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Let $A$, $B$ be finite free $\mathbb{Z}$-algebras such that $\operatorname{Spec}(A)$ and $\operatorname{Spec}(B)$ are both connected. Is $\operatorname{Spec}(A\otimes_{\mathbb{Z}} B)$ connected?

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i tried with $A=B=\mathbb{Z}[x]/(x^2-2)$, but that does not seem to work and i dont know how to go ahead with this problem. –  messi Jun 20 '12 at 16:43
    
How can a $\mathbb{Z}$-algebra be both finite and free? Also, you mention the algebra $\mathbb{Z}[x]/(x^2-2)$, but that is neither finite nor free. –  Jim Belk Jun 20 '12 at 22:30
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@Jim Belk, when people say an algebra "finite", it usually does not mean it has finitetly many elements, but means it is finitely generated as a module. This convention applies in a similar way to "free". So $\mathbb{Z}[x]/(x^2-2)=\mathbb{Z}[\sqrt{2}]=1\mathbb{Z}+\sqrt{2}\mathbb{Z}$ is free and finite. –  Ch Zh Jun 21 '12 at 5:28
    
@jim belk This is standard convention, @ zhe chen thanks for the clarification. one can take any such quotient, and you find that you are running into some kind of problem, it seems that the polynomial in the quotient should have degree greater than 3. –  messi Jun 21 '12 at 7:39
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Is this known if we replace Z by a field? –  Martin Brandenburg Jul 1 '12 at 12:29

1 Answer 1

Late edit. The answer is $\mathrm{Spec}(A\otimes_{\mathbb Z}B)$ is always connected.

One reduces to the case $A=B=\mathcal O_F$ for some finite Galois extension $F/\mathbb Q$ as below. Let $G=\mathrm{Gal}(F/\mathbb Q)$. Let $X=\mathrm{Spec}(\mathcal O_F)$ and $S=\mathrm{Spec}(\mathbb Z)$. For any $g\in G$, consider the surjective map $$ \mathcal O_F\otimes_{\mathbb Z}\mathcal O_F\to \mathcal O_F, \quad b\otimes c\mapsto bg(c).$$ It induces a closed immersion $i_g: X\to X\times_S X$. It is not hard to see that the $i_g(X)$, $g\in G$, are the irreducible components of $X\times_S X$, and that $i_g(X)\cap i_h(X)\ne\emptyset$ if (and in fact only if) $gh^{-1}$ belongs to the inertia subgroup $I_x$ of $G$ at some $x\in X$ (if $g=\theta h$ with $\theta\in I_x$, then $i_g(x)=i_h(x)\in i_g(X)\cap i_h(X)$). Using the fact the $I_x$'s, when $x$ varies, generate $G$ ($\mathbb Q$ has no nontrivial unramified extension), we easily get the connectedness.


I don't have a solution, but just some remarks.

  1. It is enough to deal with the case when $A, B$ are ring of integers of number fields $K, L$. Proof: Denote by $X=\mathrm{Spec}(A), Y=\mathrm{Spec}(B)$ and $S=\mathrm{Spec}(\mathbb Z)$. As $X, Y$ are finite over $S$ and are connected, each of their irreducible components $X_1,\dots, X_n, Y_1, \dots, Y_m$ are finite and surjective over $S$. If we can prove that $X_i\times_S Y_j$ is connected for all $i, j$, then simple topological arguments show that $X\times_S Y_j$ is connected. Similary, $X\times_S Y$ is connected. So we are reduced to the case $X, Y$ irreducible. As the connectedness property is purely topological, we can replace $X, Y$ by their maximal reduced subschemes and suppose that they are reduced (hence integral). Let $X', Y'$ be their normalizations. Then $X'\times_S Y'\to X\times_S Y$ is surjective. If $X'\times_S Y'$ is connected, then so is $X\times_S Y$. Therefore it is enough to treat the case when $X, Y$ are integral and normal, so their are defined by ring of integers of number fields.

  2. It is enough to deal with the case $A=B=O_F$ for some finite Galois extension $F/\mathbb Q$. Proof: let $K, L$ be as above, let $F$ be a finite Galois extension containing both $K, L$. Let $Z=\mathrm{Spec}(O_F)$. Then $Z\times_S Z\to X\times_S Y$ is surjective and we are done as in (1).

  3. Let $X=\mathrm{Spec}(O_F)$ with $G=\mathrm{Gal}(F)$. Then $X\times_S X$ splits into union of copies of $X$ parametrized by $G$. This union can be written explicitely. In the very special case where $O_F=\mathbb Z[t]$, we have $X\times_S X=\cup_{\sigma\in G} \mathrm{Spec}(O_F[s]/(s-\sigma(t)))$.

    EDIT Two irreducible components $\mathrm{Spec}(O_F[s]/(s-\sigma(t)))$ and $\mathrm{Spec}(O_F[s]/(s-\tau(t)))$ meet at $\mathrm{Spec}(O_F/(\sigma(t)-\tau(t)))$. This intersection is non empty iff $\sigma(t)-\tau(t)$ is not an unit in $O_F$.

  4. Suppose $\mathbb {Z}\to O_F$ is totally ramified at some $p$ (e.g $[F:\mathbb Q]$ is prime), then $(X\times_S X)_p$ consist in one point $q$, hence all its irreducible components meet each other at $q$, and $X\times_S X$ is connected.

  5. I tried some explicite examples, all of them are connected. I think it is not too hard to show the connectedness if the Galois group $G$ is $2$-transitive on itself. In general, I don't know whether connectedness holds !

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thanks, your remarks are surely helpful, but i am not sure yet, if i can accpet this answer. I will wait for some more time to see if i can solve it myself or i.f someone else can solve the problem –  messi Jul 1 '12 at 14:53
    
@messi: if you find a solution, please let us know. No problem to not accepting my remarks as a solution. –  user18119 Jul 2 '12 at 19:29

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