Mixing
Computation of tensor product
$begingroup$
Let $k$ be a field.
Consider the map $k[X,X^{-1},Y, Y^{-1}] to k[T, T^{-1}] $ where $X to T$ and
$Y to T$.
I'm interested in computing the tensor product
$ k[T,T^{-1}] otimes_{k[X,X^{-1},Y, Y^{-1}]} k[X^{1/n},Y^{1/n},X^{-1/n},Y^{-1/n}] $.
I'm not sure how to proceed with this computation. Any help would be appreciated
abstract-algebra tensor-products
asked Jan 4 at 12:15
grontimgrontim
310110
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field.
Consider the map $k[X,X^{-1},Y, Y^{-1}] to k[T, T^{-1}] $ where $X to T$ and
$Y to T$.
I'm interested in computing the tensor product
$ k[T,T^{-1}] otimes_{k[X,X^{-1},Y, Y^{-1}]} k[X^{1/n},Y^{1/n},X^{-1/n},Y^{-1/n}] $.
I'm not sure how to proceed with this computation. Any help would be appreciated
abstract-algebra tensor-products
asked Jan 4 at 12:15
grontimgrontim
310110
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field.
Consider the map $k[X,X^{-1},Y, Y^{-1}] to k[T, T^{-1}] $ where $X to T$ and
$Y to T$.
I'm interested in computing the tensor product
$ k[T,T^{-1}] otimes_{k[X,X^{-1},Y, Y^{-1}]} k[X^{1/n},Y^{1/n},X^{-1/n},Y^{-1/n}] $.
I'm not sure how to proceed with this computation. Any help would be appreciated
abstract-algebra tensor-products
asked Jan 4 at 12:15
grontimgrontim
310110
$endgroup$
Let $k$ be a field.
Consider the map $k[X,X^{-1},Y, Y^{-1}] to k[T, T^{-1}] $ where $X to T$ and
$Y to T$.
I'm interested in computing the tensor product
$ k[T,T^{-1}] otimes_{k[X,X^{-1},Y, Y^{-1}]} k[X^{1/n},Y^{1/n},X^{-1/n},Y^{-1/n}] $.
I'm not sure how to proceed with this computation. Any help would be appreciated
abstract-algebra tensor-products
abstract-algebra tensor-products
asked Jan 4 at 12:15
grontimgrontim
310110
asked Jan 4 at 12:15
grontimgrontim
310110
asked Jan 4 at 12:15
grontimgrontim
310110
asked Jan 4 at 12:15
grontimgrontim
310110
asked Jan 4 at 12:15
grontimgrontim
310110
310110
add a comment |
add a comment |
1 Answer
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$begingroup$
Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.
Capital variables are very easy to mistype, so I'll use lower case.
Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.
Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
$$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
$$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
$$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
$$cong k[a,b,u]/(a^n-b^n,au-1)$$
answered Jan 4 at 19:07
jgonjgon
13.5k22041
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.
Capital variables are very easy to mistype, so I'll use lower case.
Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.
Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
$$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
$$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
$$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
$$cong k[a,b,u]/(a^n-b^n,au-1)$$
answered Jan 4 at 19:07
jgonjgon
13.5k22041
$endgroup$
add a comment |
$begingroup$
Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.
Capital variables are very easy to mistype, so I'll use lower case.
Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.
Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
$$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
$$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
$$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
$$cong k[a,b,u]/(a^n-b^n,au-1)$$
answered Jan 4 at 19:07
jgonjgon
13.5k22041
$endgroup$
add a comment |
$begingroup$
Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.
Capital variables are very easy to mistype, so I'll use lower case.
Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.
Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
$$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
$$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
$$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
$$cong k[a,b,u]/(a^n-b^n,au-1)$$
answered Jan 4 at 19:07
jgonjgon
13.5k22041
$endgroup$
Use the lemma $R/I otimes_R S cong S/IS$, and then simplify.
Capital variables are very easy to mistype, so I'll use lower case.
Let $R=k[x,y,x^{-1},y^{-1}]cong k[x,y,z,w]/(xz-1,yw-1)$, then $k[t,t^{-1}]cong R/(x-y)$, since then $z-w = (xzw-w)-(yzw-z)$ is contained in the ideal automatically. Also $S:=k[x^{1/n},y^{1/n},x^{-1/n},y^{-1/n}]cong R[a,b]/(a^n-x,b^n-y)$.
Thus $$k[t,t^{-1}]otimes_R Scong R/(x-y)otimes_R R[a,b]/(a^n-x,b^n-y)$$
$$cong R[a,b]/(a^n-x,b^n-y,x-y)$$
$$cong k[x,y,z,w,a,b]/(xz-1,yw-1,a^n-x,b^n-y,x-y)$$
$$cong k[x,z,a,b]/(xz-1,a^n-x,b^n-x)$$
$$cong k[a,b,u]/(a^n-b^n,au-1)$$
answered Jan 4 at 19:07
jgonjgon
13.5k22041
answered Jan 4 at 19:07
jgonjgon
13.5k22041
answered Jan 4 at 19:07
jgonjgon
13.5k22041
answered Jan 4 at 19:07
jgonjgon
13.5k22041
13.5k22041
add a comment |
add a comment |
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