Proving a tensor product has dimension 2 over $mathbb{Q}$












1












$begingroup$


Can u guys help me prove that $$mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)$$has dimension 2 as a vector space over $mathbb{Q}$. I'm not having good ideas about it, thanks.






abstract-algebra modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you compute the dimension of the two spaces on each side of the tensor product? Dimension is multiplicative for tensor products.
    $endgroup$
    – quarague
    Jan 24 at 10:55










  • $begingroup$
    No i did not, i only knew that worked for free modules. But yeah that makes sense.
    $endgroup$
    – Pedro Santos
    Jan 24 at 10:57






  • 3




    $begingroup$
    @quarague This were true if you would tensor over a field, here the tensor product is over a commutative ring which is not a field. Clearly in this question the right tensor factor is infinite dimensional, so there is nothing to gain just computing the dimension.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:06






  • 1




    $begingroup$
    $mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)=mathbb{Q}[x,y]/(x^3+y,x^2+y,x^2-y)$
    $endgroup$
    – Mustafa
    Jan 24 at 16:35
















1












$begingroup$


Can u guys help me prove that $$mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)$$has dimension 2 as a vector space over $mathbb{Q}$. I'm not having good ideas about it, thanks.






abstract-algebra modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you compute the dimension of the two spaces on each side of the tensor product? Dimension is multiplicative for tensor products.
    $endgroup$
    – quarague
    Jan 24 at 10:55










  • $begingroup$
    No i did not, i only knew that worked for free modules. But yeah that makes sense.
    $endgroup$
    – Pedro Santos
    Jan 24 at 10:57






  • 3




    $begingroup$
    @quarague This were true if you would tensor over a field, here the tensor product is over a commutative ring which is not a field. Clearly in this question the right tensor factor is infinite dimensional, so there is nothing to gain just computing the dimension.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:06






  • 1




    $begingroup$
    $mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)=mathbb{Q}[x,y]/(x^3+y,x^2+y,x^2-y)$
    $endgroup$
    – Mustafa
    Jan 24 at 16:35














1












1








1


1



$begingroup$


Can u guys help me prove that $$mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)$$has dimension 2 as a vector space over $mathbb{Q}$. I'm not having good ideas about it, thanks.






abstract-algebra modules







share|cite|improve this question











$endgroup$




Can u guys help me prove that $$mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)$$has dimension 2 as a vector space over $mathbb{Q}$. I'm not having good ideas about it, thanks.






abstract-algebra modules




abstract-algebra modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 10:55









J.G.

28.9k22845




28.9k22845










asked Jan 24 at 10:52









Pedro SantosPedro Santos

1519




1519












  • $begingroup$
    Did you compute the dimension of the two spaces on each side of the tensor product? Dimension is multiplicative for tensor products.
    $endgroup$
    – quarague
    Jan 24 at 10:55










  • $begingroup$
    No i did not, i only knew that worked for free modules. But yeah that makes sense.
    $endgroup$
    – Pedro Santos
    Jan 24 at 10:57






  • 3




    $begingroup$
    @quarague This were true if you would tensor over a field, here the tensor product is over a commutative ring which is not a field. Clearly in this question the right tensor factor is infinite dimensional, so there is nothing to gain just computing the dimension.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:06






  • 1




    $begingroup$
    $mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)=mathbb{Q}[x,y]/(x^3+y,x^2+y,x^2-y)$
    $endgroup$
    – Mustafa
    Jan 24 at 16:35


















  • $begingroup$
    Did you compute the dimension of the two spaces on each side of the tensor product? Dimension is multiplicative for tensor products.
    $endgroup$
    – quarague
    Jan 24 at 10:55










  • $begingroup$
    No i did not, i only knew that worked for free modules. But yeah that makes sense.
    $endgroup$
    – Pedro Santos
    Jan 24 at 10:57






  • 3




    $begingroup$
    @quarague This were true if you would tensor over a field, here the tensor product is over a commutative ring which is not a field. Clearly in this question the right tensor factor is infinite dimensional, so there is nothing to gain just computing the dimension.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:06






  • 1




    $begingroup$
    $mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)=mathbb{Q}[x,y]/(x^3+y,x^2+y,x^2-y)$
    $endgroup$
    – Mustafa
    Jan 24 at 16:35
















$begingroup$
Did you compute the dimension of the two spaces on each side of the tensor product? Dimension is multiplicative for tensor products.
$endgroup$
– quarague
Jan 24 at 10:55




$begingroup$
Did you compute the dimension of the two spaces on each side of the tensor product? Dimension is multiplicative for tensor products.
$endgroup$
– quarague
Jan 24 at 10:55












$begingroup$
No i did not, i only knew that worked for free modules. But yeah that makes sense.
$endgroup$
– Pedro Santos
Jan 24 at 10:57




$begingroup$
No i did not, i only knew that worked for free modules. But yeah that makes sense.
$endgroup$
– Pedro Santos
Jan 24 at 10:57




3




3




$begingroup$
@quarague This were true if you would tensor over a field, here the tensor product is over a commutative ring which is not a field. Clearly in this question the right tensor factor is infinite dimensional, so there is nothing to gain just computing the dimension.
$endgroup$
– Julian Kuelshammer
Jan 24 at 11:06




$begingroup$
@quarague This were true if you would tensor over a field, here the tensor product is over a commutative ring which is not a field. Clearly in this question the right tensor factor is infinite dimensional, so there is nothing to gain just computing the dimension.
$endgroup$
– Julian Kuelshammer
Jan 24 at 11:06




1




1




$begingroup$
$mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)=mathbb{Q}[x,y]/(x^3+y,x^2+y,x^2-y)$
$endgroup$
– Mustafa
Jan 24 at 16:35




$begingroup$
$mathbb{Q}[x,y]/(x^3+y,x^2+y) bigotimes_{mathbb{Q}[x,y]} mathbb{Q}[x,y]/(x^2-y)=mathbb{Q}[x,y]/(x^3+y,x^2+y,x^2-y)$
$endgroup$
– Mustafa
Jan 24 at 16:35










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint: Use that for a commutative ring $R$ you have $R/Iotimes_R R/Jcong R/(I+J)$, see e.g. this math.stackexchange question.



Then try to find a basis of $R/(I+J)$ by using the conditions in $I+J$. Further hint: In $R/(I+J)$ you have $x^2=y$ and $x^2=-y$. What does that say about $y$?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah thats my biggest problem , is identifying the elements in $R/(I+J)$
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:49












  • $begingroup$
    @PedroSantos I provided a further hint to get you started.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:55










  • $begingroup$
    The class of $y$ and $x^2$ are going to be zero , so thats why i get the 2 dimension vector space ? cause an element will be of the form $ax+b$?
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:58










  • $begingroup$
    @PedroSantos Yes, exactly (for a complete answer you also have to prove that the residue classes of $1$ and $x$ are linearly independent in the quotient).
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 12:14











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Hint: Use that for a commutative ring $R$ you have $R/Iotimes_R R/Jcong R/(I+J)$, see e.g. this math.stackexchange question.



Then try to find a basis of $R/(I+J)$ by using the conditions in $I+J$. Further hint: In $R/(I+J)$ you have $x^2=y$ and $x^2=-y$. What does that say about $y$?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah thats my biggest problem , is identifying the elements in $R/(I+J)$
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:49












  • $begingroup$
    @PedroSantos I provided a further hint to get you started.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:55










  • $begingroup$
    The class of $y$ and $x^2$ are going to be zero , so thats why i get the 2 dimension vector space ? cause an element will be of the form $ax+b$?
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:58










  • $begingroup$
    @PedroSantos Yes, exactly (for a complete answer you also have to prove that the residue classes of $1$ and $x$ are linearly independent in the quotient).
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 12:14
















2












$begingroup$

Hint: Use that for a commutative ring $R$ you have $R/Iotimes_R R/Jcong R/(I+J)$, see e.g. this math.stackexchange question.



Then try to find a basis of $R/(I+J)$ by using the conditions in $I+J$. Further hint: In $R/(I+J)$ you have $x^2=y$ and $x^2=-y$. What does that say about $y$?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah thats my biggest problem , is identifying the elements in $R/(I+J)$
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:49












  • $begingroup$
    @PedroSantos I provided a further hint to get you started.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:55










  • $begingroup$
    The class of $y$ and $x^2$ are going to be zero , so thats why i get the 2 dimension vector space ? cause an element will be of the form $ax+b$?
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:58










  • $begingroup$
    @PedroSantos Yes, exactly (for a complete answer you also have to prove that the residue classes of $1$ and $x$ are linearly independent in the quotient).
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 12:14














2












2








2





$begingroup$

Hint: Use that for a commutative ring $R$ you have $R/Iotimes_R R/Jcong R/(I+J)$, see e.g. this math.stackexchange question.



Then try to find a basis of $R/(I+J)$ by using the conditions in $I+J$. Further hint: In $R/(I+J)$ you have $x^2=y$ and $x^2=-y$. What does that say about $y$?






share|cite|improve this answer











$endgroup$



Hint: Use that for a commutative ring $R$ you have $R/Iotimes_R R/Jcong R/(I+J)$, see e.g. this math.stackexchange question.



Then try to find a basis of $R/(I+J)$ by using the conditions in $I+J$. Further hint: In $R/(I+J)$ you have $x^2=y$ and $x^2=-y$. What does that say about $y$?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 24 at 11:54

























answered Jan 24 at 11:45









Julian KuelshammerJulian Kuelshammer

7,56232666




7,56232666












  • $begingroup$
    Yeah thats my biggest problem , is identifying the elements in $R/(I+J)$
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:49












  • $begingroup$
    @PedroSantos I provided a further hint to get you started.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:55










  • $begingroup$
    The class of $y$ and $x^2$ are going to be zero , so thats why i get the 2 dimension vector space ? cause an element will be of the form $ax+b$?
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:58










  • $begingroup$
    @PedroSantos Yes, exactly (for a complete answer you also have to prove that the residue classes of $1$ and $x$ are linearly independent in the quotient).
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 12:14


















  • $begingroup$
    Yeah thats my biggest problem , is identifying the elements in $R/(I+J)$
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:49












  • $begingroup$
    @PedroSantos I provided a further hint to get you started.
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 11:55










  • $begingroup$
    The class of $y$ and $x^2$ are going to be zero , so thats why i get the 2 dimension vector space ? cause an element will be of the form $ax+b$?
    $endgroup$
    – Pedro Santos
    Jan 24 at 11:58










  • $begingroup$
    @PedroSantos Yes, exactly (for a complete answer you also have to prove that the residue classes of $1$ and $x$ are linearly independent in the quotient).
    $endgroup$
    – Julian Kuelshammer
    Jan 24 at 12:14
















$begingroup$
Yeah thats my biggest problem , is identifying the elements in $R/(I+J)$
$endgroup$
– Pedro Santos
Jan 24 at 11:49






$begingroup$
Yeah thats my biggest problem , is identifying the elements in $R/(I+J)$
$endgroup$
– Pedro Santos
Jan 24 at 11:49














$begingroup$
@PedroSantos I provided a further hint to get you started.
$endgroup$
– Julian Kuelshammer
Jan 24 at 11:55




$begingroup$
@PedroSantos I provided a further hint to get you started.
$endgroup$
– Julian Kuelshammer
Jan 24 at 11:55












$begingroup$
The class of $y$ and $x^2$ are going to be zero , so thats why i get the 2 dimension vector space ? cause an element will be of the form $ax+b$?
$endgroup$
– Pedro Santos
Jan 24 at 11:58




$begingroup$
The class of $y$ and $x^2$ are going to be zero , so thats why i get the 2 dimension vector space ? cause an element will be of the form $ax+b$?
$endgroup$
– Pedro Santos
Jan 24 at 11:58












$begingroup$
@PedroSantos Yes, exactly (for a complete answer you also have to prove that the residue classes of $1$ and $x$ are linearly independent in the quotient).
$endgroup$
– Julian Kuelshammer
Jan 24 at 12:14




$begingroup$
@PedroSantos Yes, exactly (for a complete answer you also have to prove that the residue classes of $1$ and $x$ are linearly independent in the quotient).
$endgroup$
– Julian Kuelshammer
Jan 24 at 12:14


















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