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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?2011-09-12
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    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.2011-09-12
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    @jspecter: I didn't know that result about diagenicness (?) Can you point to a reference on that?2011-09-12
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    Regarding the above, see also http://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 http://math.uga.edu/~pete/integral.pdf.2011-09-12
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    Hmmm. Looks like I'm wrong. Sorry. How do you get the count for the number maximal ideals above 2?2011-09-12
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    @jspecter, re your first comment; thanks for the help. Even in the monogenic case though, I don't understand the last isomorphism. It looks like you're using the chinese remainder theorem, but in general the factors of $irr(a, S)$ in $R[x]$ won't generate comaximal ideals. This will happen even for $R = Z[\sqrt{d}]$ - am I missing something here?2011-09-12
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    @jspecter: Dear jspecter, As anon notes in the preceding comment, Even in the monogenic case, one does typically not get a non-trivial factorization of the kind you claim. One sees this already with the example $R = \mathbb Z[i]$. Regards,2011-09-12
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    @Pete L. Clark. God knows this is probably wrong, but I don't think I agree with your bound on the number of ideals above 2. For example take $K = \mathbb{Q}(\zeta_{3\cdot5\cdot7})$ and $k = \mathbb{Q}.$ Then $\mathcal{O}_K = \mathbb{Z}[\zeta_{3\cdot5\cdot7}]$ is a monogenic $\mathbb{Z}$-algebra so by your estimate there should be at most $2$ ideals lying above $2.$ But I don't think this is the case. To see this observe that the extension is unramified at $2$ and that there is no cyclic subgroup of...2011-09-13
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    $Gal(K/k) \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/6\mathbb{Z}$ of index $1$ or $2.$ Thus the number of primes above 2 which is equal to the index of decomposition group of $K/k$ at $2$ inside $Gal(K/k),$ must be greater than $2.$2011-09-13
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    That said. What I've written in my first comment is almost entirely incorrect. Anon's observation is correct and I was too hasty in writing my comment. This seems to be a common thread among many of my posts to this site. Sorry for this.2011-09-13
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    @jspecter: Yes, I think my estimate was slightly off: I forgot (!) that $\mathbb{Z}$ itself has a homomorphism to $\mathbb{Z}/2\mathbb{Z}$. So the number of distinct ring homomorphisms from $\mathbb{Z}[t_1,\ldots,t_n]$ to $\mathbb{Z}/2\mathbb{Z}$ is at most $2^{n+1}$ (since such a homomorphism is determined by the image of $1,t_1,\ldots,t_n$ and there are at most $2$ choices for each).2011-09-13
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    @Pete L. Clark. I don't think I agree with that either. $Gal(K/k)$ in my above example doesn't have a cyclic subgroup of index $3$ or $4$ either. I think where you are making a mistake is assuming reducing modulo a prime above $(2)$ lands you in $\mathbb{F}_2.$ The number of primes above $2$ is bounded by $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}[t_1,...,t_n],\mathbb{\overline{F}}_2)$ not $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}[t_1,...,t_n],\mathbb{F}_2)$2011-09-13
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    @jspecter: apologies again. I mean to assume that $2$ splits completely in $K$.2011-09-13
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    @Pete L. Clark. I see. Very nice. I see though you have deleted your comment and now this conversation makes little sense. Should I do the same?2011-09-13

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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.2011-09-13