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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 5}
\centerline{October 11, 2007}
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\noindent
{\it Section 5.1}
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\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:
\[
[0],[1],[x],[x+1],[x^2],[x^2+1],[x^2+x],[x^2+x+1].
\]
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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.
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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.
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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 $a_i$.  



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