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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 6}
\centerline{October 18, 2007}
\vskip 10pt
\noindent
{\it Section 6.1}
\vskip 10pt
\noindent
2.  We let 
\[
I = \{p(x) = a_0 + a_1 x + \cdots \in \mathbb{Z}[x] : a_0 \text{ is even}\}.
\]
$I$ is clearly nonempty since $0,2, 4 + x^2 \in I$ and so forth.  Moreover,
if $p(x) = a_0 + a_1 x + \cdots $ and $q(x) = b_0 + b_1 x + \cdots$ are in $I$, 
then $p(x) - q(x) = a_0 - b_0 + (a_1 - b_1) x + \cdots$.  Since $a_0, b_0$
are divisible by $2$, so is $a_0 - b_0$.  Hence $I$ is closed under differences.
If $r(x) = c_0 + c_1 x + \cdots$ is an arbitrary polynomial in $\mathbb{Z}[x]$, 
and $p(x)$ as above is in $I$, 
then $r(x) p(x) = a_0 c_0 + (a_0c_1 + a_1 c_0) x + \cdots $.  Since
$a_0$ is divisible by $2$, so is $a_0 c_0$.  Hence $I$ is closed under
multiplication by arbitrary $r(x) \in \mathbb{Z}[x]$.  This shows 
$I$ is an ideal by the definition.
\vskip 10pt
\noindent
6.
\begin{enumerate}  
\item[a.]  The set of nonunits in $\mathbb{Z}_8$ is $I = \{0,2,4,6\}$.  
This (nonempty!) set is closed under differences and products by arbitrary
elements in $\mathbb{Z}_8$.  This is true because these are the classes 
of the even numbers mod 8.  Since $8$ is also even, 
$a \equiv 2k \bmod 8$ and $b \equiv 2\ell \bmod 8$
implies $a - b \equiv 2(k - \ell) \bmod 8$.  If $2(k - \ell) \equiv c \bmod 8$, then
$2(k - \ell) - c = 8 m$ for some integer $m$.  But then $c = 2(k - \ell) - 8m$
is also even.  Similarly if $r$ is an arbitrary class mod 8, and $a \equiv 2k \bmod 8$, 
then if  $2 r k \equiv c \bmod 8$, we have $2k - c = 8m$ for some integer $m$, so
$c$ is also divisible by $2$.
\item[b.]  The set of nonunits in $\mathbb{Z}_9$ is $I = \{0,3,6\}$.  This
is an ideal for a similar reason to what happened in part a -- this is
the set of classes of numbers divisible by $3$ in $\mathbb{Z}_9$.
\end{enumerate}
\vskip 10pt
\noindent
7. 
\begin{enumerate}
\item[a.]  In $\mathbb{Z}_5$, $(0) = \{0\}$, and $(1) = (2) = (3) = (4) = \mathbb{Z}_5$.
\item[b.]  In $\mathbb{Z}_9$, $(0) = \{0\}$, $(1) = (2) = (4) = (5) = (7) = (8) = \mathbb{Z}_9$, and $(3) = (6) = \{0,3,6\}$.
\item[c]  In $\mathbb{Z}_{12}$, $(0) = \{0\}$, $(1) = (5) = (7) = (11) = \mathbb{Z}_{12}$,
$(2) = (10) = \{0,2,4,6,8,10\}$, $(3) = (9) = \{0,3,6,9\}$, $(4) = (8) = \{0,4,8\}$,
and $(6) = \{0,6\}$.
\end{enumerate} 
\vskip 10pt
\noindent
12.  
\begin{enumerate}
\item[a]  An example is $(2) = (-2)$.  This follows since every even integer $n = 2k$
is also $n = (-2)(-k)$, and vice versa.  (This is the {\it only} way the equality
$(c) = (d)$ can happen if $c \ne d$ in $\mathbb{Z}$, since $c \in (d)$ says 
$c = dm$ for some integer $m$.  Conversely, $d \in (c)$ says $d = cn$ for some
integer $n$.  Then substituting, we see $c = c(nm)$, so $nm = 1\in \mathbb{Z}$.
This is only possible if $n = m = 1$ or $n = m = -1$.) 
\item[b] To show the equality, we show inclusions in both directions.
 $\subseteq$:  Let $n \in (4,6)$.  Then $n = 4a + 6b$ for some 
integers $a,b$.  But then $n = 2(2a + 3b)$, and $2a + 3b \in \mathbb{Z}$, so $n \in (2)$.  $\supseteq$:
Note that $2 = 4(-1) + 6(1) \in (4,6)$.  Then for all $n \in \mathbb{Z}$, 
$2n = 4(-n) + 6(n) \in (4,6)$.  Hence $(2) \subseteq (4,6)$.  
\item[c]  Similar to part b.  First we show $\subseteq$:  Let $n \in (6,9,15)$.
Then $n = 6a + 9b + 12c = 3(2a + 3b + 4c)$ for some $a,b,c\in \mathbb{Z}$.
This shows $(6,9,15) \subseteq (3)$.  On the other hand, to show 
$\supseteq$: since $3 = 6(-1) + 9(1)$, we
have $3n = 6(-n) + 9(n) + 12(0) \in (6,9,15)$.  {\it Comment:} Note that
this argument actually shows somewhat more than the problem asked.  In 
fact $(6,9) = (3)$, and the third generator is unnecessary.
\end{enumerate}
\vskip 10pt
\noindent
13.
\begin{enumerate}
\item[a.]  If $1_R \in I$ and $I$ is an ideal, then for all $r \in R$, $r\cdot 1_R = r \in I$.
Hence $R \subseteq I$.  Since $I\subseteq R$ by definition, this shows $I = R$.
\item[b.]  If $u \in I$ and $u$ is a unit, then say $a \in R$ satisfies $au = 1_R$.
By definition, since $I$ is an ideal, $au \in I$, so $1_R \in I$, and $I = R$ by part a.
\end{enumerate}
\vskip 10pt
\noindent
14.  Let $I$ be an ideal in a field $F$, and let $I \ne (0_F) = \{0_F\}$.  Then there is some 
$a \ne 0_F$ in $I$.  But then since $F$ is a field, 
$a$ is a unit in $F$, and then $I = F$ by problem 13.
\vskip 10pt
\noindent
21.
\begin{enumerate}
\item[a.] $I = \{0,3\}$ is clearly nonempty, and closed under differences and products
by arbitrary elements in $\mathbb{Z}_6$.  (A formal proof would proceed as in 6 a above.)
The distinct cosets are $0 + I = I = 3 + I$, $1 + I = \{1,4\} = 4 + I$, and 
$2 + I = \{2,5\} = 5 + I$.
\item[b.] $I = \{0,3,6,9,12\}$ is clearly nonempty, and closed under differences and products
by arbitrary elements in $\mathbb{Z}_{15}$ (since this subset consists
of the classes of multiples of 3 mod 15).  (A formal proof would proceed as in 6 a above.)
The distinct cosets are $0 + I = I$, $1 + I = \{1,4,7,10,13\}$, and 
$2 + I = \{2,5,8,11,14\}$.
\end{enumerate}
\vskip 10pt
\noindent
23.  $I \ne \emptyset$ since $r = 0_R$ satisfies $0_R t = 0_R$ for all $t \in J$.
Hence $0_R \in I$.
Next, let $a,b \in I$.  Then $a t = 0_R$ and $b t = 0_R$ for all $t \in J$.  
But then for all $t \in J$, $(a - b)t = at - bt = 0_R - 0_R = 0_R$.  This shows
that $a - b \in I$.  Finally let $a \in I$ and $r \in R$.  By associativity, we have
$(sa)t = s(at) = s\cdot 0_R = 0_R$ for all $t \in J$.  Hence $sa \in I$.  Similarly
for all $t \in J$, $(as)t = a(st)$.  Now $J$ is an ideal in $R$ and $s \in R$
so $st \in J$.  Hence $a(st) = 0_R$.  This shows $as\in I$ also.  
By definition, $I$ is an ideal in $R$.
\vskip 10pt
\noindent
24.  $K \ne \emptyset$ since $a = 0_R$ satisfies $r 0_R = 0_R$ for all $r \in R$.
$J$ is an ideal so $0_R \in J$, and hence $r 0_R \in J$ for all $r \in R$.
Hence $0_R \in K$.  
Next, let $a,b \in K$.  Then $r a \in J$ and $r b \in J$ for all $r \in R$.  
But then for all $r \in R$, $r (a - b) = r a - r b \in J$, since $J$ is 
closed under differences.  This shows
that $a - b \in K$.  Finally let $a \in K$ and $s \in R$.  By associativity, we have
$r (sa) = (rs) a \in J$ since $rs \in R$.  Hence $sa \in K$.  Similarly
for all $r \in R$, $r(as) = (ra) s$.  Now $ra \in J$, and $J$ is an ideal in $R$
so $(ra)s \in J$.  Hence $as \in K$.  By the definition, $K$ is an ideal in $R$.
\vskip 10pt
\noindent
38.  If $I = \{0\}$, then $I = (0)$ is principal.  Hence we may assume now
that $I \ne \{0\}$.  We claim that $I$ always contains positive elements
in this case.  This is true since $I$ must contain some $n \ne 0$.  If
$n > 0$, then we are done.  If $n < 0$, then $(-1)(n) = -n > 0$ and $-n \in I$
as well.  Now, by the Well-Ordering Principle, there is a smallest strictly positive
element in $I$, call it $c$.  We will show that $I = (c)$, so $I$ is principal.
The $\supseteq$ inclusion is automatic since $c \in I$ implies $ck \in I$
for all $k\in \mathbb{Z}$.  For the other ($\subseteq$) inclusion, let $n \in I$, and
divide $c$ into $n$ by integer division:  $n = qc + r$, where $0 \le r < c$.
However this equation shows $r = n - qc$.  Since $n, c\in I$, this implies
$r \in I$ as well.  However, $c$ was the smallest strictly positive element
in $I$, so the possibility is $r = 0$.  This shows $n = qc \in (c)$ so 
$I \subseteq (c)$.  Therefore $I = (c)$ is a principal ideal.
\vskip 10pt 
\noindent
39. 
\begin{enumerate}
\item[a.]  $S$ is nonempty since $0 = \frac{0}{1} \in S$.  If $\frac{a}{b}, \frac{c}{d} \in S$,
then the difference $\frac{a}{b} - \frac{c}{d} = \frac{da - bc}{bd}$.  To put this in
lowest terms, we would compute the gcd of the top and bottom, say $n = (da - bc, bd)$.
Then we have $\frac{da - bc}{bd} = \frac{nA}{nB} = \frac{A}{B}$ where $(A,B) = 1$.  Note that 
since $B|(bd)$ and $b,d$ are odd, then $B$ must also be odd.
Hence $\frac{a}{b} - \frac{c}{d} \in S$.
Similarly, $\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$.  To put this in
lowest terms, we would compute the gcd of the top and bottom, say $n = (ac, bd)$.
Then we have $\frac{ac}{bd} = \frac{nA}{nB} = \frac{A}{B}$ where $(A,B) = 1$.  Note that 
$B|(bd)$ must be odd since $b,d$ are odd.  Hence $\frac{a}{b} \cdot \frac{c}{d} \in S$.
\item[b.]  Let $I$ be the set of elements of $S$ with even numerators.
$I$ is nonempty since $0 = \frac{0}{1} \in I$.  $I$ is closed under differences, since
if $\frac{2k}{b}, \frac{2\ell}{d} \in I$, then 
$\frac{2k}{b} - \frac{2\ell}{d} = \frac{2(dk - b\ell)}{bd}$.  If this is not
already in lowest terms, then we might have to cancel some common factors.  However
note that $bd$ is odd so $2 \not|\ (bd)$.  Hence, because of unique
factorization in $\mathbb{Z}$, the factor of $2$ on the top will 
remain after any common factors are cancelled.  This shows that $I$ is closed
under differences.  Similarly, if $\frac{2k}{b}\in I$, and $\frac{c}{d} \in S$, then
$\frac{2k}{b}\cdot \frac{c}{d} = \frac{2kc}{bd}$.  If this is not
already in lowest terms, then we might have to cancel some common factors.  However
note that $bd$ is odd so $2 \not|\ (bd)$.  Hence the factor of $2$ on the top will 
remain after any common factors are cancelled.  Thus $I$ is an  ideal in $S$.
\item[c.]  We claim that $1 + I$ is the only coset of $I$ different from $I$ itself,
and we will show that by showing that $1 + I$ coincides with the set of all elements of $S$
where the numerator is {\it odd}.  (Since every element of $S$ either has an even 
numerator or an odd numerator, there cannot be any additional cosets.)
Note first that $1 + \frac{2k}{b} = \frac{b + 2k}{b}$ has an odd numerator so $1 + I$
is contained in the set of elements in $S$ with odd numerators.  Conversely, 
if $\frac{2k + 1}{b}$ has an odd numerator, then we can rewrite this as
$\frac{2k + 1}{b} = 1 + \frac{2k + 1 - b}{b}$  Since $b$ is odd, $2k + 1 - b$ is
even so $\frac{2k + 1 - b}{b}$ is in $I$.  Hence $\frac{2k+1}{b} \in 1 + I$.  
\vskip 10pt
\noindent
40.  Notice that 39 is the special case of 40 where $p = 2$.  The proofs of the 
more general statements here are similar.  To prepare, note that the fact used
in the proof of 39 that $b,d$ odd implies $bd$ odd carries over to this setting
because of the contrapositive form of Euclid's Lemma in $\mathbb{Z}$:  If $p$ 
is prime and $p\not|\ b$ and $p\not|\ d$, then $p\not|\ (bd)$.  
\item[a.]  $T$ is nonempty since $0 = \frac{0}{1} \in S$.  If $\frac{a}{b}, \frac{c}{d} \in T$,
then the difference $\frac{a}{b} - \frac{c}{d} = \frac{da - bc}{bd}$.  To put this in
lowest terms, we would compute the gcd of the top and bottom, say $n = (da - bc, bd)$.
Then we have $\frac{da - bc}{bd} = \frac{nA}{nB} = \frac{A}{B}$ where $(A,B) = 1$.  Note that 
since $B|(bd)$ and $b$ and $d$ are not divisible by $p$, 
then $bd$ and hence $B$ must also be not divisible by $p$.
Hence $\frac{a}{b} - \frac{c}{d} \in T$.
Similarly, $\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$.  To put this in
lowest terms, we would compute the gcd of the top and bottom, say $n = (ac, bd)$.
Then we have $\frac{ac}{bd} = \frac{nA}{nB} = \frac{A}{B}$ where $(A,B) = 1$.  Note that 
$B|(bd)$ must be not divisible by $p$ since $b,d$ are not
divisible by $p$.  Hence $\frac{a}{b} \cdot \frac{c}{d} \in T$.
\item[b.]  Let $I$ be the set of elements of $T$ with numerators divisible by $p$.
$I$ is nonempty since $0 = \frac{0}{1} \in I$.  $I$ is closed under differences, since
if $\frac{pk}{b}, \frac{p\ell}{d} \in I$, then 
$\frac{pk}{b} - \frac{p\ell}{d} = \frac{p(dk - b\ell)}{bd}$.  If this is not
already in lowest terms, then we might have to cancel some common factors.  However
note that $b$ and $d$ are not divisible by $p$, so $p \not|\ (bd)$.  Hence, because of unique
factorization in $\mathbb{Z}$, the factor of $p$ on the top will 
remain after any common factors are cancelled.  This shows that $I$ is closed
under differences.  Similarly, if $\frac{pk}{b}\in I$, and $\frac{c}{d} \in T$, then
$\frac{pk}{b}\cdot \frac{c}{d} = \frac{pkc}{bd}$.  If this is not
already in lowest terms, then we might have to cancel some common factors.  However
note that $bd$ is divisible by $p$ so $p \not|\ (bd)$.  Hence the factor of $p$ on the top will 
remain after any common factors are cancelled.  Thus $I$ is an  ideal in $T$.
\item[c.]  We claim that $0 + I, 1 + I, 2 + I, \ldots, (p - 1) + I $ are the only cosets 
of $I$.
We will show that by showing that every element of $T$ belongs to one of these
cosets.   This is significantly trickier than the 
corresponding statement in 39, though(!)  To see it, we need to recall from 
Algebraic Structures that an integer $b$ has a multiplicative inverse
mod $p$ exactly when $(b,p) = 1$ or $p \not|\ b$.  So let $\frac{a}{b} \in T$.
We claim that this element belongs to the coset $s + I$ where 
$s \in \{0,1,\ldots,p-1\} \equiv ua \bmod p$, where $u$ is the multiplicative
inverse of $b$ mod $p$: $ub \equiv 1 \bmod p$.  To see this, write $a = qp + i$
with $i \in \{0,1,\ldots,p-1\}$, and $ub = 1 + kp$.  Then 
\[
\frac{a}{b} = \frac{i + qp}{b} = \frac{i(ub - kp) + qp}{b} = ui + \frac{p(q - ki)}{b} \in ui + I.
\]
But now $ui \equiv ua \equiv s \mod p$, so $ui = s + \ell p$ for some $\ell \in \mathbb{Z}$
and hence
\[
\frac{a}{b} = s + \left(\frac{p\ell}{1} + \frac{p(q-ki)}{b}\right) \in s + I.
\]
This establishes our claim.
\end{enumerate}
\vskip 10pt
\noindent
{\it Section 6.2}
\vskip 10pt
\noindent
2.  By the First Isomorphism Theorem, if $\varphi : F \to S$ is any ring homomorphism, then
${\rm im}(\varphi) \simeq F/\ker(\varphi)$.  Moreover $\ker(\varphi)$ is an ideal in $F$.
By 6.1/14, this says $\ker(\varphi) = (0_F)$ or $\ker(\varphi) = F$.  In the 
first case ${\rm im}(\varphi) \simeq F/(0_F) \simeq F$ (each element of $F$ defines
a distinct coset).  In the second case, 
${\rm im}(\varphi) \simeq F/F = \{0\}$, the zero ring (all elements of $F$ belong to 
the same coset).  
\vskip 10pt
\noindent
3.  Continuing the reasoning from 2, if $f: F \to R$ is a surjective homomorphism and
$R$ is not the zero ring, there
must be at least two elements in $R$ ($0_R$ and something else).  Hence
we cannot be in the second case above since $f$ surjective says ${\rm im}(f) = R$.  
It follows that ${\rm im}(f) \simeq F/(0_F) \simeq F$.  Since $\ker(f) = (0_F)$, 
$f$ is injective as well as surjective, hence is an isomorphism.
\vskip 10pt
\noindent 
6.  
\begin{enumerate}
\item[a.]  See Section 6.1/7c above.  This is the same question(!)
\item[b.]  We will show the case of $I = (6) = \{0,6\}$.  The others are similar, 
and simpler, since there are fewer distinct cosets.  Here the distinct cosets
are $0 + I$, $1 + I$, $2 + I$, $3 + I$, $4 + I$, $5 + I$.  The tables 
for $\mathbb{Z}_{12}/I$ are as follows (writing $i$ for $i + I$ in all cases
to save typing!)
\[
\begin{array}{cccccccc}
+ 
  & \vline & 0 & 1 & 2 & 3 & 4 & 5\\ \hline
0 & \vline & 0 & 1 & 2 & 3 & 4 & 5\\
1 & \vline & 1 & 2 & 3 & 4 & 5 & 0\\
2 & \vline & 2 & 3 & 4 & 5 & 0 & 1\\
3 & \vline & 3 & 4 & 5 & 0 & 1 & 2\\
4 & \vline & 4 & 5 & 0 & 1 & 2 & 3\\
5 & \vline & 5 & 0 & 1 & 2 & 3 & 4\\
\end{array}
\]
and
\[
\begin{array}{cccccccc}
\cdot 
  & \vline & 0 & 1 & 2 & 3 & 4 & 5\\ \hline
0 & \vline & 0 & 0 & 0 & 0 & 0 & 0\\
1 & \vline & 0 & 1 & 2 & 3 & 4 & 5\\
2 & \vline & 0 & 2 & 4 & 0 & 2 & 4\\
3 & \vline & 0 & 3 & 0 & 3 & 0 & 3\\
4 & \vline & 0 & 4 & 2 & 0 & 4 & 2\\
5 & \vline & 0 & 5 & 4 & 3 & 2 & 1\\
\end{array}
\]
For instance, $(2 + I) + (5 + I) = 7 + I = 1 + I$ and 
$(2 + I)(5 + I) = 10 + I = 4 + I$.  
By inspection, we see that this quotient ring is isomorphic
to $\mathbb{Z}_6$.  Similarly, $\mathbb{Z}_{12}/(0) \simeq \mathbb{Z}_{12}$,
$\mathbb{Z}_{12}/(1) \simeq \{0\}$ (the zero ring), $\mathbb{Z}_{12}/(2) \simeq \mathbb{Z}_2$, 
$\mathbb{Z}_{12}/(3) \simeq \mathbb{Z}_3$, $\mathbb{Z}_{12}/(4) \simeq \mathbb{Z}_4$.
\item[c.]  This follows from the First Isomorphism Theorem.
Every ring homomorphism $\varphi : \mathbb{Z}_{12} \to S$ has 
${\rm im}(\varphi) \simeq \mathbb{Z}_{12}/I$ for $I = \ker(\varphi)$.  Every
ideal $I$ in $\mathbb{Z}_{12}$ is one of the six principal ideals found
in 6.1/7c.  Hence by part b, every homomorphic image of $\mathbb{Z}_{12}$
is isomorphic to one of the rings listed.
\end{enumerate}
\vskip 10pt
\noindent 
8.  These can be done by the same method illustrated in the solution for 6b above.
To see a less computational method for part a, notice first that $I$
is an ideal by Section 6.1/21a. We have
$\mathbb{Z}_6/([3]) = \{[0] + I, [1] + I, [2] + I\}$.  Then we define a mapping 
$\varphi : \mathbb{Z}_6 \to \mathbb{Z}_3$ by $\varphi([i]_6) = [i]_3$
(the added subscripts indicate which ring the congruence class lives in, 
for clarity).
This is well-defined since if $[i]_6 = [j]_6$, then $6|(i - j)$ which 
implies $3|(i - j)$, so $[i]_3 = [j]_3$.  We have that $\varphi$
is a ring homomorphism since
\[
\varphi([i]_6 + [j]_6) = \varphi([i+j]_6) = [i+j]_3 = [i]_3 + [j]_3 = 
\varphi([i]_6) + \varphi([j]_6), 
\]
and
\[
\varphi([i]_6 \cdot [j]_6) = \varphi([ij]_6) = [ij]_3 = [i]_3 \cdot [j]_3 = 
\varphi([i]_6) \cdot \varphi([j]_6).
\]
We see that $\varphi$ is surjective since every element in $\mathbb{Z}_3$
is in the image.  Finally, $\ker(\varphi) = \{[0]_6,[3]_6\}$.  Hence
by the First Isomorphism Theorem, 
\[
\mathbb{Z}_3 = {\rm im}(\varphi) \simeq \mathbb{Z}_6/\ker(\varphi) = \mathbb{Z}_6/([3]_6).
\]
The proof of part b is exactly analogous and will be omitted.
\vskip 10pt
\noindent
28.  Define $\varphi : S \to \mathbb{Z}_2$ by 
$\varphi\left(\frac{a}{b}\right) = [a]\in \mathbb{Z}_2$.
Since the denominator $b$ is odd, we have $[b] = [1]$ in $\mathbb{Z}_2$.
Hence
\[
\varphi\left(\frac{a}{b}+\frac{c}{d}\right) = 
\varphi\left(\frac{ad + bc}{bd}\right) =  [ad + bc] = [a][d] + [b][c] = [a] + [c] \in \mathbb{Z}_2.
\]
Since this equals $\varphi\left(\frac{a}{b}\right) + \varphi\left(\frac{c}{d}\right)$,
the first homomorphism property holds.  Similarly,
\[
\varphi\left(\frac{a}{b}\cdot\frac{c}{d}\right) = 
\varphi\left(\frac{ac}{bd}\right) =  [ac] = [a] \cdot [c] \in \mathbb{Z}_2.
\]
Since this equals $\varphi\left(\frac{a}{b}\right) \cdot \varphi\left(\frac{c}{d}\right)$,
the second homomorphism property holds.  Now, $\ker(\varphi) = I$ since
it is exactly the elements in $S$ with even numerators that map to $[0]$
in $\mathbb{Z}_2$.  $\varphi$ is also surjective since the elements in $S$
with odd numerators map to $[1]$.  So by the First Isomorphism Theorem,
\[
S/\ker(\varphi) = S/I \simeq {\rm im}(\varphi) = \mathbb{Z}_2.
\]
\vskip 10pt
\noindent
29.  The idea here is similar to what we did in 28 above. However, as
noted in the solution to 6.1/40, this is somewhat trickier.  A mapping 
$\varphi : T \to \mathbb{Z}_p$ that works here is to define
$\varphi\left(\frac{a}{b}\right) = [a][b]^{-1} \in \mathbb{Z}_p$.
(Recall that $[b]$ has a multiplicative inverse in $\mathbb{Z}_p$ exactly
when $(p,b) = 1$ and that is the condition for $\frac{a}{b} \in T$.)
This is a ring homomorphism since
\begin{eqnarray*}
\varphi\left(\frac{a}{b} + \frac{c}{d}\right) &=& \varphi\left(\frac{ad + bc}{bd}\right)\\
&=& [ad + bc][bd]^{-1}\\
&=& ([a][d] + [b][c])[b]^{-1} [d]^{-1}\\
&=& [a][b]^{-1} + [c][d]^{-1}\\
&=& \varphi\left(\frac{a}{b}\right) + \varphi\left(\frac{c}{d}\right)
\end{eqnarray*}
and
\begin{eqnarray*}
\varphi\left(\frac{a}{b} \cdot \frac{c}{d}\right) &=& \varphi\left(\frac{ac}{bd}\right)\\
&=& [ac][bd]^{-1}\\
&=& [a][c][b]^{-1} [d]^{-1}\\
&=& ([a][b]^{-1})([c][d]^{-1})\\
&=& \varphi\left(\frac{a}{b}\right) \cdot \varphi\left(\frac{c}{d}\right).
\end{eqnarray*}
The kernel of $\varphi$ is exactly the set of $\frac{a}{b}\in T$ with 
$[a] = [0] \in \mathbb{Z}_p$, which is $I$.  The image of $\varphi$
is all of $\mathbb{Z}_p$ since $\varphi\left(\frac{i}{1}\right) = [i]$
for $i = 0, 1, \ldots, p - 1$.  Hence by the First Isomorphism Theorem, 
$\mathbb{Z}_p = {\rm im}(\varphi) \simeq T/\ker(\varphi) = T/I$.
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