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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 5}
\centerline{October 11, 2007}
\vskip 10pt
\noindent
{\it Section 5.1}
\vskip 10pt
\noindent
3.  The congruence classes in $\mathbb{Z}_2[x]/(x^3 + x + 1)$
are in 1-1 correspondence with the remainders on division by 
$x^3 + x + 1$ in $\mathbb{Z}_2[x]$.  These are all the polynomials
of degree 2 or less:  $a x^2 + b x + c$, with $a,b,c \in \mathbb{Z}_2$.
This gives $2\cdot 2\cdot 2 = 8$ classes in all:
\[
[0],[1],[x],[x+1],[x^2],[x^2+1],[x^2+x],[x^2+x+1].
\]
\vskip 10pt
\noindent
4.  This is similar to problem 3 above, except that now the coefficients
in the remainders $a x^2 + b x + c$ are in $\mathbb{Z}_3$.  
There are $3$ choices for each of $a,b,c$, so there are
$3\cdot 3\cdot 3 = 27$ congruence classes in all.
\vskip 10pt
\noindent
5.  The congruence classes in $\mathbb{Q}[x]/(x^2 - 2)$
are in 1-1 correspondence with the remainders on division by 
$x^2 - 2$ in $\mathbb{Q}[x]$.  These are all the polynomials
of degree 1 or less: $a x + b$ with $a,b\in \mathbb{Q}$.  Since
the set $\mathbb{Q}$ is infinite and distinct linear polynomials
give distinct congruence classes (that is, $a x + b \equiv c x + d \!\bmod
x^2 - 2$ only when $a = c$ and $b = d$), there are infinitely many
distinct congruence classes in this case.
\vskip 10pt
\noindent
7.  If $p(x)$ has degree $k$ in $\mathbb{Z}_p[x]$, the congruence
classes mod $p(x)$ are in 1-1 correspondence with the remainders
on division by $p(x)$.  These are all polynomials of degree $< k$
and $0$:
$$r(x) = a_{k-1} x^{k-1} + \cdots + a_1 x + a_0$$
with $a_i \in \mathbb{Z}_p$ for $i = 0, \ldots, k - 1$.  This
gives $p^k$ different elements since there are $p$ choices
for each of the $a_i$.  
\vskip 10pt
\noindent
8.  The statement is true and here is a proof:  Since $p(x)$ is relatively
prime to $k(x)$, there exist polynomials $u(x)$ and $v(x)$ such
that $u(x) p(x) + v(x) k(x) = 1$, and as a result,
$v(x)k(x) \equiv 1 \!\bmod p(x)$.
Therefore, if $f(x)k(x) \equiv g(x)k(x) \!\bmod p(x)$, then 
using Theorem 5.2 part 2, 
$v(x)f(x)k(x) \equiv v(x)g(x)k(x) \!\bmod p(x)$, and
hence by commutativity and associativity of multiplication,
\[
f(x)(v(x)k(x)) \equiv f(x) \cdot 1 \equiv g(x) \cdot 1 \equiv g(x)(v(x)k(x)) \!\bmod p(x).
\]
This shows $f(x) \equiv g(x)\!\bmod p(x)$.  
\vfil\eject
\noindent  
{\it Section 5.2}
\vskip 10pt
\noindent
1.  The tables for $R = \mathbb{Z}_2[x]/(x^3+x+1)$  ($[,]$ omitted
for simplicity of typing and reading).  The addition table:
\footnotesize
\[
\begin{array}{cccccccccc}
+_R &\vline & 0 & 1 & x  & x+1 & x^2  & x^2+1 & x^2+x  & x^2+x+1\\ \hline
0 &\vline & 0 & 1 & x & x+1 & x^2 & x^2+1 & x^2+x  & x^2+x+1\\
1 &\vline & 1 & 0 & x+1& x & x^2+1 & x^2 & x^2+x+1 & x^2+x\\
x &\vline & x & x+1 & 0 & 1 & x^2+x & x^2+x+1 & x^2 & x^2 + 1\\
x+1 & \vline & x+1 & x & 1 & 0 & x^2+x+1 & x^2 + x & x^2 + 1 & x^2\\
x^2 & \vline & x^2 & x^2+1 & x^2+x & x^2+x+1 & 0 & 1 & x & x + 1\\
x^2+1 & \vline & x^2+1 & x^2 & x^2+x+1 & x^2+x & 1 & 0 & x+1 & x\\
x^2+x & \vline & x^2+x & x^2+x+1 & x^2 & x^2+1 & x & x + 1 & 0 & 1\\
x^2+x+1 & \vline & x^2+x+1 & x^2+x & x^2+1 & x^2 & x+1 & x & 1 & 0\\
\end{array}
\]
\normalsize
The multiplication table:
\footnotesize
\[
\begin{array}{cccccccccc}
+_R &\vline & 0 & 1 & x  & x+1 & x^2  & x^2+1 & x^2+x  & x^2+x+1\\ \hline
0 &\vline & 0 & 0 & 0 & 0 & 0 & 0 & 0  & 0\\
1      &\vline & 0 & 1       & x       & x+1    & x^2     & x^2+1    & x^2+x   & x^2+x+1\\
x      &\vline & 0 & x       & x^2     & x^2+x  & x+1     & 1        & x^2+x+1 & x^2+1\\
x+1    &\vline & 0 & x+1     & x^2+x   & x^2+1  & x^2+x+1 & x^2      & 1       & x\\
x^2    &\vline & 0 & x^2     & x+1     & x^2+x+1& x^2+x   & x        & x^2+1   & 1\\
x^2+1  &\vline & 0 & x^2+1   & 1       & x^2    & x       & x^2+x+1  & x+1     & x^2+x\\
x^2+x  &\vline & 0 & x^2+x   & x^2+x+1 &     1  & x^2+1   & x+1      & x       & x^2\\
x^2+x+1&\vline & 0 & x^2+x+1 & x^2+1   & x      &   1     & x^2+x    & x^2     & x+1\\
\end{array}
\]
\normalsize
By examining the multiplication table, we see that every nonzero element of 
this congruence class ring does have a multiplicative inverse.  Hence
$R$ is a field.  
\vskip 10pt
\noindent
3.  The tables for $R = \mathbb{Z}_2[x]/(x^2+1)$  ($[,]$ omitted
for simplicity of typing and reading).  The addition table:
\footnotesize
\[
\begin{array}{cccccc}
+_R &\vline & 0 & 1 & x & x+1\\ \hline
0 &\vline & 0 & 1 & x & x+1 \\
1 &\vline & 1 & 0 & x+1& x \\
x &\vline & x & x+1 & 0 & 1 \\
x+1 & \vline & x+1 & x & 1 & 0 \\
\end{array}
\]
\normalsize
The multiplication table:
\footnotesize
\[
\begin{array}{cccccc}
\cdot_R &\vline & 0 & 1 & x  & x+1\\ \hline
0   &\vline & 0 & 0 & 0 & 0 \\
1   &\vline & 0 & 1 & x & x+1 \\
x   &\vline & 0 & x & 1 & x+1 \\
x+1 &\vline & 0 & x+1 & x+1 & 0 \\
\end{array}
\]
\normalsize
By examining the multiplication table, we see that $x + 1$ is a zero-divisor
in this ring $R$.  Hence
$R$ is not a field.
\vskip 10pt
\noindent
6.  The addition is $[a x + b] + [c x + d] = [(a+c)x + (b+d)]$.  The
multiplication is done by multiplying $(ax + b)(cx + d) = ac x^2 + (bc + ad)x + bd$,
then taking the remainder on division by $x^2 - 2$, which is $(bc + ad)x + bd + 2ac$.
Hence $[ax + b][cx + d] = [(bc + ad)x + bd + 2ac]$.
\vskip 10pt
\noindent
14.  One general method for these is to use the criterion from Theorem 5.9:
$[f(x)]$ is a unit in $F[x]/(p(x))$ when $(f(x),p(x)) = 1$ (gcd).  You 
can find the inverse using the Euclidean algorithm as you learned in 
Algebraic Structures.  An alternate method (simpler) is to solve directly for the 
coefficients in the multiplicative inverse as follows.
\begin{enumerate}
\item[a]  $[2x - 3][ax + b] = [(-3a + 2b)x + (4a - 3b)]$ by Exercise 6 above.
This equals $[1]$ if $-3a + 2b = 0$ and $4a - 3b = 1$.  Solving simultaneously,
we obtain $a = -2$, $b = -3$.  So $[2x - 3]^{-1} = [-2x - 3]$.
\item[b] In $\mathbb{Z}_3[x]/(x^2+1)$, $[x^2 + x + 1] = [x]$  (remainder on division
by $x^2 + 1$).  Hence $[x][2x] = [2x^2] = [1]$, so the multiplicative inverse
of $[x]$ is $[2x]$.  
\item[c]  In $\mathbb{Z}_2[x]/(x^3 + x + 1)$, $[x^2+x+1][x^2] = [1]$ (see multiplication
table in answer to question 1 above.  Hence $[x^2+x+1]^{-1} = [x^2]$.
\end{enumerate}
\vskip 10pt
\noindent
{\it Section 5.3}
\vskip 10pt
\noindent
1.  We will use the criterion from Theorem 5.10 -- to show $F[x]/(p(x))$ is
a field, it is equivalent to show $p(x)$ is irreducible in $F[x]$.  For
polynomials of degree $3$ or less, we know that reducibility is equivalent
to the polynomial having a root in $F$.  
\begin{enumerate}
\item[a]  Let $f(x) = x^3 + 2x^2 + x + 1\in \mathbb{Z}_3[x]$.  We have
$f(0) = 1$, $f(1) = 2$ and $f(2) = 2 + 2 + 2 + 1 = 1$ in $\mathbb{Z}_3$.
Hence $f(x)$ has no roots in $\mathbb{Z}_3$ and hence is irreducible 
in $\mathbb{Z}_3[x]$.
\item[b]  Similarly, if $g(x) = 2x^3 - 4x^2 + 2x + 1\in \mathbb{Z}_5[x]$, then
$g(2) = 16 - 16 + 4 + 1 \equiv 0 \!\bmod 5$.  Since $g(x)$ has a root in $\mathbb{Z}_5$,
so $g(x)$ is not irreducible and
the ring $\mathbb{Z}_5[x]/(g(x))$ is {\it not a field}. 
\item[c]  There are no roots of $h(x) = x^4 + x^2 + 1$ in $\mathbb{Z}_2$.  However, 
it is easy to check that $x^4 + x^2 + 1 = (x^2 + x + 1)^2$.  So $h(x)$ is not
irreducible and
$\mathbb{Z}_2[x]/(h(x))$ is {\it not a field}. 
\end{enumerate}
\vskip 10pt
\noindent
2. 
\begin{enumerate}
\item[a] $\mathbb{Q}(\sqrt{2})$ is clearly nonempty ($r,s \in \mathbb{Q}$ are arbitrary).
It is also closed under differences since
\[
(r + s\sqrt{2}) - (r' + s'\sqrt{2}) = (r - r') + (s - s')\sqrt{2}\in \mathbb{Q}(\sqrt{2}).
\]
It is also closed under products since
\[
(r + s\sqrt{2})\cdot (r' + s'\sqrt{2}) = (rr' + 2ss') + (rs' + r's)\sqrt{2}\in \mathbb{Q}(\sqrt{2}).
\]
This shows by Theorem 3.6 that $\mathbb{Q}(\sqrt{2})$ is a subring of $\mathbb{R}$.
Since multiplication in $\mathbb{R}$ is commutative, the same is true in the subring.
Moreover $1 = 1 + 0\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ so $\mathbb{Q}(\sqrt{2})$
is a commutative ring with identity.  Hence the only remaining thing to show
to see that it is a field is to show that every nonzero element has a multiplicative
inverse in $\mathbb{Q}(\sqrt{2})$.   To see this, note that if $r + s\sqrt{2} \ne 0$, 
then $(r + s\sqrt{2})(r - s\sqrt{2}) = r^2 - 2s^2$.  This cannot be zero, since
otherwise, we would have a square root of $2$ in $\mathbb{Q}$.  Hence
\[
(r + s\sqrt{2})^{-1} = \frac{r}{r^2 - 2s^2} + \left(\frac{-s}{r^2 - 2s^2}\right)\sqrt{2}
\]
is a multiplicative inverse.  
\item[b]  Define 
\begin{eqnarray*}
\varphi : \mathbb{Q}(\sqrt{2}) & \to & \mathbb{Q}[x]/(x^2 - 2)\\
                  r + s\sqrt{2}& \mapsto & [r + sx]
\end{eqnarray*}
Then $\varphi$ is clearly 1-1 and onto.  Moreover, 
\begin{eqnarray*}
\varphi( (r + s\sqrt{2}) + ( r' + s'\sqrt{2})) & = & \varphi((r + r') + (s + s')\sqrt{2})\\
&=& [(r + r') + (s + s')x] \\
&=& [r + sx] + [r' + s'x]\\
&=& \varphi(r + s\sqrt{2}) + \varphi( r' + s'\sqrt{2}).
\end{eqnarray*}
Moreover, 
\begin{eqnarray*}
\varphi( (r + s\sqrt{2})\cdot ( r' + s'\sqrt{2})) & = & \varphi((rr'+2ss') + (rs' + r's)\sqrt{2})\\
&=& [(rr'+2ss') + (rs' + r's')x] \\
&=& [r + sx] \cdot [r' + s'x]\\
&=& \varphi(r + s\sqrt{2}) \cdot \varphi( r' + s'\sqrt{2}).
\end{eqnarray*}
This shows that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}[x]/(x^2 - 2)$ 
are isomorphic.
\end{enumerate}
\vskip 10pt
\noindent
5. 
\begin{enumerate}
\item[a] $\mathbb{Q}[\sqrt{3}]$ is clearly nonempty ($r,s \in \mathbb{Q}$ are arbitrary).
It is also closed under differences since
\[
(r + s\sqrt{3}) - (r' + s'\sqrt{3}) = (r - r') + (s - s')\sqrt{3}\in \mathbb{Q}(\sqrt{3}).
\]
It is also closed under products since
\[
(r + s\sqrt{3})\cdot (r' + s'\sqrt{3}) = (rr' + 3ss') + (rs' + r's)\sqrt{3}\in \mathbb{Q}(\sqrt{3}).
\]
This shows by Theorem 3.6 that $\mathbb{Q}(\sqrt{3})$ is a subring of $\mathbb{R}$.
Since multiplication in $\mathbb{R}$ is commutative, the same is true in the subring.
Moreover $1 = 1 + 0\sqrt{3} \in \mathbb{Q}(\sqrt{3})$ so $\mathbb{Q}(\sqrt{3})$
is a commutative ring with identity.  Hence the only remaining thing to show
to see that it is a field is to show that every nonzero element has a multiplicative
inverse in $\mathbb{Q}(\sqrt{3})$.   To see this, note that if $r + s\sqrt{3} \ne 0$, 
then $(r + s\sqrt{3})(r - s\sqrt{3}) = r^2 - 3s^2$.  This cannot be zero, since
otherwise, we would have a square root of $3$ in $\mathbb{Q}$.  Hence
\[
(r + s\sqrt{3})^{-1} = \frac{r}{r^2 - 3s^2} + \left(\frac{-s}{r^2 - 3s^2}\right)\sqrt{3}
\]
is a multiplicative inverse.  
\item[b]  Define 
\begin{eqnarray*}
\varphi : \mathbb{Q}(\sqrt{3}) & \to & \mathbb{Q}[x]/(x^2 - 3)\\
                  r + s\sqrt{3}& \mapsto & [r + sx]
\end{eqnarray*}
Then $\varphi$ is clearly 1-1 and onto.  Moreover, 
\begin{eqnarray*}
\varphi( (r + s\sqrt{3}) + ( r' + s'\sqrt{3})) & = & \varphi((r + r') + (s + s')\sqrt{3})\\
&=& [(r + r') + (s + s')x] \\
&=& [r + sx] + [r' + s'x]\\
&=& \varphi(r + s\sqrt{3}) + \varphi( r' + s'\sqrt{3}).
\end{eqnarray*}
Moreover, 
\begin{eqnarray*}
\varphi( (r + s\sqrt{3})\cdot ( r' + s'\sqrt{3})) & = & \varphi((rr'+3ss') + (rs' + r's)\sqrt{3})\\
&=& [(rr'+ 3ss') + (rs' + r's')x] \\
&=& [r + sx] \cdot [r' + s'x]\\
&=& \varphi(r + s\sqrt{3}) \cdot \varphi( r' + s'\sqrt{3}).
\end{eqnarray*}
This shows that $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}[x]/(x^2 - 3)$ 
are isomorphic.
\end{enumerate}
\vskip 10pt
\noindent
9 a.  This follows from what we did in problem 1 of Section 5.2.  b.  I will write
the equation as $z^3 + z + 1 = 0$.  Using the multiplication and addition tables
from problem 1 in Section 5.2, we have $z = [x], [x^2], [x^2 + x]$ are all roots
of $z^3 + z + 1$.  So there are three roots in this field.
\vskip 10pt
\noindent
10.  We argue by contradiction.  If there were such an isomorphism $\varphi$, 
then we note first that since $\varphi$ is surjective, 
$\varphi([1]) = [1]$ by one of our general properties
of ring homomorphisms.  Hence by the homomorphism properties.  
$\varphi([n]) = [n]$ for all $n\in \mathbb{Z}$.  Now,   
there would have to be some $[a x + b] \in \mathbb{Q}[x]/(x^2 - 2)$ that
maps to $[x] \in \mathbb{Q}[x]/(x^2 - 3)$.  This implies $\varphi([ax + b]^2) = [x]^2 = [3]$.
But then, 
$[a x + b][a x + b] = [2ab x + b^2 + 2a^2] = [3]$ in $\mathbb{Q}[x]/(x^2 - 2)$ also.
We claim this cannot happen for the following reason.  
We must have $2ab = 0$ and $b^2 + 2a^2 = 3$.  These equations have no solutions
in the rational numbers: the first says $a = 0$ or $b = 0$, and then either
way there is no solution of the second there are no rational numbers
$r$ with $r^2 = 3$ or $r^2 = \frac{3}{2}$.  
\end{document} 
