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Math 541 \\
Problem Set 11
\begin{enumerate}


\item[9.2.2.] It suffices to show that there are $q^n$ polynomials of degree at most $n-1$, since by exercise 9.2.1, the elements of $F[x]/(f)$ are in bijection with such polynomials.  A polynomial of degree at most $n-1$ is of the form
\[ a_0 + a_1 x + \dotsb + a_{n-1} x^{n-1} \]
with $a_0, a_1, \dotsc, a_{n-1} \in F$.  There are $|F| = q$ possibilities for $a_0$, $q$ possibilities for $a_1$, and so on, so there are $q^n$ possibilities.

\item[9.2.3.] $F[x]/(f)$ is a field if and only if $(f)$ is maximal.  If $(f)$ is maximal then it is prime; conversely, if $(f)$ is prime then it is maximal since $F[x]$ is a P.I.D.  Now $(f)$ is prime if and only if $f$ is prime, but since $F[x]$ is a U.F.D., primes and irreducibles are the same.


\item[9.3.1.] We prove the contrapositive.  Suppose that $R$ is a U.F.D.  By Gauss' Lemma, there are $r, s \in F$ such that $ra(x), sb(x) \in R[x]$ and $ra(x) \cdot sb(x) = p(x)$.  Since $a(x)$ and $b(x)$ are monic, the leading coefficients of $ra(x)$ and $sb(x)$ are $r$ and $s$, so $r, s \in R$.  Comparing the leading coefficients of $p(x)$ and $ra(x) \cdot sb(x)$, we find that $rs = 1$, so $a(x) = s \cdot ra(x)$, so $a(x) \in R[x]$.

Observe that the field of fractions of $\ZZ[2\sqrt2]$ is $\QQ[\sqrt2]$.  Now $x^2-2 = (x+\sqrt2)(x-\sqrt2)$; everything in sight is monic, but the left-hand side is monic and lies in $\ZZ[2\sqrt2][x]$, while the factors of the left-hand side do not.


\item[9.4.2.] We use Eisenstein's Criterion.
\begin{enumerate}
\item The coefficients $-4$ and 6 are both divisible by 2, but 6 is not divisible by $2^2$.
\item The coefficients 30, $-15$, 6, and $-120$ are all divisible by 3, but $-120$ is not divisible by $3^2$.
\item We take the hint: $(x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 2(x-1) + 1 = x^4 - 2x + 2$.  The coefficients $-2$ and 2 are divisible by 2, but 2 is not divisible by $2^2$.
\item By the binomial theorem,
\[ (x+2)^p = x^p + p \cdot 2x^{p-1} + {p \choose 2} 2^2 x^{p-2} + {p \choose 3} 2^3 x^{p-3} + \dotsb + {p \choose p-2} 2^{p-2} x^2 + p \cdot 2^{p-1} x + 2^p. \]
Observe that $p \choose n$ is divisible by $p$ if $1 \le n \le p-1$, since
\[ {p \choose n} = \frac{p(p-1)\dotsb(p-n+1)}{n!} \]
and if $n < p$ then $p$ does not divide $n!$, so the $p$ in the numerator cannot be canceled.  Thus in
\[ \frac{(x+2)^p - 2^p}{x} = x^{p-1} + p \cdot 2x^{p-2} + {p \choose 2} 2^2 x^{p-2} + {p \choose 3} 2^3 x^{p-3} + \dotsb + {p \choose p-2} 2^{p-2} x + p \cdot 2^{p-1} \]
all the coefficients are divisible by $p$, but the constant term $p \cdot 2^{p-1}$ is not divisible by $p^2$.
\end{enumerate}

\item[9.4.3.] Let $f(x) = (x-1)(x-2)\dotsb(x-n)-1$, and suppose that $f(x) = a(x)b(x)$, where $\deg a, \deg b < n$.  Then $a(1)b(1) = -1$, so $a(1) = \pm 1$ and $b(1) = \mp 1$, so $a(1) + b(1) = 0$.  Similarly, $a(2) + b(2) = a(3) + b(3) = \dotsb = a(n) + b(n) = 0$.  Thus $a(x) + b(x)$ is has degree less than $n$ but has $n$ distinct roots, so $a(x) + b(x) = 0$, so $f(x) = -a(x)^2$.  But $f(n+2) > 0$, whereas $-a(n+2)^2 \le 0$, which is a contradiction.


\end{enumerate}
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