R$, as claimed. \Qed
So suppose that $a\in R$ is not invertible. Then either $\sqrt{a}$ or
$\sqrt
{-a}$ belongs to $R$, and cannot be invertible, because $a$ cannot be a
product
of invertible elements. So $a$ is a product of non-invertible elements
and is not irreducible.
We can do this without the axiom of choice by taking $R$ to be the
smallest
subring of $\Bbb R$ containing $1$ and such that ${1\over{1-2b}}\in R$
for
every $b\in R$, $\sqrt{a}\in R$ for every positive $a\in R$.
Now $R$ cannot be a unique factorisation domain because $2$ cannot be
expressed
as a product of irreducible elements.
}
\def\ansAI{(a)
An {\it integral domain} is a ring $R$ such that
\qquad multiplication in $R$ is commutative;
\qquad
$R$ has a multiplicative identity $1\ne 0$;
\qquad if $ab=0$ then at least one of $a$, $b$ is $0$.
A {\it principal ideal domain} is an integral domain $R$ such that
every ideal in $R$ is of the form $aR$ for some $a\in R$.
Let $R$ be an integral domain. Set $X=R\times(R\setminus\{0\})$ and
define a relation
$\sim$ on $X$ by saying that $(a,b)\sim(c,d)$ if $ad=bc$. Because $R$
is an integral domain, this is an equivalence relation on $X$. It is
easy
to check that if $(a,b)\sim(a',b')$ and $(c,d)\sim(c'd')$ then
$(ad+bc,bd)
\sim(a'd'+b'c',b'd')$ and $(ac,bd)\sim(a'c',b'd')$; consequently we may
define addition and mutliplication on the set $F$ of equivalence classes
in $X$ by setting
$(a,b)^{\bullet}+(c,d)^{\bullet}=(ad+bc,bd)^{\bullet}$,
$(a,b)^{\bullet}(c,d)^{\bullet}=(ac,bd)^{\bullet}$. It is
straightforward
to verify that $F$ is a ring under these operations, with zero
$0_F=(0_R,1_R)
^{\bullet}$. $F$ has a multiplicative identity
$1_F=(1_R,1_R)^{\bullet}$
which is not equal to $0_F$ because $1_R1_R=1_R\ne 0_R=1_R0_R$.
Finally
$F$ is a field, with the multiplicative inverse of $(a,b)^{\bullet}$
being
$(b,a)^{\bullet}$ whenever $a$, $b\in R\setminus\{0\}$.
\medskip
(b) If $R$ is a principal ideal domain then
($\alpha$)
every non-zero, non-invertible element of $R$ is expressible as a
product of
irreducible elements
$(\beta)$ if $p_1,\ldots,p_m$, $q_1,\ldots,q_n$ are irreducible
and $p_1p_2\ldots p_m=q_1q_2\ldots q_n$ then $m=n$ and there are
invertible
elements $e_1,\ldots,e_n$ and a permutation $\pi$ of $\{1,\ldots,n\}$
such
that $q_i=e_ip_{\pi(i)}$ for every $i$.
\medskip
(c) ({\bf Apology:} for `distinct' read `non-associate'.)
If $r=1$ this is trivial.
So take $r>1$. For each $i\le r$ let $a_i=p_1^{k_1}\ldots p_{i-
1}^{k_{i-1}}
p_{i+1}^{k_{i+1}}\ldots p_r^{k_r}$.
Because $R$ is a principal ideal domain there are $b_1,\ldots,b_r\in R$
such that $d=b_1a_1+\ldots +b_ra_r$ is a factor of every $a_i$.
Any irreducible factor of $d$ would have to be an associate of some
$p_i$,
for $i>1$, because $d$ is a factor of $a_1$; but it would also have to
be
an associate of some $p_j$, for $j\ne i$, because $d$ is a factor of
$a_j$;
and this is impossible because $p_i$ and $p_j$ are not associates.
Accordingly $d$ is invertible and we have $q=qd^{-1}d=q'd$ say. Now
$x=q'\sum_{i\le r}b_ia_i/p_1^{k_1}\ldots p_r^{k_r}=\sum_{i\le
r}q_i/p_i^{k_i}$
where $q_i=q'b_i$ for each $i$.
\medskip
(d) We have $60=2^2.3.5$ while $15=3.5$, $20=2^2.5$ and $12=2^2.3$
are coprime; now we have $1=3.15-1.20-2.12$ so that $17=51.15-17.20-
34.12$
and ${{17}\over{60}}={{51}\over 4}-{{17}\over 3}-{{34}\over 5}$.
}
\def\ansBJ{No answer written out.}
\def\ansBA{A {\it Euclidean domain} is an integral domain $R$ with a {\it Euclidean
valuation} $\nu:R\setminus\{0\}\to\{0,1,2,\ldots\}$ such that
\qquad $\nu(a)\le\nu(ab)$ for all $a$, $b\in R\setminus\{0\}$;
\qquad for all $a\in R$, $b\in R\setminus\{0\}$ there are $q$, $r\in R$
such
that $a=bq+r$ and either $r=0$ or $\nu(r)<\nu(b)$.
A {\it principal ideal domain} is an integral domain $R$ such that
every ideal in $R$ is of the form $aR$ for some $a\in R$.
If $R$ is a principal ideal domain then ($\alpha$)
every non-zero, non-invertible element of $R$ is expressible as a
product of
irreducible elements
$(\beta)$ if $p_1,\ldots,p_m$, $q_1,\ldots,q_n$ are irreducible
and $p_1p_2\ldots p_m=q_1q_2\ldots q_n$ then $m=n$ and there are
invertible
elements $e_1,\ldots,e_n$ and a permutation $\pi$ of $\{1,\ldots,n\}$
such
that $q_i=e_ip_{\pi(i)}$ for every $i$.
\medskip
Set $\nu(a)=|a|^2$ for $a\in R\setminus\{0\}$. Then
$\nu(m+ni)=m^2+n^2\in
\Bbb N$ for $m+ni\in R\setminus\{0\}$. If $a$, $b\in R\setminus\{0\}$
then $\nu(ab)=\nu(a)\nu(b)\ge\nu(a)$. If $a\in R$, $b\in
R\setminus\{0\}$
then ${a\over b}\in\Bbb C$ must be within a distance ${1\over{\sqrt 2}}$
of
some $q\in R$; now $r=a-bq$ has $|r|\le{1\over{\sqrt{2}}}|b|$ so
either $r=0$ or $\nu(r)\le{1\over 2}\nu(b)<\nu(b)$, while $a=bq+r$.
Thus $\nu$ is a Euclidean valuation on $R$ and $R$ is a Euclidean
domain.
Suppose $m+ni$ is irreducible and $m^2+n^2$ is not a prime in $\Bbb Z$;
say
$m^2+n^2=rs$ where $r$, $s\in Z$ and neither is $\pm 1$. Note that if
there were a non-trivial factorisation of $m-ni$ as $ab$, then $m+ni$
would be
$\bar a\bar b$, which is not so; so $m-ni$ is also irreducible. We
have
$m^2+n^2=(m+ni)(m-ni)$, so by the unique factorisation theorem in $R$,
$r$ and
$s$ must be associates of $m+ni$ and $m-ni$; that is, multiples of
these
by $1$, $-1$, $i$ or $-i$. Accordingly $m+ni$ is either real or purely
imaginary, that is, one of $m$, $n$ is $0$; and of course the other
must
be of the form $\pm p$ for some prime $p\in\Bbb N$.
Finally, in this case, we cannot have $p=r^2+s^2$, as then it would have
a
factorisation $(r+is)(r-is)$ in $R$ which would give a factorisation of
$m+ni$.
Now for the converse. If $\nu(m+ni)=m^2+n^2$ is a prime then it cannot
be
expressed as $\nu(a)\nu(b)$ unless one of $\nu(a)$, $\nu(b)$ is $1$, so
that $m+ni$ cannot be expressed as $ab$ unless one of $a$, $b$ is
invertible
in $R$. If $p$ is a prime in $\Bbb N$ which is not the sum of two
squares,
but $p=(r+is)(r'+is')$ in $R$, then $\nu(p)=p^2=\nu(r+is)\nu(r'+is')$,
so
$\nu(r+is)$ is either $1$ or $p$ or $p^2$. In the first case $r+is$ is
invertible and in the third case $r'+is'$ is invertible; and the middle
case is impossible because $p$ is not the sum of two squares.
So $p$ is irreducible in $R$. Accordingly its associates $-p$, $ip$
and $-ip$ are also irreducible in $R$.
}
\def\ansBB{(a) A {\it Euclidean domain} is an integral domain $R$ with a {\it
Euclidean
valuation} $\nu:R\setminus\{0\}\to\{0,1,2,\ldots\}$ such that
\qquad $\nu(a)\le\nu(ab)$ for all $a$, $b\in R\setminus\{0\}$;
\qquad for all $a\in R$, $b\in R\setminus\{0\}$ there are $q$, $r\in R$
such
that $a=bq+r$ and either $r=0$ or $\nu(r)<\nu(b)$.
If $R$ is an integral domain, an element $p$ of $R$ is {\it irreducible}
if it is not $0$, not invertible and whenever $p=ab$ then one of $a$,
$b$ is invertible.
If $R$ is an integral domain, an element of $R$ is {\it prime} if it is
not $0$, not invertible and whenever $p$ is a factor of a product $ab$
then it is a factor of at least one of $a$, $b$.
A {\it principal ideal domain} is an integral domain $R$ such that
every ideal in $R$ is of the form $aR$ for some $a\in R$.
\medskip
(b)(i) Let $R$ be an integral domain and $p\in R$ a prime. Then of
course $p$ is
non-zero and
non-invertible. If $p=ab$, then $p$ divides $ab$, so divides one of
$a$, $b$; suppose the former, so that $a=pc$ and $p=ab=pcb$. Because
$R$ is an integral domain and $p\ne 0$ we may cancel $p$ and get $1=cb$,
so that $b$ is invertible. As $a$ and $b$ are arbitrary, $p$ is
irreducible.
\medskip
\quad(ii) Now suppose that $R$ is a principal ideal domain and that $p$
is irreducible. Then of course $p$ is
non-zero and
non-invertible. Let $a$, $b$ be such that $p$ divides $ab$. Then
$pR+aR$ is an ideal of $R$, so is of the form $dR$; we have $d=pr+as$
for appropriate $r$ and $s$, and also $d$ is a common divisor of $a$ and
$p$. Let $q$ be such that $p=dq$; then one of $d$ and $q$ must be
invertible.
If $d$ is invertible, we have $b=bd^{-1}d=bd^{-1}(pr+as)
=pbd^{-1}r+abd^{-1}s$ which is a multiple of $p$ because $ab$ is.
If $q$ is invertible then $d=pq^{-1}$ is a multiple of $p$ while $a$ is
a multiple of $d$, so $a$ is a multiple of $p$.
Because $a$, $b$ are arbitrary, $p$ is prime.
\medskip
(c) If $R$ is a principal ideal domain, it is a {\it unique
factorisation domain}, that is, it is an integral domain such that
($\alpha$)
every non-zero, non-invertible element of $R$ is expressible as a
product of
irreducible elements
$(\beta)$ if $p_1,\ldots,p_m$, $q_1,\ldots,q_n$ are irreducible
and $p_1p_2\ldots p_m=q_1q_2\ldots q_n$ then $m=n$ and $p_1,\ldots,p_m$
can be re-ordered in such a way as to make each $p_i$ an associate of
the corresponding $q_i$.
\medskip
(d) Because $\omega^2=-1-\omega$, $R$ is a subring of $\Bbb C$. (It is
obvious that $R$ is closed under addition and additive inverses and
contains $0$; to see that it is closed under multiplication, we note
that $(m+n\omega)(m'+n'\omega)=(mm'-nn')+(mn'+m'n-nn')\omega\in R$
whenever $m+n\omega$, $m'+n'\omega\in R$.) As it also contains $1$ and
$\Bbb C$ is an integral domain, $R$ is an integral domain. Now define
$\nu:R\setminus\{0\}\to\Bbb N$ by setting $\nu(m+n\omega)=|m+n\omega|^2
=m^2-mn+n^2$ for $m+n\omega\in R\setminus\{0\}$. Then we have
$\nu(ab)=\nu(a)\nu(b)\ge\nu(a)$ whenever $a$, $b\in R\setminus\{0\}$.
Now suppose that $a$, $b\in R$ and that $b\ne 0$. Examining the
arrangement of the points of $R$ in $\Bbb C$, we see that they are the
vertices of a lattice made up of equilateral triangles of side $1$. So
every complex number is within a distance ${1\over{\sqrt 3}}$ of some
member of $R$. In particular there is a $q\in R$ such that
$|{a\over b}-q|\le {1\over{\sqrt 3}}$. Now $|a-bq|\le{1\over{\sqrt
3}}|b|$ so setting $r=a-bq$ we have $a=bq+r$ and either $r=0$ or
$\nu(r)\le{1\over 3}\nu(b)<\nu(b)$.
We have $6=2.3=2(2+\omega)(1-\omega)$.
}
\def\ansBC{(a) $0=a^2-b^2=(a+b)(a-b)$ (because $R$ is commutative); so
one of $a+b$, $a-b$ must be $0$, that is, either $a=-b$ or $a=b$.
\medskip
(b) The only solutions of $x^2=1=1^2$ are $x=\pm 1$. (It can, of course,
happen that $1=-1$, but it doesn't matter.) For all other
non-zero $x$, $x\ne x^{-1}$. So if we multiply all the non-zero
elements of $F$ together, we have a collection of products $x.x^{-1}$,
all equal to $1$, together with $-1$ and (if $1\ne -1$) $1$ itself, and
the product of the whole lot is either $-1$ or $-1.1$; in either case
the answer is $-1$.
\medskip
(c) In the field $\Bbb Z_p$, $1\times_p 2\times_p\ldots
\times_p p-1=-1$, by (b); so in $\Bbb Z$ we must have $1.2.\ldots .p-1
\equiv -1$ mod $p$, as claimed.
\medskip
(d) The point is that $q$ is even. Now $(p-1)!=q!(p-q)(p-q+1)\ldots
(p-1)$; modulo $p$, we have $p-k\equiv -k$ for each $k$, so that $-1
\equiv (p-1)!\equiv
q!(-1)^q q!=(q!)^2$.
\medskip
(e) Yes, because it is a Euclidean domain (a valuation is $\nu(a)=
|a|^2$); and every Euclidean domain is a principal ideal domain,
every principal ideal domain is a unique factorisation domain.
\medskip
(f) Setting $q=(p-1)/2$, we have $(q!)^2\equiv -1$ mod $p$, that is,
$(q!)^2+1$ is a multiple of $p$; accordingly, setting $n=q!$, there
is a $k$ such that $(n-i)(n+i)=kp$. But of course $p$ cannot divide
either $n+i$ or $n-i$ (because $p(m+m'i)=pm+pm'i$ will always have
imaginary part divisible by $p$). So $p$ cannot be a prime.
$p$ is therefore not irreducible (because $R$ is a unique factorisation
domain). Suppose that $p=rr'$ where $r$, $r'\in R$ are
non-invertible. Now $p^2=|r|^2|r'|^2$ so $|r|^2$ is either $1$ or $p$ or
$p^2$. The first is impossible because the only elements of $R$ of modulus
$1$ are $\pm 1$, $\pm i$ and all of these are invertible. Similarly
$|r'|\ne 1$ and $|r|^2\ne p^2$. So $p=|r|^2$. But if $r=a+ib$ then
$p=a^2+b^2$ is a sum of two squares.
}
\def\ansBD{No answer written out.}
\def\ansBE{No answer written out.}
\def\ansBF{(a) Let $f\in R[x]$ be non-constant and irreducible in $R[x]$.
\Quer\ Suppose, if possible, that $f$ is reducible in $F[x]$; say
$f=g\times h$ in $F[x]$, where neither $g$ nor $h$ is invertible
in $F[x]$. Then we can express $g$ as $\gamma g_1$
where $\gamma\in F$ and $g_1\in R[x]$ is primitive in $R[x]$. Now
$g_1$ is primitive in $R[x]$ and is a factor of $f$ in $F[x]$, so must
be a factor of $f$ in $R[x]$; say $f=g_1\times h_1$ in $R[x]$.
But now observe that because $f\ne 0$, neither $g$ nor $h$ is zero;
and as $F$ is a field, it follows that neither is constant, so that
the degree of $g$ is strictly between $0$ and deg$(f)$, and the
degree of $g_1$ is the same. So neither $g_1$ nor $h_1$ can be
constant, and therefore neither can be invertible in $R[x]$; and
$f$ is not irreducible. \Bang
(The argument above relies on the following two facts: {\bf (i)}
If $R$ is a unique factorisation domain and $F$ its field of fractions,
then every non-constant polynomial in $F[x]$ can be expressed in the
form $\gamma g$ where $\gamma\in F$ and $g\in R[x]$ is primitive;
{\bf (ii)} If $F$ is a field and $R\subseteq F$ is a subring which is
(in itself) a unique factorisation domain, and if $f\in R[x]$ is
primitive, then any $g\in R[x]$ which is a multiple of $f$ in $F[x]$
is a multiple of $f$ in $R[x]$. It is good style to write out such
basic facts clearly, even if not explicitly asked. In some moods an
examiner might expect you to write out the proof of one or more of
these.)
\medskip
(b)(i) The polynomial $2x^2+x+2$ is
irreducible in $Z[x]$
(because the only possible non-trivial factors are of the
form $mx+n$ where $m$, $n$ are $\pm 1$ or $\pm 2$, and none of these
work, by the Remainder Theorem); by the result of (a), $2x^2+x+2$
is irreducible in $\Bbb Q[x]$; now its associate $x^2+{1\over 2}x
+1$ must also be irreducible.
\medskip
\quad(ii) $4x^3+3x+2=(2x+1)(2x^2-x+2)$ so $x^3+{3\over 4}x+{1\over 2}
=(x+{1\over 2})(x^2-{1\over 2}x+1)$.
\medskip
\quad(iii) In $\Bbb Z[x]$ the only possible factors of degree $1$ are
$\pm x\pm 1$ (none of which work) and the only possible quadratic factors
are of the form $\pm x^2+mx\pm 1$. If we try multiplying two of
these together to get the required polynomial we have
$$\eqalign{x^4+x^3+x^2+x+1&=(\pm x^2+mx\pm 1)(\pm x^2+nx\pm 1)
\cr
&=\pm x^4+(\pm m\pm n)x^3+(\pm 1\pm 1\pm mn)x^2+(\pm m\pm n)x\pm 1.\cr
}$$
\noindent Examining the
coefficients of $x^3$, $x$ we see that one of $m$, $n$ must be even;
but examining the coefficient of $x^2$ we see that $mn$ must be odd.
So this won't work either.
Accordingly $x^4+x^3+x^2+x+1$ is irreducible in $\Bbb Z[x]$ and therefore
in $\Bbb Q[x]$.
\medskip
None of these polynomials is irreducible in $\Bbb C[x]$, by the fundamental
theorem of algebra.
}
\def\ansBG{No answer written out.}
\def\ansBH{No answer written out.}
\def\ansBI{No answer written out.}
\def\ansCJ{No answer written out.}
\def\ansCA{(a) If $R$ is an integral domain, an element $p$ of $R$ is
{\it irreducible} if it is non-zero, non-invertible and not expressible
as $ab$ for any pair of non-invertible elements $a$, $b$ of $R$.
In $\Bbb Z_2[x]$ the irreducible quadratic is $x^2+x+1$ and the
irreducible cubics are $x^3+x+1$, $x^3+x^2+1$.
\medskip
(b) $x^4+x^2+1=(x^2+x+1)^2$ is not irreducible. $x^4+x+1$ is irreducible
because by the Remainder Theorem it has no linear factors and it is
not equal to $(x^2+x+1)^2$, which is the only reducible quartic without
linear factors.
}
\def\ansCB{(a) (Not written out)
\medskip
(b) $x^2+0=x\times x$ and $x^2+2=(x+1)(x+2)$ are not irreducible; while
$x^2+1$ is irreducible because (by the Remainder Theorem) it has no linear
factor.
\medskip
(c) Setting $\xi=x^{\bullet}$, the equivalence class of $x$ in the quotient
$F=R/pR$, and $1=1_R^{\bullet}$, $0=0_R^{\bullet}$, we get
\centerline{$F=\{0,1,2,\xi,\xi+1,\xi+2,2\xi,2\xi+1,
2\xi+2\}$.}
Computing powers in $F$, we have $\xi^2=2$, $\xi^3=2\xi$,
$\xi^4=1$; now $(\xi+1)^2=2\xi$, $(\xi+1)^3=2\xi+1$,
$(\xi+1)^4=2$, $(\xi+1)^5=2\xi+2$, $(\xi+1)^6=\xi$,
$(\xi+1)^7=\xi+2$, $(\xi+1)^8=1$.
So $\xi+1$ (or $2\xi+1$ or $2\xi+2$ or $\xi+2$) is a
generator of $F\setminus\{0\}$.
}
\def\ansCC{No answer written out.}
\def\ansCD{No answer written out.}
\def\ansCE{No answer written out.}