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Consider $A$ and $B$ two $\mathbb C$-algebras such that $A\otimes_{\mathbb C}B$ is finitely generated as a $\mathbb C$-algebra. Does this imply that $A$ and $B$ are finitely generated?

I know that for general algebras, this is false. Indeed $\mathbb Q$ is infinitely generated over $\mathbb Z$ but the tensor product $ \mathbb Q\otimes_\mathbb Z \mathbb Z_2 =0$. For $\mathbb C$-algebras however, I just can't seem to find a counter-example.

Eric Wofsey
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user306194
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1 Answers1

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Yes, it does, as long as $A$ and $B$ are both not the zero ring (obviously $A\otimes 0=0$ is finitely generated for any $A$). Choose a finite set of generators of $A\otimes_\mathbb{C} B$; each of these is a finite sum of tensors $a\otimes b$. Let $A_0\subseteq A$ be the subalgebra generated by all the $a$'s appearing in these tensors. Then $A_0$ is finitely generated, and we see that the natural map $A_0\otimes_\mathbb{C} B\to A\otimes_\mathbb{C} B$ is surjective (since its image contains all of the tensors $a\otimes b$ in our generators). This means $A/A_0\otimes_\mathbb{C} B=0$, so as long as $B\neq 0$, we must have $A/A_0=0$ and so $A_0=A$. Thus $A$ is finitely generated. By the same argument, $B$ is also finitely generated.

This argument clearly works with $\mathbb{C}$ replaced by any field. Much more generally, a similar argument shows that if $R$ is any base ring and $A$ and $B$ are $R$-algebras such that $B$ is faithfully flat over $R$, then if $A\otimes_R B$ is finitely generated as a $B$-algebra (in particular, if it is finitely generated as an $R$-algebra), then $A$ is finitely generated as an $R$-algebra.

Maxime Ramzi
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Eric Wofsey
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    The fact that $B\neq 0$ implies $A_0 = A$ seems to be the crux of the argument, since that's where the proof would fail for $\mathbb{Q} \otimes _{\mathbb{Z}} \mathbb{Z}_2$. Although this seems obvious to me, I feel like I'm on slightly shaky ground. – Callus - Reinstate Monica Jan 19 '16 at 03:07
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    We can choose a vector subspace $A_1\subseteq A$ such that $A=A_0\oplus A_1$ as vector spaces, and so $A_0\otimes B\to A\otimes B=A_0\otimes B\oplus A_1\otimes B$ is surjective iff $A_1\otimes B=0$. If $B\neq 0$, this is true iff $A_1=0$, which means $A_0=A$. – Eric Wofsey Jan 19 '16 at 03:11
  • yeah, that definitely takes care of the vector space case, which is the OP's question. – Callus - Reinstate Monica Jan 19 '16 at 03:25
  • Why do we have if $B\neq 0$ , $A_1\otimes B=0$ iff $A_1=0$? I sense that this is where the field argument comes in but it is still not quite clear. – user306194 Jan 19 '16 at 13:13
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    @user306194: If $S$ is a basis for $A_1$ and $T$ is a basis for $B$, then $S\times T$ is a basis for $A_1\otimes B$. A cartesian product of sets can be empty only if one of the factors is empty. – Eric Wofsey Jan 19 '16 at 20:19
  • Nice and elementary. There is this more general result http://math.stackexchange.com/questions/93228/what-does-a-zero-tensor-product-imply?rq=1 – Callus - Reinstate Monica Jan 19 '16 at 23:23
  • hmm why can we write $A=A_0\oplus A_1$ though? Subalgebras are not necessarily subvectorspaces. – user306194 Jan 20 '16 at 23:09
  • Why not? they are closed under multiplication and addition, right? – Callus - Reinstate Monica Jan 21 '16 at 00:32
  • The very definition of subalgebra is "vector subspace which is also a subring". – Eric Wofsey Jan 21 '16 at 02:08