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$Aotimes_mathbb CB$ is finitely generated as a $mathbb C$-algebra. Does this imply that $A$ and $B$ are finitely generated?

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14

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$begingroup$

Consider $A$ and $B$ two $mathbb C$-algebras such that $Aotimes_mathbb CB$ 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 Qotimes_mathbb Z mathbb Z_2 =0$. For $mathbb C$-algebras however, I just can't seem to find a counter-example.

commutative-algebra tensor-products

share|cite|improve this question

edited Jan 19 '16 at 0:43

Eric Wofsey

189k14216347

asked Jan 18 '16 at 22:55

user306194user306194

1428

$endgroup$

add a comment | 

14

5

$begingroup$

Consider $A$ and $B$ two $mathbb C$-algebras such that $Aotimes_mathbb CB$ 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 Qotimes_mathbb Z mathbb Z_2 =0$. For $mathbb C$-algebras however, I just can't seem to find a counter-example.

commutative-algebra tensor-products

share|cite|improve this question

edited Jan 19 '16 at 0:43

Eric Wofsey

189k14216347

asked Jan 18 '16 at 22:55

user306194user306194

1428

$endgroup$

add a comment | 

$begingroup$

Consider $A$ and $B$ two $mathbb C$-algebras such that $Aotimes_mathbb CB$ 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 Qotimes_mathbb Z mathbb Z_2 =0$. For $mathbb C$-algebras however, I just can't seem to find a counter-example.

commutative-algebra tensor-products

share|cite|improve this question

edited Jan 19 '16 at 0:43

Eric Wofsey

189k14216347

asked Jan 18 '16 at 22:55

user306194user306194

1428

$endgroup$

share|cite|improve this question

edited Jan 19 '16 at 0:43

Eric Wofsey

189k14216347

asked Jan 18 '16 at 22:55

user306194user306194

1428

share|cite|improve this question

edited Jan 19 '16 at 0:43

Eric Wofsey

189k14216347

asked Jan 18 '16 at 22:55

user306194user306194

1428

edited Jan 19 '16 at 0:43

Eric Wofsey

189k14216347

edited Jan 19 '16 at 0:43

Eric Wofsey

189k14216347

asked Jan 18 '16 at 22:55

user306194user306194

1428

asked Jan 18 '16 at 22:55

user306194user306194

1428

1 Answer
1

active

oldest

votes

9

$begingroup$

Yes, it does, as long as $A$ and $B$ are both not the zero ring (obviously $Aotimes 0=0$ is finitely generated for any $A$). Choose a finite set of generators of $Aotimes_mathbbC B$; each of these is a finite sum of tensors $aotimes b$. Let $A_0subseteq A$ 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 $A_0otimes_mathbbC Bto Aotimes_mathbbC B$ is surjective (since its image contains all of the tensors $aotimes b$ in our generators). This means $A/A_0otimes_mathbbC B=0$, so as long as $Bneq 0$, 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 $mathbbC$ 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 $Aotimes_R B$ 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.

share|cite|improve this answer

edited Mar 12 at 9:04

Max

15.3k11143

answered Jan 18 '16 at 23:43

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The fact that $Bneq 0$ implies $A_0 = A$ seems to be the crux of the argument, since that's where the proof would fail for $mathbbQ otimes _mathbbZ mathbbZ_2$. Although this seems obvious to me, I feel like I'm on slightly shaky ground.
  $endgroup$
  – Callus
  Jan 19 '16 at 3:07
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We can choose a vector subspace $A_1subseteq A$ such that $A=A_0oplus A_1$ as vector spaces, and so $A_0otimes Bto Aotimes B=A_0otimes Boplus A_1otimes B$ is surjective iff $A_1otimes B=0$. If $Bneq 0$, this is true iff $A_1=0$, which means $A_0=A$.
  $endgroup$
  – Eric Wofsey
  Jan 19 '16 at 3:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, that definitely takes care of the vector space case, which is the OP's question.
  $endgroup$
  – Callus
  Jan 19 '16 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do we have if $Bneq 0$ , $A_1otimes B=0$ iff $A_1=0$? I sense that this is where the field argument comes in but it is still not quite clear.
  $endgroup$
  – user306194
  Jan 19 '16 at 13:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user306194: If $S$ is a basis for $A_1$ and $T$ is a basis for $B$, then $Stimes T$ is a basis for $A_1otimes B$. A cartesian product of sets can be empty only if one of the factors is empty.
  $endgroup$
  – Eric Wofsey
  Jan 19 '16 at 20:19
  

  
  
  
  

 | 
show 4 more comments

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9

$begingroup$

Yes, it does, as long as $A$ and $B$ are both not the zero ring (obviously $Aotimes 0=0$ is finitely generated for any $A$). Choose a finite set of generators of $Aotimes_mathbbC B$; each of these is a finite sum of tensors $aotimes b$. Let $A_0subseteq A$ 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 $A_0otimes_mathbbC Bto Aotimes_mathbbC B$ is surjective (since its image contains all of the tensors $aotimes b$ in our generators). This means $A/A_0otimes_mathbbC B=0$, so as long as $Bneq 0$, 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 $mathbbC$ 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 $Aotimes_R B$ 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.

share|cite|improve this answer

edited Mar 12 at 9:04

Max

15.3k11143

answered Jan 18 '16 at 23:43

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The fact that $Bneq 0$ implies $A_0 = A$ seems to be the crux of the argument, since that's where the proof would fail for $mathbbQ otimes _mathbbZ mathbbZ_2$. Although this seems obvious to me, I feel like I'm on slightly shaky ground.
  $endgroup$
  – Callus
  Jan 19 '16 at 3:07
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We can choose a vector subspace $A_1subseteq A$ such that $A=A_0oplus A_1$ as vector spaces, and so $A_0otimes Bto Aotimes B=A_0otimes Boplus A_1otimes B$ is surjective iff $A_1otimes B=0$. If $Bneq 0$, this is true iff $A_1=0$, which means $A_0=A$.
  $endgroup$
  – Eric Wofsey
  Jan 19 '16 at 3:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, that definitely takes care of the vector space case, which is the OP's question.
  $endgroup$
  – Callus
  Jan 19 '16 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do we have if $Bneq 0$ , $A_1otimes B=0$ iff $A_1=0$? I sense that this is where the field argument comes in but it is still not quite clear.
  $endgroup$
  – user306194
  Jan 19 '16 at 13:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user306194: If $S$ is a basis for $A_1$ and $T$ is a basis for $B$, then $Stimes T$ is a basis for $A_1otimes B$. A cartesian product of sets can be empty only if one of the factors is empty.
  $endgroup$
  – Eric Wofsey
  Jan 19 '16 at 20:19
  

  
  
  
  

 | 
show 4 more comments

9

$begingroup$

Yes, it does, as long as $A$ and $B$ are both not the zero ring (obviously $Aotimes 0=0$ is finitely generated for any $A$). Choose a finite set of generators of $Aotimes_mathbbC B$; each of these is a finite sum of tensors $aotimes b$. Let $A_0subseteq A$ 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 $A_0otimes_mathbbC Bto Aotimes_mathbbC B$ is surjective (since its image contains all of the tensors $aotimes b$ in our generators). This means $A/A_0otimes_mathbbC B=0$, so as long as $Bneq 0$, 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 $mathbbC$ 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 $Aotimes_R B$ 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.

share|cite|improve this answer

edited Mar 12 at 9:04

Max

15.3k11143

answered Jan 18 '16 at 23:43

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The fact that $Bneq 0$ implies $A_0 = A$ seems to be the crux of the argument, since that's where the proof would fail for $mathbbQ otimes _mathbbZ mathbbZ_2$. Although this seems obvious to me, I feel like I'm on slightly shaky ground.
  $endgroup$
  – Callus
  Jan 19 '16 at 3:07
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We can choose a vector subspace $A_1subseteq A$ such that $A=A_0oplus A_1$ as vector spaces, and so $A_0otimes Bto Aotimes B=A_0otimes Boplus A_1otimes B$ is surjective iff $A_1otimes B=0$. If $Bneq 0$, this is true iff $A_1=0$, which means $A_0=A$.
  $endgroup$
  – Eric Wofsey
  Jan 19 '16 at 3:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, that definitely takes care of the vector space case, which is the OP's question.
  $endgroup$
  – Callus
  Jan 19 '16 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do we have if $Bneq 0$ , $A_1otimes B=0$ iff $A_1=0$? I sense that this is where the field argument comes in but it is still not quite clear.
  $endgroup$
  – user306194
  Jan 19 '16 at 13:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user306194: If $S$ is a basis for $A_1$ and $T$ is a basis for $B$, then $Stimes T$ is a basis for $A_1otimes B$. A cartesian product of sets can be empty only if one of the factors is empty.
  $endgroup$
  – Eric Wofsey
  Jan 19 '16 at 20:19
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Yes, it does, as long as $A$ and $B$ are both not the zero ring (obviously $Aotimes 0=0$ is finitely generated for any $A$). Choose a finite set of generators of $Aotimes_mathbbC B$; each of these is a finite sum of tensors $aotimes b$. Let $A_0subseteq A$ 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 $A_0otimes_mathbbC Bto Aotimes_mathbbC B$ is surjective (since its image contains all of the tensors $aotimes b$ in our generators). This means $A/A_0otimes_mathbbC B=0$, so as long as $Bneq 0$, 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 $mathbbC$ 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 $Aotimes_R B$ 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.

share|cite|improve this answer

edited Mar 12 at 9:04

Max

15.3k11143

answered Jan 18 '16 at 23:43

Eric WofseyEric Wofsey

189k14216347

$endgroup$

share|cite|improve this answer

edited Mar 12 at 9:04

Max

15.3k11143

answered Jan 18 '16 at 23:43

Eric WofseyEric Wofsey

189k14216347

edited Mar 12 at 9:04

Max

15.3k11143

edited Mar 12 at 9:04

Max

15.3k11143

answered Jan 18 '16 at 23:43

Eric WofseyEric Wofsey

189k14216347

answered Jan 18 '16 at 23:43

Eric WofseyEric Wofsey

189k14216347