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I came across the following claim in K Conrad's notes: Let $L/K$ be a finite extension of number fields, For fractional ideals $\mathfrak{a}, \mathfrak{b}$, and $\mathfrak{c}$ of $\mathcal{O}_L$, with $\mathfrak{a} \supseteq \mathfrak{b}$, it holds that $\mathfrak{ac}/\mathfrak{bc} \cong \mathfrak{a}/\mathfrak{b}$ as $\mathcal{O}_K$-modules.

When $\mathfrak{c}=(c)$ is principal the isomorphism is simply multiplication by $c$. In general I am not able to see how to get the isomorphism. Can someone help me? Thanks a lot.

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You should be able to localize the problem: there are many numbers you can invert without changing the quotients. If you invert enough numbers, your rings become principal ideal domains. (all but finitely many rational primes is guaranteed to be enough)

It may help to first reduce the problem to one of integral ideals.

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  • $\begingroup$ First one has to construct the map $ac/bc \to a/b$. Then one can check locally if it is an isomorphism. $\endgroup$ – Martin Brandenburg Nov 5 '13 at 11:26
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Define $\mathfrak{ac}\to\mathfrak{a/b}$ by inclusion $x\mapsto x+\mathfrak{b}$. Note then that the kernel of this map is precisely $\mathfrak{ac}\cap\mathfrak{b}$. But, since we're in a Dedekind domain $\mathfrak{ac}\cap\mathfrak{b}=\text{lcm}(\mathfrak{ac},\mathfrak{b})=\mathfrak{bc}$.

EDIT: As anon points out, I forgot to mention surjectivity. To see this merely note that we must show that the sets $\mathfrak{ac/bc}$ and $\mathfrak{a/b}$ have the same cardinality--recall that all indexes in sight are finite. But, we know that

$$[\mathfrak{a}:\mathfrak{b}]=\|\mathfrak{a}\|\|\mathfrak{b}\|^{-1}$$

and

$$[\mathfrak{ac}:\mathfrak{bc}]=\|\mathfrak{ac}\|\|\mathfrak{bc}\|^{-1}=\|\mathfrak{a}\|\|\mathfrak{c}\|\|\mathfrak{b}\|^{-1}\|\mathfrak{c}\|^{-1}=\|a\|\|\mathfrak{b}\|^{-1}$$

where $\|\mathfrak{r}\|=[\mathcal{O}_K:\mathfrak{r}]$ is the absolute norm of an ideal.

I assume above that no ideal is zero, but if that's true this is trivial.

EDIT: The above only works when $\text{gcd}(\mathfrak{b},\mathfrak{c})=(1)$--let me think about this for a bit, it would be nice to do it without having to localize!

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  • $\begingroup$ @anon How's that look? $\endgroup$ – Alex Youcis Nov 5 '13 at 5:55
  • $\begingroup$ Isn't this the zero map if $\mathfrak{c} = \mathfrak{b}$? $\endgroup$ – Hurkyl Nov 5 '13 at 6:00
  • $\begingroup$ I imagine what you want is a number such that $(x) = \mathfrak{c} \mathfrak{d}$, with $\mathfrak{d}$ relatively prime to everything relevant, and then the OP's proof works. $\endgroup$ – Hurkyl Nov 5 '13 at 6:22
  • $\begingroup$ @Hurkyl By OP, do you mean me? $\endgroup$ – Alex Youcis Nov 5 '13 at 6:29
  • $\begingroup$ I meant Zeyu by OP. Your argument might be needed to deal properly with $\mathfrak{d}$, though: I haven't thought it through. $\endgroup$ – Hurkyl Nov 5 '13 at 6:31

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