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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 2}
\centerline{September 15, 2007}
\vskip 10pt
\noindent
{\it Section 3.3}
\vskip 10pt
\noindent
4.  The tables for {\it addition} mod 10 in $S$ and {\it addition}
mod 5 in ${\mathbb Z}_5$ do correspond under the given mapping 
$f : {\mathbb Z}_5\to S$.  However, the tables for multiplication
mod 10 in $S$ and multiplication in ${\mathbb Z}_5$ do not.
In other words for the given mapping $f : {\mathbb Z}_5 \to S$
it is not the case that $f(a\cdot_5 b) = f(a)\cdot_{10} f(b)$ for
all $a,b \in {\mathbb Z}_5$.  The tables are:
\[
\begin{array}{ccccccc}
\cdot_{5} & \vline & 0 & 1 & 2 & 3 & 4\\ \hline
0 & \vline & 0 & 0 & 0 & 0 & 0\\
1 & \vline & 0 & 1 & 2 & 3 & 4\\
2 & \vline & 0 & 2 & 4 & 1 & 3\\
3 & \vline & 0 & 3 & 1 & 4 & 2\\
4 & \vline & 0 & 4 & 3 & 2 & 1\\
\end{array}
\]
and 
\[
\begin{array}{ccccccc}
\cdot_{10} & \vline & 0 & 2 & 4 & 6 & 8\\ \hline
0 & \vline & 0 & 0 & 0 & 0 & 0\\
2 & \vline & 0 & 4 & 8 & 2 & 6\\
4 & \vline & 0 & 8 & 6 & 4 & 2\\
6 & \vline & 0 & 2 & 4 & 6 & 8\\
8 & \vline & 0 & 6 & 2 & 8 & 4\\
\end{array}
\]
For example, using Hungerford's notation, note that 
$f(\overline{2}\cdot_5 \overline{3}) = f(\overline{1}) = 2$, but 
$f(\overline{2}) = 4$, and $f(\overline{3}) = 6$, so 
$f(\overline{2})\cdot_{10} f(\overline{3}) = 4$.
\vskip 10pt
\noindent
5.  The mapping $f$ is injective since $f(a) = f(a')$ implies
$(a,0_S) = (a',0_S)$, so $a = a'$.  It is also surjective
since every $(a,0_S)$ in $\overline{R}$ is $f(a)$ for the $a\in R$
appearing in the first component.  Then
\[
f(a + b) = (a+b,0_S) = (a,0_S) + (b,0_S) = f(a) + f(b)
\]
and
\[
f(a \cdot b) = (a\cdot b,0_S) = (a,0_S) \cdot (b,0_S) = f(a) \cdot f(b)
\]
by the definition of the operations in $R\times S$.  Therefore $f$
is an isomorphism.
\vskip 10pt
\noindent
10.  
\begin{enumerate}
\item[a)] This is not an homomorphism because
$f(a\cdot b) = -(a\cdot b) \ne (-a)\cdot (-b) = f(a)\cdot f(b)$.
\item[b)] This is an isomorphism, hence a homomorphism.
The reason is that in fact $-x = x$ for all
$x \in {\mathbb Z}_2$.  This is the same as the identity mapping which
is clearly an isomorphism.
\item[c)]  This is not a homomorphism since, for instance $g(0) = 1 \ne 0$
in ${\mathbb Q}$ shows that the property of Theorem 3.12, part (1) fails for this $f$.
\item[d)] This is not a homomorphism since 
\[
h(ab) = \left( \begin{array}{cr} -(ab) & 0 \\ ab & 0\end{array}\right) \ne
\left( \begin{array}{rr} -a & 0 \\ a & 0\end{array}\right)
\left( \begin{array}{rr} -b & 0 \\ b & 0\end{array}\right) = h(a)h(b).
\]
\item[e)]  This is a homomorphism since by the definitions of the 
ring operations in ${\mathbb Z}_{12}$ and ${\mathbb Z}_4$, 
\[
f([x]_{12} + [y]_{12}) = f([x+y]_{12}) = [x + y]_4 = [x]_4 + [y]_4 = f([x]_{12}) + f([y]_{12}),
\]
and 
\[
f([x]_{12} \cdot [y]_{12}) = f([x\cdot y]_{12}) = [x \cdot y]_4 = [x]_4 \cdot [y]_4
= f([x]_{12})\cdot f([y]_{12}).
\]
\end{enumerate}
\vskip 10pt
\noindent
17.  Note that $S$ consists of the classes $[x] \in {\mathbb Z}_{28}$ for $x$ 
divisible by $4$.  This set is closed under differences and multiplication 
in ${\mathbb Z}_{28}$, so by the criterion of Theorem 3.6, it is a subring.
The mapping $f([x]_7) = [8x]_{28}$ is seen to be injective and surjective by computing 
the values:  $f([0]_7) = [0]_{28}$, and
\[
f([1]_7) = [8]_{28}, f([2]_7) = [16]_{28}, f([3]_7) = [24]_{28},
f([4]_7) = [4]_{28}, f([5]_7) = [12]_{28}, f([6]_7) = [20]_{28}.
\]
The mapping $f$ is a homomorphism since by the definitions of the 
ring operations in ${\mathbb Z}_{7}$ and ${\mathbb Z}_{28}$, 
\[
f([x]_7 + [y]_7) = f([x+y]_7) = [8(x + y)]_{28} = [8x]_{28} + [8y]_{28} = f([x]_7) + f([y]_7),
\]
and since $[8\cdot 8]_{28} = [64]_{28} = [8]_{28}$:
\[
f([x]_7 \cdot [y]_7) = f([x\cdot y]_7) = [8(x \cdot y)]_{28} = [8x]_{28} \cdot [8y]_{28}
= f([x]_7)\cdot f([y]_7).
\]
\noindent
19 and 3.1/18.  
First we show that $\mathbb{Z}^*$ is a ring under the $\oplus,\odot$ operations.
(part of 3.1/18).
Closure under $\oplus$, and $\odot$ is clear since the elements of $\mathbb{Z}^*$ 
are just the ordinary integers and $a \oplus b = a + b - 1$, $a \odot b = a + b - ab$.
$\oplus$ is commutative since $a \oplus b = a + b - 1 = b + a - 1 = b\oplus a$.
$\oplus$ is associative since $(a \oplus b) \oplus c = (a + b - 1) + c - 1 = a + b + c - 2$, 
while $a \oplus (b \oplus c) = a + (b + c - 1) - 1 = a + b + c - 2$ also.
The integer $0_{\mathbb{Z}^*} = 1$ is an identity element for the $\oplus$ operation since
$a \oplus 1 = a + 1 - 1 = a$ for all $a$.  Each integer $a$ has an inverse $-(a) = -a + 2$ under
the $\oplus$ operation since $a \oplus (-a + 2) = a - a + 2 - 1 = 1 = 0_{\mathbb{Z}^*}$.
$\odot$ is associative since
$(a\odot b)\odot c = (a + b - ab) + c - (a + b - ab)c = a + b + c - ab - ac - bc + abc$, 
while 
$a\odot (b\odot c) = a + (b + c - bc) - a(b + c - bc) = a + b + c - ab - ac - bc + abc$
also.  Finally, we have a distributive law since
$a \odot (b \oplus c) = a + (b + c - 1) - a(b + c - 1) = (a + b - ab) + (a + c - ac) - 1
 = (a\odot b) \oplus (a \odot c)$.  It is not necessary to prove the other distributive
property because it easy to see that $\odot$ is commutative as well:
$a \odot b = a + b - ab = b + a - ba = b \odot a$ for all $a,b$.  

The ring $\mathbb{Z}^*$ is an integral domain because it is a commutative ring
by the previous paragraph, it has a multiplicative identity $1_{\mathbb{Z}^*} = 0$
(because $a \odot 0 = a + 0 - a\cdot 0 = a$ for all $a \in \mathbb{Z}$), and if
$a \odot b = 0_{\mathbb Z}^*$, then $a + b - ab = 1$, so $(a - 1)(b - 1) = 0$,
and hence by properties of the usual multiplication in $\mathbb{Z}$, $a = 1 = 0_{\mathbb{Z}^*}$,
or $b = 1 = 0_{\mathbb{Z}^*}$.  This completes 3.1/18.

Now, consider the mapping $f : \mathbb{Z} \to \mathbb{Z}^*$ defined by $f(x) = 1 - x$.
$f$ is an injective and surjective mapping because if $y \in {\mathbb Z}^*$ (also an
ordinary integer, then there exists exactly
one $x\in \mathbb{Z}$ with $f(x) = 1 - x = y$, namely $x = 1 - y$. 
Next, we have 
\[
f(a + b) = 1 - (a + b) = (1 - a) + (1 - b) - 1 = f(a) \oplus f(b).
\]
Finally, 
\[
f(a\cdot b) = 1 - ab = (1 - a) + (1 - b) - (1 - a)(1 - b) = f(a) \otimes f(b).
\]
Therefore the mapping $f$ is a ring homomorphism, and $\mathbb{Z}$ and $\mathbb{Z}^*$
are isomorphic as rings.
\vskip 10pt
\noindent
24.  The mapping is $g : \mathbb{R} \to M_{2\times 2}(\mathbb{R})$ with 
\[
g(r) = \left( \begin{array}{cc} 0 & 0 \\ -r & r\end{array}\right).
\]
This is injective because if $g(r) = g(s)$ for $r,s\in \mathbb{R}$, then 
\[
\left( \begin{array}{cc} 0 & 0 \\ -r & r\end{array}\right) = 
\left( \begin{array}{cc} 0 & 0 \\ -s & s\end{array}\right),
\]
which implies $r = s$ (look at the lower right entries).  However $g$ is
not surjective since not every $2\times 2$ matrix is $g(r)$ for some $r$
(take any matrix $A$ with nonzero entries in the first row, for instance).
\vskip 10pt
\noindent
27.  Let $f : R\to S$ be an isomorphism.  Then $f$ is injective and
surjective, so it has an inverse function $g : S \to R$.
The inverse function of an injective and surjective mapping is itself
injective and surjective by ``general nonsense'' (i.e. this is a direct
consequence of the definitions -- $f$ is an inverse function for $g$,
so $g$ must also be injective and surjective.)  To show that $g$ satisfies
the other properties of an isomorphism of rings, let $x,y \in S$.  Since
$f$ is surjective, there are $a,b\in R$ with $x = f(a)$ and $y = f(b)$, so
$g(a) = x$ and $g(b) = y$.  Then using the homomorphism properties for $f$:
\[
g(a + b) = g(f(x) + f(y)) = g(f(x + y)) = x + y = g(a) + g(b),
\]
and 
\[
g(a \cdot b) = g(f(x) \cdot f(y)) = g(f(x \cdot y)) = x \cdot y = g(a) \cdot g(b).
\]
Hence $g$ is also an isomorphism of rings.
\vskip 10pt
\noindent
28.  We will use the subring criterion of Theorem 3.6.  First note that
$K$ is always nonempty by part 1 of Theorem 3.12: $f(0_R) = 0_S$, so 
$O_R \in K$.  Next, let $a,b\in K$.  Then 
since $f$ is a homomorphism,
\[
f(a - b) = f(a) - f(b) = 0_S - 0_S = 0_S.
\]
This shows $a - b\in K$, so $K$ is closed under differences.
Similarly,
\[
f(a \cdot b) = f(a) \cdot f(b) = 0_S \cdot 0_S = 0_S.
\]
This shows $a\cdot b\in K$ so $K$ is closed under products.  Therefore 
$K$ is a subring of $R$.
\vskip 10pt
\noindent
29.  This very similar to 28.
We again use the subring criterion of Theorem 3.6.  First note that
$P$ is always nonempty by part 1 of Theorem 3.12: $f(0_R) = 0_S \in T$, so 
$O_R \in P$.  Next, let $a,b\in P$, so $f(a),f(b) \in T$.  Then 
since $f$ is a homomorphism,
\[
f(a - b) = f(a) - f(b) \in T,
\]
since $f(a), f(b) \in T$ and $T$ is a subring of $S$. 
This shows $a - b\in P$, so $P$ is closed under differences.
Similarly,
\[
f(a \cdot b) = f(a) \cdot f(b) \in T,
\]
since $T$ is a subring of $S$. 
This shows $a\cdot b\in P$ so $P$ is closed under products.  Therefore 
$P$ is a subring of $R$.
\vskip 10pt
\noindent
{\it Section 4.1}
\vskip 10pt
\noindent
5.
\begin{enumerate}
\item[a)]  $q(x) = 3x^2-5x+8$, and $r(x) = -4x - 6$.
\item[c)]  $q(x) = x^3+3x^2+2x+3$ and $r(x) = 4$.
\end{enumerate}
\vskip 10pt
\noindent
14.
\begin{enumerate}
\item[a)]  Consider what happens at each step of the division process.
If the leading term of the intermediate dividend polynomial
is $ax^s$ and the leading term of $g$ is $u x^r$, where $u$ is a
unit in $R$, then the term that goes into the quotient is
$u^{-1}a x^{s - r}$, and the process proceeds exactly as in the 
case of polynomials with coefficients in a field $F$.
\item[b)]  Consider $f = x + 1$ and $g = 2 x + 1$.  There are
no integer polynomials $q,r$ as in the statement of the division 
algorithm because if $q$ is any nonzero polynomial with integer coefficients,
the leading term of $qg$ will have a coefficient that is divisible by $2$.
Hence $f$ cannot be written as $f = qg + r$ with $\deg(r) < \deg(g)$ or $r = 0$. 
{\it Comment:}  When $q \ne 0$, 
the leading term in $qg + r$ will come from $qg$ in the situation of 
the division algorithm because of the condition  $\deg(r) < \deg(g)$, or $r = 0$.
This follows because $\deg(qg) \ge \deg(g)$.  
\end{enumerate}
\vskip 10pt
\noindent
17.  $\varphi$ is surjective because every $[a_m] x^m + \cdots + [a_0] \in \mathbb{Z}_n[x]$
is $\varphi(a_m x^m + \cdots + a_0)$, for the polynomial $a_m x^m + \cdots + a_0 \in \mathbb{Z}[x]$.  We next show $\varphi$ is a homomorphism.   First, if 
$f = a_m x^m + \cdots + a_0$ and $g = b_k x^k + \cdots + b_0$, then 
$f + g$ is the polynomial where for each $\ell \ge 0$,
the coefficient of $x^\ell$ is $a_\ell + b_\ell$. 
 Hence the coefficient of $x^\ell$ in $\varphi(f + g)$
is $[a_\ell + b_\ell] = [a_\ell] + [b_\ell] \in \mathbb{Z}_n$.  This shows
$\varphi(f + g) = \varphi(f) + \varphi(g)$.  Similarly, in the polynomial
$f\cdot g$, for each $\ell \ge 0$, 
the coefficient of $x^\ell$ is $\sum_{i = 0}^\ell a_i b_{\ell - i}$.
Hence using the definition of the sum and product in $\mathbb{Z}_n$,
the coefficient of $x^\ell$ in $\varphi(f\cdot g)$ is
\[
\left[ \sum_{i = 0}^\ell a_i b_{\ell - i}\right] = \sum_{i=0}^\ell [a_i][b_{\ell - i}],
\]
which is the coefficient of $x^\ell$ in $\varphi(f)\cdot \varphi(g)$.  This shows
$\varphi(f\cdot g) = \varphi(f)\cdot \varphi(g)$.
\vskip 10pt
\noindent
18.  $D$ is neither a homomorphism of rings, nor an isomorphism. This
can be seen because $D$ does not satisfy the multiplicative property of 
homomorphisms.  For example $D(x\cdot x) = D(x^2) = 2x \ne D(x)\cdot D(x) = 1$.
{\it Comment:}  Of course, $D$ also fails to be injective, since 
$D(1) = 0 = D(0)$, but $1 \ne 0$.
\vskip 10pt
\noindent
{\it Section 4.2}
\vskip 10pt
\noindent
5 b)  Here is a Maple session carrying out the computations of the 
remainders in the Euclidean Algorithm:
\begin{verbatim}
> f := x^5+x^4+2*x^3-x^2-x-2;
                   5    4      3    2        
                  x  + x  + 2 x  - x  - x - 2
> g := x^4+2*x^3+5*x^2+4*x+4;
                    4      3      2          
                   x  + 2 x  + 5 x  + 4 x + 4
> r1 := rem(f, g, x);
                                3    
                           2 - x  - x
> r2 := rem(g, r1, x);
                                2      
                         8 + 4 x  + 4 x
> r3 := rem(r1, r2, x);
                               0

\end{verbatim}
The last nonzero remainder, made monic, is the gcd, so 
\[
\gcd(f,g) = x^2 + x + 2.
\]  
\vskip 10pt
\noindent
14.  Since $f,g$ are relatively prime, $\gcd(f,g) = 1$.  Hence
there are polynomials $u,v\in F[x]$ such that $1 = uf + vg$.  Multiplying
by $h$ on both sides, we get 
\begin{equation}
\label{eq}
h = ufh + vgh.
\end{equation}  
Now if $f|h$, then
$h = kf$ for some polynomial $k\in F[x]$. Similarly, if $g|h$ then
$h = \ell g$ for some polynomial $\ell \in F[x]$.  Substitute $h = \ell g$ for 
the $h$ in the first $h$ in (\ref{eq}) and $h = kf$ for the second.  We
get $h = uf\ell g + vgkf = h$.  Hence $h = (u\ell + vk)(fg)$.  Since $u\ell + vk\in F[x]$,
this shows $(fg)|h$.
\vskip 10pt
\noindent
15.  We assume $\gcd(f,g) = 1$.  Hence there are $u,v \in F[x]$ with 
$uf + vg = 1$.  Now if $h|f$, then $f = hk$ for some $k \in F[x]$.  But
then $(uk)h + vg = 1$.  This shows that $1 \in S = \{Uh + Vg : U,V \in F[x]\}$.
By the proof of Theorem 4.5, it follows that $1 = \gcd(h,g)$, so 
$h$ and $g$ are relatively prime. 
\end{document} 
