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\noindent
Math 541 \\
Problem Set 10
\begin{enumerate}


\item[7.6.5.] 
\begin{enumerate}
\item Let $\phi: \ZZ \to \ZZ/n_1\ZZ \times \ZZ/n_2\ZZ \times \dotsb \times \ZZ/n_k\ZZ$ be the natural map.  Solving the simultaneous congruences $x \equiv a_i \mod n_i$ is the same as finding an $x \in \ZZ$ such that $\phi(x) = (a_1, a_2, \dotsc, a_n)$.  Since the $n_i$ are pairwise relatively prime, the ideals $n_i\ZZ$ are pairwise comaximal, so by the Chinese Remaidner Theorem, $\phi$ is surjective, so there is such an $x$.  Suppose that $y$ is another solution.  Then $\phi(y) = \phi(x)$, so $\phi(x-y) = 0$, so $x-y \in \ker \phi$.  But the Chinese Remainder Theorem tells us that $\ker \phi = n_1 n_2 \dotsm n_k \ZZ$, so $x \equiv y \mod{n_1 n_2 \dotsm n_k}$.
\item By construction, $t_i n_i' \equiv 1 \mod n_i$.  If $j \ne i$ then $n_i$ divides $n_j'$, so $t_j n_j' \equiv 0 \mod n_i$.  Thus
\[ a_1 t_1 n_1' + \dotsb + a_i t_i n_i' + \dotsb + a_k t_k n_k'\ \equiv\ a_1 \cdot 0 + \dotsb + a_i \cdot 1 + \dotsb + a_k \cdot 0 \mod n_i \]
as required.
\item Take $n_1 = 8$, $n_2 = 25$, and $n_3 = 81$.  Then $n_1' = 25\cdot81 \equiv 1\cdot1 \mod 8$, $n_2' = 8 \cdot 81 \equiv 8\cdot6 = 48 \equiv -2 \mod{25}$, and $n_3' = 8\cdot25 \equiv 38 \mod{81}$.  Thus $t_1 = 1$, $t_2 = -13$, and $t_3 = 32$, so
\[ x \equiv 1\cdot1\cdot81\cdot25 - 2\cdot13\cdot81\cdot8 + 3\cdot32\cdot8\cdot25 = 4377 \mod{16200}. \]
\end{enumerate}


\item[8.1.4.]
\begin{enumerate}
\item Since $(a,b) = 1$, there are $x, y \in R$ such that $ax + by = 1$, so $acx + bcy = c$.  Now $a$ divides $acx$ and $bcy$, so $a$ divides $c$.

More generally, there are $x, y \in R$ such that $ax + by = (a,b)$, so $acx + bcy = (a,b)c$.  Now $a$ divides $acx$ and $bcy$, so $a$ divides $(a,b)c$, so $a/(a,b)$ divides $c$.
\item We showed in Exercise 0.2.4 that these are solutions; now we show that these are the only solutions.  If $ax + by = N$ then $ax + by = ax_0 + by_0$, so
\begin{equation}\label{eq1}
a(x-x_0) = b(y_0-y).
\end{equation}
Thus $a$ divides $b(y_0-y)$, so $a/(a,b)$ divides $y_0-y$, so there is an $m \in \ZZ$ such that
\begin{equation}\label{eq2}
y_0-y = m \frac{a}{(a,b)},
\end{equation}
so $y = y_0 - ma/(a,b)$.  Substituting \eqref{eq2} into \eqref{eq1}, we have
\begin{align*}
a(x-x_0) &= b m \frac{a}{(a,b)} \\
x-x_0 &= m \frac{b}{(a,b)}
\end{align*}
so $x = x_0 + m b/(a,b)$.
\end{enumerate}


\item[8.2.5.]
\begin{enumerate}
\item That $I_3$ is not principal is proved in example 2 on page 273.  For $I_3'$, the proof is almost identical.  It remains to show that $I_2 = (2, 1+\sqrt{-5})$ is not principal.  Let $N$ be the field norm $N(a+b\sqrt{-5}) = a^2+5b^2$.  Observe that for all $r \in R$, $N(r) \ge 0$, $N(r) = 1$ if and only if $r = \pm 1$, and $N(r) \ne 2$.

Suppose that $I_2$ is generated by one element $r \in R$.  Then $r$ divides $2$ and $1+\sqrt{-5}$, so $N(r)$ divides $N(2) = 4$ and $N(1+\sqrt{-5}) = 6$, hence is either 1 or 2.  $N(r) \ne 2$, so $N(r) = 1$, so $r = \pm 1$, so $I_2 = (1)$.  Thus there are $x, y \in R$ such that $2x + (1+\sqrt{-5})y = 1$.  If we multiply through by $1-\sqrt{-5}$ we get $2(1-\sqrt{-5})x - 4y = 1-\sqrt{-5}$, but this is impossible since $1-\sqrt{-5}$ is not a multiple of 2.
\end{enumerate}


\newpage
\item[8.3.5.] Let $N$ be the field norm $N(a+b\sqrt{-n}) = a^2 + nb^2$.  Again observe that for all $r \in R$, $N(r) \ge 0$, $N(r) = 1$ if and only if $r = \pm 1$, and $N(r) \ne 2$ since $n \ge 3$.
\begin{enumerate}
\item First we show that 2 is irreducible.  Suppose that $2 = rs$ with $r, s \in R$.  Taking norms, we have $4 = N(r)N(s)$, so $N(r)$ is 1, 2, or 4.  If $N(r) = 1$ then $r$ is a unit.  $N(r) = 2$ is impossible.  If $N(r) = 4$ then $N(s) = 1$, so $s$ is a unit.

Next we show that  $\sqrt{-n}$ is irreducible.  Suppose that $\sqrt{-n} = (a + b\sqrt{-n})(c + d\sqrt{-n})$.  Taking norms,
\[ n = (a^2 + nb^2)(c^2 + nd^2). \]
Thus $b = 0$ or $d = 0$, for otherwise $(a^2 + nb^2)(c^2 + nd^2) \ge n^2$.  If $b = 0$ then $a^2 = 1$ since $n$ is squarefree, so $a + b\sqrt{-n} = \pm 1$ is a unit.  Similarly, if $d = 0$ then $c + d\sqrt{-n}$ is a unit.

Last we show that $1+\sqrt{-n}$ is irreducible.  Suppose that $1+\sqrt{-n} = (a + b\sqrt{-n})(c + d\sqrt{-n})$.  Since $N(1+\sqrt{-n}) = 1+n < n^2$, we have $b = 0$ or $d = 0$ as before.  If $b = 0$ then $1+\sqrt{-n} = ac + ad\sqrt{-n}$, so $ad = 1$, so $a= \pm 1$, so $a + b\sqrt{-n}$ is a unit.  Similarly, if $d = 0$ then $c + d\sqrt{-n}$ is a unit.

\item If $n$ is odd then $(1+\sqrt{-n})(1-\sqrt{-n}) = 1 + n$ is a multiple of 2, but neither factor is a multiple of 2, so 2 is not prime.

If $n$ even then there is an $m > 1$ with $n = 2m$ since $n \ge 3$.  Then $(\sqrt{-n})^2 = -2m$, but $\sqrt{-n}$ does not divide either factor, as follows.  If $\sqrt{-n}$ divides $m$ then there are $a, b \in \ZZ$ with $m = (a + b\sqrt{-n})(\sqrt{-n}) = -bn + a\sqrt{-n}$, so $-bn = m$, but this is impossible since $m < n$.  Similarly, $\sqrt{-n}$ does not divide $2$.  Thus $\sqrt{-n}$ is not prime.

In either case $R$ contains an irreducible that is not prime, hence is not a U.F.D.

\item If $n$ is odd then $I = (2, 1+\sqrt{-n})$ is not principal, as follows.  Suppose on the contrary that $I = (r)$.  Then there is an $s \in R$ with $2 = rs$, so one of $r$ and $s$ is a unit as we saw above.  If $s$ is a unit then $r = 2s^{-1}$, but $1+\sqrt{-n}$ is a multiple of $r$ and not of 2, so this is impossible.  If $r$ is a unit then $I = (1)$, so there are $x, y \in R$ such that $2x + (1+\sqrt{-n})y = 1$.  If we multiplying through by $1-\sqrt{-n}$ we get $2(1-\sqrt{-n})x + (1+n)y = 1-\sqrt{-n}$, but this is impossible since $1-\sqrt{-n}$ is not a multiple of 2.

If $n$ is even then $I = (2, \sqrt{-n})$ is not principal, as follows.  Suppose on the contrary that $I = (r)$.  Then $N(
r)$ divides $N(2) = 4$ and $N(\sqrt{-n}) = n$, which is squarefree, so $N(r)$ is 1 or 2, so $r = \pm 1$.  Thus $I = (1)$, so there are $x, y \in R$ such that $2x + \sqrt{-n}y = 1$.  If we Multiply through by $\sqrt{-n}$ we get $2\sqrt{-n}x - ny = \sqrt{-n}$, but this is impossible since $\sqrt{-n}$ is not a multiple of 2.
\end{enumerate}


\item[9.1.4.] Define a map $\QQ[x,y] \to \QQ[y]$ by $x \mapsto 0$ and $y \mapsto y$; that is,
\[ \sum_{i,j} a_{ij} x^i y^j \mapsto \sum_j a_{0j} y^j. \]
This map is surjective onto the integral domain $\QQ[y]$, so its kernel $(x)$ is prime.

Define another map $\QQ[x,y] \to \QQ$ by $x \mapsto 0$ and $y \mapsto 0$; that is,
\[ \sum_{i,j} a_{ij} x^i y^j \mapsto a_{00}. \]
This map is surjective onto the field $\QQ$, so its kernel $(x, y)$ is maximal (and hence prime).

Since $(x) \subseteq (x,y)$, $(x)$ is not maximal.


\end{enumerate}
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