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?
Connectedness of the spectrum of a tensor product.
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algebraic-geometry
commutative-algebra
algebraic-number-theory
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0i 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. – 2012-06-20
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0How 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. – 2012-06-20
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6@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. – 2012-06-21
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0@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. – 2012-06-21
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0@ZheChen Thanks for the clarification. That makes sense. – 2012-06-21
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2Is this known if we replace Z by a field? – 2012-07-01