# Thursday, September 08

## Dedekind domains

:::{.remark}
Last time: facts about Dedekind domains, e.g. $R \in \Dedekind$ iff every fractional ideal $I\in \Frac(R)$ is invertible iff every fractional ideal factors uniquely into primes (by a of Matusita).
Check (exercise) that $v_{p_i}(x) \geq v_{p_i}(I)$ for $x\in I$.
Recall that $\supp I = \st{p \in \mspec R \st v_p(I) \neq 0}$, and Weil=Cartier since these are nonsingular affine curves (invertibility of fractional ideals $\approx$ projectivity here).
Recall that larger powers of ideals get smaller.
Ideals are projective modules in a Dedekind domain.
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:::{.lemma title="Strong approximation and the moving lemma"}
Let $R\in \Dedekind$ and $S \da \ts{p_1,\cdots, p_n} \subseteq \mspec R$.

- (Strong approximation) If $I\in \Frac(R)$ then there is an $x\in I$ such that $v_{p_i}(x) = v_{p_i}(I)$ where $I = \prod_{p} p^{ v_{p_i}(I) }$.

- (Moving lemma) If $A\in \Frac(R)$ then there exists an integral ideal $b\normal R$ such that $\supp(b)$ is disjoint from $S$ and $[a] = [b]$ in $\Pic(R)$.

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:::{.corollary title="?"}
If $R\in \Dedekind$, then

- Exactly one of the following holds:
  - $R$ is a PID.
  - There are infinitely many nonprincipal prime ideals. 

- If $R$ is semilocal ($\size \mspec R < \infty$) then $R$ is a PID by finiteness.

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:::{.proof title="?"}
Suppose $R$ is not a PID, pick a nonprincipal prime $p$ and suppose the set $S$ of such nonprincipal primes is nonempty and finite.
By the moving lemma, if $A\in \Frac(R)$ it can avoid $S$, so pick $[b] = [A] \in \Pic(R)$ with disjoint support from $S$.
Therefore $b$ is a product of principal primes and thus principal, so $[b] = 1 = [A] \in \Pic(R)$ and $A$ must be principal.
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:::{.exercise title="This fails for non-Dedekind domains"}
Let $R = \ZZ\adjoin{\sqrt{-3}}$.
Show this is not integrally closed (since $-3\equiv 1\mod 4$) and thus not a Dedekind domain.
Moreover, show there exists a unique nonprincipal prime ideal, namely the one containing 3.

> This yields a nonmaximal order, and 3 is a singular prime (the local ring is not a DVR, so dimensions of tangent spaces are $2>1$).
> See equivalent characterizations for a 1-dimensional Noetherian ring to be a DVR.

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:::{.proof title="of lemma, first part"}
**Step 1**:
Suppose $I\normal R$ is integral, write it as $I = \prod_{i=1}^r p_i^{a_i} \prod_{j=1}^s q_i^{b_j}$ with $a_i\in \ZZ_{\geq 0}$ and the $q_j$ are the other primes in the support of $I$, so the $b_j \in \ZZ_{> 0}$ -- i.e. formally include primes not in the support, just allowing exponent zero.
By CRT, $R \surjects \prod_{i=1}^r R/p_i^{a_i + 1} \times \prod_{j=1}^s R/q_j^{b_j + 1}$ and so there is some $x \in p_i^{a_i} \sm p_i^{a_i+1}$ for all $i$ and in $q_j^{b_j} \sm qq_j^{b_j + 1}$ for all $j$ by minimality of valuations.
Thus $v_{p_i}(x) = a_i = v_{p_I}(I)$ and similarly for $q_j$ and $b_j$, so $x\in I$.
This proves the first part when $a$ is an integral ideal.

**Step 2**:
For $I\in \Frac(R)$, try to clear denominators.
Write $I = {J\over b}$ with $J\normal R, b\in R\smz$, and by step 1 pick $x\in R$.
Then for every $p\divides J$ we have $v_p(x) = v_p(J)$ and thus $x\in J$.
We claim ${x\over b}$ works -- check ${x\over b}\in {J\over B} = I$, and for all $i$, 
\[
v_{p_i}(x/b) = v_{p_i}(x) - v_{p_i}(b) = v_{p_i}(J) - v_{p_i}(b) = v_{p_i}(J/\gens{b}) = v_{p_i}(I)
.\]

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:::{.proof title="of lemma, second part"}
Apply the first part to $I \da a\inv$.
Pick $x\in a\inv$ such that $v_{p_i}(x) = v_{p_i}(a\inv)$ for all $i$ and thus $v_{p_i}(a\inv x\inv) = 0$.
So $\supp(a\inv x\inv)$ is disjoint from $S$, and $a\inv x\inv \contains a\inv(a\inv)\inv = R$ -- the exact opposite of being integral.
Thus taking inverses, $ax \subseteq R$, and note that inverting an ideal doesn't change the support (just negate the exponents) so $\supp(ax)$ is still disjoint from $S$.
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## Modules over Dedekind domains

:::{.remark}
We'll focus on finitely-generated modules -- see [@CA, Ch.2] for more general modules.
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:::{.theorem title="Structure theorem"}
Let $R\in \Dedekind$ and $M\in\rmod$, then

a. The quotient $P\da M/M_\tors$ is not free, but rather finitely-generated and projective of some finite rank $r$.
b. If $r=0$ then $M = M_\tors$. 
  If $r\geq 1$ then $\exists I\normal R$ nonzero such that $M\cong M_\tors \oplus P \cong M_\tors \oplus R^{r-1} \oplus I$ for some non-principal ideal $I$ -- more directly, $M \cong M_\tors \oplus P$ and $P \cong R^{r-1} \oplus I$ (since rank 1 projectives are isomorphic to ideals).
c. The *Steinitz class* $\St(I) \da [I] \in \Pic(R)$ is an invariant of $M$.
d. If $T\da M_\tors \neq 0$ then $M_\tors \cong \bigoplus _{i=1}^N R/p_i^{n_i}$ in a uniquely determined way.
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\todo[inline]{
Check: free would imply principal for ideals?
}

:::{.remark}
For the last part of the theorem,  if $T$ is finitely-generated torsion, factor $\Ann_R(T) = \prod _{i=1}^r p_i^{a_i}$ and canonically $T\in \mods{R_{p_1,\cdots, p_r}}$ which is a semilocal Dedekind domain and hence a PID.
But $R/p_i^{a_i} \iso R_{p_1,\cdots, p_r}/ (p_i R_{p_1,\cdots, p_r} )^{a_i}$.

Note that $\Pic(R)$ is the only major complication in the structure theorem.
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:::{.exercise title="?"}
Show $I \oplus J \cong R \oplus IJ$.
:::