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\noindent{\bf Math 431 homework Discussion}\hfill Due Friday, October 26, 2001

\ex
Let $I = (2,1+\sqrt{-3}) \subset \Z[\sqrt{-3}]$.  Show
that $I^2 = 2I$.  Why does this not contradict unique factorization
of ideals in a Dedekind domain?

\A
The product of two ideals is generated by all possible products of
pairs of generators.  So
$$I^2 = (4,2(1+\sqrt{-3}),-2+2\sqrt{-3}) = (4,2+2\sqrt{-3}) = 2I.$$
This does not contract unique factorization in a Dedekind ring
because $\Z[\sqrt{-3}]$ isn't a Dedekind domain.

\ex
Let $F = \Q(\alpha)$, where $\alpha^3-\alpha^2-2\alpha-8 = 0$.
Show that $\Z_F = \spn_\Z(1,\alpha,(\alpha^2+\alpha)/2)$.
Find three nontrivial ring homomorphisms from $\Z_F$ to $\F_2$.

\A
The fab $L := \spn_\Z(1,\alpha,(\alpha^2+\alpha)/2)$ satisfies
$$ \Z[\alpha] \subset_{\ne} L \subset \Z_F.$$
From the formula
$$\disc(1,\alpha,\alpha^2) = \left[\Z_F:\Z[\alpha]\right] \disc(\Z_F)$$
and a calculation that the discriminant of the minimal polynomial
for $\alpha$ is $-2^2\cdot 503$ (note: the formula for the discriminant
of a general monic cubic is in the class notes on resultants and
discriminants), we see that the index of $\Z[\alpha]$ in $\Z_F$ must
be 2, and that $L = \Z_F$.

If we let $\beta = (\alpha^2+\alpha)/2$, then $\Z_F = \spn_\Z(1,\alpha,\beta)$.
As can be checked by computing $\alpha\beta$ and $\beta^2$, the 
ring structure is given by the rules 
$$\alpha^2 = -\alpha+2\beta, \quad
\alpha\beta = 2\beta+4, \quad \beta^2 = 2\alpha+3\beta+6.$$
In order to define a homomorphism from $\Z_F$ to $\F_2 = \{0,1\}$ we have
to map rational integers to their residues mod~2, and map
map each of $\alpha$ and $\beta$ to 0 or~1.  So there are nominally
four possibilities.  A quick glance at the multiplication rules shows that
mapping $\alpha$ and $\beta$ both to 1 would contradiction the multiplication
rule for $\alpha\beta$, but that every other possibility gives a
homomorphism.
Thus there are three homomorphisms all together.  Note that the kernel
of each of this is a prime ideal of index (norm) equal to~2.

Digression: This explains why the field $\Q(\alpha)$ has no power
basis.  The prime 2 splits into three prime ideals of degree~1.  
\emph{If} there was a power basis, then the minimal polynomial
of the generator would have to factor into the product of three
irreducible polynomials of degree~1 modulo~2.  Unfortunately,
there are only two such irreducible polynomials, $x$ and $x+1$.

\ex
Show that any Dedekind domain with finitely many prime ideals is a
PID.  (Hint: Use the CRT to find a generator of an ideal.)

\A One way to do this is to show that each prime ideal $P$ is principal.
Choose an element~$\beta$ in $P$ but not in $P^2$.  (This is possible
since factorization into primes is unique, so $P$ and $P^2$ are
distinct ideals.)  Now use the Chinese Remainder Theorem to chose
an element~$\alpha$ that is congruent to $\beta$ modulo~$P^2$ and
congruent to~1 modulo all of the (finitely many) other prime ideals.
By construction the principal ideal $(\alpha)$ is equal to~$P$, and
$P$ is principal, as desired.

\ex
Show that if $I$ is any ideal in a number ring $\Z_F$ then
$\Z_F/I$ is a PID.

\A There is a one-to-one correspondence between ideals $\overline J$ in the
quotient ring and ideals $J$ in $\Z_F$ that contain~$I$.  Given such an
ideal $\overline J$, and a corresponding ideal~$J$, chose a nonzero
element of~$I$, and then apply our ``two-generator'' theorem from class
to find a $\beta$ such that $J = (\alpha,\beta)$.  Since $\alpha$ is
in $I$, we see that in the quotient $\alpha$ vanishes, and
$\overline J = (\overline \beta)$ so that $\overline J$ is principal
as desired.

The ring $\Z_F/I$ need \emph{not} be a Dedekind domain, since, for instance,
it might not even be an integral domain.

\ex
Let $F = \Q(\alpha)$ where $\alpha^3+\alpha+1 = 0$.
Show that in the ring $\Z_F$
$$(3) = PQ, \mbox{ where } P = (3,\alpha+2), Q = (3,\alpha^2+\alpha+2)$$
$$(31) = P^2Q, \mbox{ where } P = (31,\alpha-14), Q = (31,\alpha-3).$$

For the primes over~3 we have
$$PQ = (9,3\alpha+6, 3\alpha^2+3\alpha+6, 3\alpha^2+3\alpha+3).$$
The difference of the last two generators is 3, and since every other
generator is a multiple of 3 we have $PQ = (3)$.

For the primes over $p = 31$ we have
$P^2 = (p^2,p(\alpha-14),\alpha^2-28\alpha+196)$ and
\begin{eqnarray}
P^2Q & = & (p^3,p^2(\alpha-14),p(\alpha^2-28\alpha+196),p^2(\alpha-3),\\
&&p(\alpha^2-17\alpha+42),p(-\alpha^2+9\alpha+19)).
\end{eqnarray}
There are three generators that are nominally divisible by~$p$.
We eliminate the $\alpha^2$ terms by
subtracting the first two of these to give $u = p(-11\alpha+154)$.
Adding the last two of these gives $v = p(-8\alpha+23)$.
To eliminate the $\alpha$ term we calculate $8u-11v = -979p$.
Since $979$ is not divisible by~$p$, the gcd of $979p$ and $p^3$
is $p$, so a suitable linear combination of those elements is equal
to~$p$.  And since every element is divisible by~$p$, we conclude
that $(p) = P^2Q$.

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