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If $K/k$ is a finite Galois extension of fields, with Galois group $G$, there's an isomorphism $$ K \ \otimes_k \ K \simeq \oplus_{\sigma_i \in G} \ K$$ given by sending $a \otimes b$ to $ (..., \sigma_i(a) b, ...)$ and extending linearly. This is even an isomorphism of $K$-vector spaces, where $a \in K$ is acting by multiplication on the first factor in $K \otimes K$, and by multiplication by $(\sigma_i(a))$ on $\oplus_{\sigma_i} K$. There's also an obvious $k[G]$-module structure that's preserved.

My question is what happens if instead of $K$, we consider $R \otimes_S R$, where $R$ and $S$ are the rings of integers of number fields $K$ and $k$ respectively. I tried playing around with quadratic extensions, and I'm pretty sure that the same map above from $R \otimes R$ to $\oplus R$ is not surjective, so the same argument doesn't work? Is there a nice description of $R \otimes R$ as either an $R$ module, or an $S[G]$ module?

Thanks.

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It is already non-trivial to describe the $S[G]$-module structure of $R$ itself, no? –  Mariano Suárez-Alvarez Sep 12 '11 at 21:11
    
In the case $R$ is a monogenic S-algebra generated by some element $a$ and the extension of fraction fields, $K/k,$ is Galois, we have $$R \otimes_S R = R \otimes_S S[X]/irr(a,S) \cong R[X]/irr(a,S) \cong \displaystyle\bigoplus_{\alpha\in\mathbb{\overline{Q}}: irr(a,S)(\alpha) = 0} R.$$ Thus, the isomorphism fails at most only in the case $R$ is not monogenic. Luckily for number fields, $R$ is always diagenic. So I would advise us both to look at our favorite pet examples of algebraic number rings which are digenic but not monogenic. –  jspecter Sep 12 '11 at 21:16
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@jspecter: I didn't know that result about diagenicness (?) Can you point to a reference on that? –  Mariano Suárez-Alvarez Sep 12 '11 at 21:20
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Regarding the above, see also modular.math.washington.edu/129-05/challenges.html. Probably what you are remembering is that any ideal in a Dedekind domain can be generated (as an ideal!) by two elements. For this see e.g. $\S 20.5$ of math.uga.edu/~pete/integral.pdf. –  Pete L. Clark Sep 12 '11 at 21:39
    
Hmmm. Looks like I'm wrong. Sorry. How do you get the count for the number maximal ideals above 2? –  jspecter Sep 12 '11 at 22:06

1 Answer 1

up vote 4 down vote accepted

Let $\Delta$ denote the discriminant of $R$ over $S$. One then has that $R[1/\Delta]\otimes_S R[1/\Delta] = \prod_{\sigma \in G} R[1/\Delta].$ On the other hand, this is false without inverting $\Delta$.

Consider e.g. the simplest case: $$\mathbb Z[i] \otimes_{\mathbb Z} \mathbb Z[i] = \mathbb Z[i][x]/(x^2+1) = \mathbb Z[i][y]/y(y-2)$$ (where for the final equality, we set $y = -ix+1$).
If we invert $2$, then we can rewrite this as $\mathbb Z[i,1/2][z]/z(z-1)$ (setting $z = y/2$), which is a product of two copies of $\mathbb Z[i,1/2]$.

But Spec $\mathbb Z[i]/y(y-2)$ is connected (it contains only one point lying over the prime $2$ of $\mathbb Z$), and hence it does not factor as a non-trivial product.

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This is perfect, thanks. –  anon Sep 13 '11 at 13:05

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