\magnification=\magstep1
\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 7}
\centerline{October 26, 2007}
\bigskip
\noindent
{\it Section 6.3}
\bigskip
\noindent
3. (a)  $\Rightarrow$:  We know that for all integers $p$, we have the isomorphism
 ${\bf Z}/(p) \simeq {\bf Z}_p$.
If $p$ is a prime, then since the integers mod $p$ is a field, $(p)$ is a maximal
ideal in ${\bf Z}$ by Theorem 6.15.  
$\Leftarrow$:  Conversely, if $(p)$ is a maximal ideal in ${\bf Z}$, 
then ${\bf Z}/(p) \simeq {\bf Z}_p$ is a field (again by
Theorem 6.15).  This only occurs when $p$ is a prime 
number since ${\bf Z}_n$ has zero-divisors whenever 
$n$ is composite.
\bigskip
\noindent
(b) This follows by a similar argument, using Theorem 6.15
and Theorem 5.10: $(p(x))$ is a maximal ideal if and only 
if $F[x]/(p(x))$ is a field, and that is true if and only if
$p(x)$ is an irreducible polynomial in $F[x]$. 
\bigskip
\noindent
4.  $\Rightarrow$:  If $R$ is an integral domain, then 
$R/(0_R) \simeq R$ is an integral domain, so $(0_R)$ is 
a prime ideal by Theorem 6.14.  $\Leftarrow$:  If
$(0_R)$ is a prime ideal in $R$, then $R \simeq R/(0_R)$
is an integral domain.  
\bigskip
\noindent
{\it Comment:}  This may also be proved by a more direct
argument as follows:  $R$ is an integral domain if and only
if $R$ is a commutative ring with identity that has no 
zero divisors.  This is equivalent to saying that whenever
$a,b \in R$ satisfy $ab = 0_R$, then $a = 0_R$ or $b = 0_R$.  
However, that is equivalent to saying that $(0_R)$ is a prime 
ideal by the definition.
\bigskip
\noindent
5.  The maximal ideals in ${\bf Z}_6$ are 
$M_1 = (2) = \{0,2,4\}$ and $M_2 = (3) = \{0,3\}$.  In each 
case including any element not already in the ideal produces
an ideal containing a unit, hence equal to ${\bf Z}_6$.  
Similarly, the maximal ideals in ${\bf Z}_{12}$ are 
$M_1 = (2) = \{0,2,4,6,8,10\}$ and $M_2 = (3) = \{0,3,6,9\}$. 
({\it Comment}:  Notice that these are exactly
the ideals generated by the classes of the prime integers dividing 
$6,12$.)
\bigskip
\noindent
6.  (a)  In ${\bf Z}_8$, the ideals are 
the zero ideal (which is not maximal since it is contained
in other ideals not equal to the whole ring below),
$(1) = (3) = (5) = (7) = {\bf Z}_8$ (which is not maximal by
definition), 
$M = (2) = \{0,2,4,6\} = (6)$, and $(4) = \{0,4\}$.  
Since $(4) \subset (2) \subset {\bf Z}_8$, $(4)$ is not
maximal.  Hence $M$ is the only maximal ideal:  If 
we take any ideal $I$ with $M \subseteq I \subseteq {\bf Z}_8$,
then $M = (2)$ or $I = {\bf Z}_8$.
\bigskip\noindent
Similarly, in ${\bf Z}_9$, the only maximal ideal is $M = (3) = (6) = \{0,3,6\}$, since
$(1) = (2) = (4) = (5) = (7) = (8) = {\bf Z}_9$.
If $M \subseteq I \subseteq {\bf Z}_9$,
then $I = M$ or $I = {\bf Z}_9$.
\bigskip
\noindent
(b)  In ${\bf Z}_{10}$, both $(2)$ and $(5)$ are maximal.
In ${\bf Z}_{15}$, both $(3)$ and $(5)$ are maximal.
This can be seen either by arguing as in part a, or by
noting that by computations with the coset addition
and multiplication tables like what we did on the last
problem set, ${\bf Z}_{10}/(2) \simeq {\bf Z}_2$, and 
${\bf Z}_{10}/(5) \simeq {\bf Z}_5$.  Both of these 
are fields, so $(2)$ and $(5)$ are maximal ideals by
Theorem 6.15.   Similarly, ${\bf Z}_{15}/(3) \simeq {\bf Z}_3$
and ${\bf Z}_{15}/(5) \simeq {\bf Z}_5$.  Both of these
are fields, so $(3)$ and $(5)$ are maximal.  {\it Comment:} The
general fact that is being hinted at in these problems is
that ${\bf Z}_n$ has just one maximal ideal if and only if
$n = p^r$ for some prime and some $r \ge 2$.  This is 
true because the maximal ideals in ${\bf Z}_n$ are in
one-to-one correspondence with the primes dividing $n$.
\bigskip
\noindent
11.  We will show that 
$${\bf Z}[x]/(x - 1) \simeq {\bf Z}.\leqno(1)$$
Since ${\bf Z}$ is an integral domain but not a field, the 
isomorphism (1)
will show that $(x - 1)$ is a prime ideal but not a 
maximal ideal using Theorems 6.14 and 6.15.  To establish
(1), consider the mapping 
$$\eqalign{\varphi : {\bf Z}[x] &\to {\bf Z}\cr
                        f(x) &\mapsto f(1)\cr}$$
Then $\varphi$ is a ring homomorphism since for all $f(x), g(x) \in {\bf Z}[x]$,
$$\varphi(f(x) + g(x)) = f(1) + g(1) = \varphi(f(x)) + \varphi(g(x))$$
and 
$$\varphi(f(x) \cdot g(x)) = f(1)\cdot g(1) = \varphi(f(x)) \cdot \varphi(g(x)).$$
The mapping $\varphi$ is onto ${\bf Z}$ since for all $n\in {\bf Z}$, $n = \varphi(f(x))$
for the constant polynomial $f(x) = n + 0x + \cdots $.  The kernel
of $\varphi$ consists of all polynomials $f(x) \in {\bf Z}[x]$ such
that $\varphi(f(x)) = f(1) = 0$.  We claim that this is precisely
the ideal $(x - 1)$.   If $f(x) \in (x - 1)$, then $f(x) = (x - 1)g(x)$
for some $g(x) \in {\bf Z}[x]$.
Then it is clear that $\varphi(f(x)) = f(1) = 0$.  So $(x-1) \subseteq 
\ker(\varphi)$.  On the other hand, if $f(x) \in \ker(\varphi)$, 
then by the Factor Theorem (Theorem 4.15), in ${\bf Q}[x]$ it follows
that $(x - 1)|f(x)$.  By Theorem 4.22 (or a direct argument), 
this implies that $f(x) = (x - 1)g(x)$ for some 
$g(x) \in {\bf Z}[x]$.  Hence $\ker(\varphi) = (x - 1)$.  But
now, ``putting everything together'' we have that 
$${\bf Z} = {\rm im}(\varphi) \simeq {\bf Z}[x]/\ker(\varphi) = 
{\bf Z}[x]/(x - 1)$$
by the First Isomorphism Theorem.  This establishes (1) and
concludes the proof.
\bigskip
\noindent
12.  Let $M = \{(pa,b)\in {\bf Z}\times {\bf Z} : a, b\in {\bf Z}\}$.
Then $M$ is an ideal since it is nonempty, closed under
differences:
$$(pa,b) - (pa',b') = (p(a-a'),b-b') \in M,$$
and closed under products by arbitrary elements in ${\bf Z}\times {\bf Z}$:
$$(c,d)\cdot(pa,b) = (p(ca),bd) \in M.$$
(Note:  ${\bf Z}\times {\bf Z}$ is a commutative ring
under the componentwise sum and product, so this suffices.)
Now to show $M$ is maximal we will establish the isomorphism
$$({\bf Z} \times {\bf Z})/M \simeq {\bf Z}_p.\leqno(2)$$
Since the integers mod $p$ is a field when 
$p$ is prime, $M$ is maximal by Theorem 6.15.  To
prove (2), note that the distinct cosets of $M$ are the
$(r,0) + M$ for $r = 0,1,\ldots, p-1$:  Every $(c,d)$
belongs to one of these cosets since when we divide 
$p$ into $c$ we get $c = qp + r$ with $0\le r \le p-1$.  So 
$$(c,d) = (r,0) + (pq,d) \in (r,0) + M.$$
Then the coset addition and multiplication operations are
$$((r,0) + M) + ((r',0) + M) = (r+r',0) + M = (r + r' \bmod p,0) + M$$
and 
$$((r,0) + M) \cdot ((r',0) + M) = (r\cdot r',0) + M = 
(r \cdot r' \bmod p,0) + M.$$
This shows (2).
\bigskip
\noindent
13.  Take $I = \{(0,b) : b \in {\bf Z}\}$ for instance.
Then $I$ is an ideal since it is nonempty, closed under
sums, and closed under products by arbitrary $(c,d)\in {\bf Z}\times {\bf Z}$.  It is prime but not maximal because 
$({\bf Z}\times {\bf Z})/I \simeq {\bf Z}$.  This follows from 
the First Isomorphism Theorem applied to the projection
$$
\eqalign{
\pi : {\bf Z}\times {\bf Z} &\to {\bf Z}\cr
            (a,b) &\mapsto a\cr}$$
This $\pi$ is a ring homomorphism since
$$\pi((a,b) + (c,d)) = \pi(a+b,c+d) = a + b = \pi(a,b) + \pi(c,d)$$
and
$$\pi((a,b) \cdot (c,d)) = \pi(ab,cd) = ab = \pi(a,b) \cdot \pi(c,d).$$
We have
$${\rm im}(\pi) = {\bf Z} \simeq ({\bf Z}\times {\bf Z})/\ker(\pi)
= ({\bf Z}\times {\bf Z})/I.$$
By Theorems 6.14 and 6.15, $I$ is prime but not maximal since
${\bf Z}$ is an integral domain but not a field.
\bigskip
\noindent
17.  (Note: the surjective hypothesis here is only needed
to show that $I \ne R$, part of the way prime ideals 
are defined in the text.  This is not always included in
the definition of a prime ideal.  The reason to do it
as Hungerford does 
is so that, for instance, ${\bf Z} = (1)$ is not counted as a prime
ideal in ${\bf Z}$.  Then you can say a nonzero ideal
$I\in {\bf Z}$ is a prime
ideal if and only if $I = (p)$ for a prime number $p$.)  
To be complete, we begin by showing that the set
$I$ is an ideal in $R$.  First, $I$ is nonempty since $f(0_R) = 0_S \in J$, so
$0_R \in I$.  If $a,b \in I$, then $f(a), f(b)\in J$.
But then $f(a) - f(b) = f(a - b)\in J$ since $J$ is an ideal.
Hence $a-b\in I$.  Similarly for all $r\in R$, $a\in I$, we have
$f(ra) = f(r)f(a)$.  Since $J$ is an ideal, $f(a) \in J$,
and $f(r)\in S$, this is in $J$.  Hence $ra \in I$.

Now suppose $J$ is a prime ideal in $S$.  According to 
Hungerford's way of defining prime ideals, this means
$J \ne S$.  But then since $f$ is surjective, it must be the case
that $I \ne R$.  If $a,b\in R$ and $ab \in I$, then 
$f(ab) = f(a)f(b) \in J$.  Since $J$ is prime, this
implies $f(a) \in J$ or $f(b) \in J$.  Hence $a\in I$
or $b \in I$.  Therefore $I$ is a prime ideal in $R$.
\bye