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I'm working on a problem in Lorenzini's book "An Invitation to Arithmetic Geometry" which asks to show that if $L = Q(\sqrt{-5}, \sqrt{-1})$ and $K = Q(\sqrt{-5})$, then the ring of integers of $L$ is unramified over that of $K$. It's question 25 on page 128. Here's what I've done thus far:

Because $L$ is a simple extension of $K$, namely $L = K(\sqrt{-1})$ with minimal polynomial of $\sqrt{-1}$ over $K$ just $f(x) = x^2 + 1$, the only maximal ideal $M$ in $L$ that could be ramified would contain $f'(i) = 2i$. Being prime it would have to contain $1+i$. In particular, this means that the only prime ideal of $Z[\sqrt{-5}]$ we need to consider is $P=(2, 1+\sqrt{-5})$, as $M \cap Z[\sqrt{-5}]$ would need to contain 2, and by looking at how the ideal $(2)$ ramifies in $Z[\sqrt{-5}]$ we get $P$.

But shouldn't $P$ ramify in $\mathfrak{O}_L$, the ring of integers of $L$? I know that $Z[\sqrt{-5}, \sqrt{-1}] \subseteq \mathfrak{O}_L$, and if we look at $f(x) \pmod{P}$ we have $(x+1)^2$.

Thus we should have a prime ideal $P' \subseteq Z[\sqrt{-5}, \sqrt{-1}]$ such that $(P')^2 \subseteq PZ[\sqrt{-5}, \sqrt{-1}]$. Would this not then imply that $(P')^2\mathfrak{O}_L \subseteq P\mathfrak{O}_L$, and as $[L:K] = 2$ that $P$ ramifies in $\mathfrak{O}_L$, i.e. $P = M^2$?

I know I've done something wrong, but I'm not sure where! I'd be grateful for any assistance in pointing out where I've messed up.

Garnet
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  • I think the argument doesn't work because $(2,1+\sqrt{-5})$ is not coprime to $\mathbb{Z}[\sqrt{-5},\sqrt{-1}]$ in $\mathfrak{O}_L$. You don't get $\frac{1+\sqrt{5}}{2}$ in their sum – Cocopuffs Sep 28 '13 at 15:27
  • A solution is given here http://math.stackexchange.com/questions/33573/ramification-in-a-tower-of-extensions – Cocopuffs Sep 28 '13 at 15:30
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    To reiterate what Cocopuffs said, the problem is that you needn't $L/K$ be a simple extension whose generator has min poly $x^2+1$, but you need $\mathcal{O}_L/\mathcal{O}_K$ to be simple with generator having min poly $x^2+1$. In fact, this is not the case. Indeed, $\mathcal{O}_L=\mathbb{Z}[i,\frac{1+\sqrt{-5}}{2}]$. – Alex Youcis Sep 28 '13 at 20:50
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    Thank you for the replies! I am a bit confused as to your comment, Alex. If $O_L = Z[i, \frac{1+\sqrt{-5}}{2}]$ would this not mean that $[L:Q] = 2$ as an integral basis is also a basis for $L$ over $Q$? – Garnet Sep 29 '13 at 00:17
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    @Garnet These are generators of $\mathcal{O}_L$ as an algebra and not as a module. If you let $a := i \cdot \frac{1+\sqrt{5}}{2}$, then ${1,a,a^2,a^3}$ is an integral basis. – Cocopuffs Sep 29 '13 at 01:37

1 Answers1

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You know that the extension is ramified at most over primes above $2$. But $\mathbb Q(i,\sqrt{-5}\,)$ is also equal to $\mathbb Q(\sqrt{-5},\sqrt5\,)$. And this extension of $\mathbb Q(\sqrt{-5}\,)$ is clearly unramified at $2$, because it involves a residue-field extension.

Lubin
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  • Thanks for the answer! This makes sense; I had not thought to consider the multiplicity of e in towers. – Garnet Sep 29 '13 at 00:20
  • Sorry, I've been thinking about it more, and while I understand the argument involving the multiplicity of e in towers, how do we know that the extension is ramified at most over primes above 2?

    My argument for it can't hold, because it assumes the ring of integers in $L$ is just $A/(x^2+1)$, which isn't the case here.

     – Garnet Sep 30 '13 at 22:03
  • I mean, is it true in general that if we have a prime $P$ of $A$ unramified over $A[x]/f(x)$ for an irreducible polynomial $f(x)$ that $P$ will be unramified in the integral closure of $A$ in $K[x]/f(x)$? – Garnet Sep 30 '13 at 22:12
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    I was simply using the fact that the extension $\mathbb Q(i)\supset \mathbb Q$ is ramified only at $2$. So when you take the compositum of these two fields with $\mathbb Q(\sqrt{-5},)$, the new extension is (at worst) ramified only at primes above $2$ (and of course there’s actually only one such prime). – Lubin Oct 01 '13 at 23:12
  • Oh! Okay! Thank you so much! – Garnet Oct 02 '13 at 00:12
  • @Lubin Could you explain why, when taking the compositum of $\mathbb{Q}(i)\supset \mathbb{Q}$ with $\mathbb{Q}(\sqrt{-5})$, you can’t have new primes in $\mathbb{Q}$ that ramify in $\mathbb{Q}(\sqrt{-5},i)$ ? – Nicholas Camacho Apr 12 '18 at 22:56
  • Don’t think of it as a compositum @NicholasCamacho, think of it as a tower, $\Bbb Q\subset\Bbb Q(\sqrt{-5},)\subset\Bbb Q(\sqrt{-5},i)$. As I said, this is at worst ramified above $2$. Maybe you feel part of my argument to be missing? (remember: as OP says in his first comment, it has to do with multiplicativity of ramification index in a tower.) – Lubin Apr 14 '18 at 01:58
  • @Lubin What's troubling me is applying the following fact to our situation: If $L/K$ and $L'/K$ are two finite separable extensions, then a prime of $K$ ramifies in $LL'$ if and only if it ramifies in $L$ or in $L'$. Since $2$ and $5$ ramify in $\mathbb{Q}(\sqrt{-5})$, and $2$ ramifies in $\mathbb{Q}(i)$, then both $2$ & $5$ ramify in $\mathbb{Q}(\sqrt{-5},i)$. So where do we get that $\mathbb{Q}(\sqrt{-5},i)$ is at worst ramified above $2$, and not at both at $2$ and $5$? Moreover, I don't see where multiplicativity of ramification index comes into play... Maybe I'm really missing something. – Nicholas Camacho Apr 14 '18 at 02:17
  • What you’re missing, @NicholasCamacho, is that we’re not talking about ramification over $\Bbb Q$, but ramification of the big field over $\Bbb Q(\sqrt{-5},)$. So the theorem you quote is not of interest. – Lubin Apr 14 '18 at 02:31
  • @Lubin Right. I suppose I was trying to force that theorem to apply to this problem (since Lorenzini gives it in the chapter where this problem is found). But it's simpler than that. Only over $2$, in the tower you mentioned, could we have ramification. But this doesn't happen. Thanks. – Nicholas Camacho Apr 14 '18 at 02:48