Are $mathbb{C}[x]/x^2 otimes mathbb{C}[x]/x^2$ and $mathbb{C}[x]/x^2$ isomorphic as...
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Are $frac{mathbb{C}[x]}{x^2} otimes frac{mathbb{C}[x]}{x^2}$ and $frac{mathbb{C}[x]}{x^2}$ isomorphic as $frac{mathbb{C}[x]}{x^2}$-modules? I believe that they are but it I know that there is a mistake in my working and this is the claim that I'm the most uncertain of.
modules tensor-products
asked Nov 20 at 16:30
Daven
30119
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up vote
0
down vote
favorite
Are $frac{mathbb{C}[x]}{x^2} otimes frac{mathbb{C}[x]}{x^2}$ and $frac{mathbb{C}[x]}{x^2}$ isomorphic as $frac{mathbb{C}[x]}{x^2}$-modules? I believe that they are but it I know that there is a mistake in my working and this is the claim that I'm the most uncertain of.
modules tensor-products
asked Nov 20 at 16:30
Daven
30119
1
Is the tensor product over $Bbb{C}$ or over $Bbb{C}[x]/(x^2)$ or something else?
– Servaes
Nov 20 at 16:37
The second; the tensor product is over $mathbb{C}[x]/x^2$
– Daven
Nov 20 at 16:41
1
$Rotimes_R M cong M$ for any $R$-module $M$. In particular, $Rotimes_R R cong R$.
– Slade
Nov 20 at 16:42
Okay thanks! I guess I'll have to find some other issue with my work...
– Daven
Nov 20 at 16:43
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Are $frac{mathbb{C}[x]}{x^2} otimes frac{mathbb{C}[x]}{x^2}$ and $frac{mathbb{C}[x]}{x^2}$ isomorphic as $frac{mathbb{C}[x]}{x^2}$-modules? I believe that they are but it I know that there is a mistake in my working and this is the claim that I'm the most uncertain of.
modules tensor-products
asked Nov 20 at 16:30
Daven
30119
Are $frac{mathbb{C}[x]}{x^2} otimes frac{mathbb{C}[x]}{x^2}$ and $frac{mathbb{C}[x]}{x^2}$ isomorphic as $frac{mathbb{C}[x]}{x^2}$-modules? I believe that they are but it I know that there is a mistake in my working and this is the claim that I'm the most uncertain of.
modules tensor-products
modules tensor-products
asked Nov 20 at 16:30
Daven
30119
asked Nov 20 at 16:30
Daven
30119
asked Nov 20 at 16:30
Daven
30119
asked Nov 20 at 16:30
Daven
30119
asked Nov 20 at 16:30
Daven
30119
30119
1
Is the tensor product over $Bbb{C}$ or over $Bbb{C}[x]/(x^2)$ or something else?
– Servaes
Nov 20 at 16:37
The second; the tensor product is over $mathbb{C}[x]/x^2$
– Daven
Nov 20 at 16:41
1
$Rotimes_R M cong M$ for any $R$-module $M$. In particular, $Rotimes_R R cong R$.
– Slade
Nov 20 at 16:42
Okay thanks! I guess I'll have to find some other issue with my work...
– Daven
Nov 20 at 16:43
add a comment |
1
Is the tensor product over $Bbb{C}$ or over $Bbb{C}[x]/(x^2)$ or something else?
– Servaes
Nov 20 at 16:37
The second; the tensor product is over $mathbb{C}[x]/x^2$
– Daven
Nov 20 at 16:41
1
$Rotimes_R M cong M$ for any $R$-module $M$. In particular, $Rotimes_R R cong R$.
– Slade
Nov 20 at 16:42
Okay thanks! I guess I'll have to find some other issue with my work...
– Daven
Nov 20 at 16:43
1
1
Is the tensor product over $Bbb{C}$ or over $Bbb{C}[x]/(x^2)$ or something else?
– Servaes
Nov 20 at 16:37
Is the tensor product over $Bbb{C}$ or over $Bbb{C}[x]/(x^2)$ or something else?
– Servaes
Nov 20 at 16:37
The second; the tensor product is over $mathbb{C}[x]/x^2$
– Daven
Nov 20 at 16:41
The second; the tensor product is over $mathbb{C}[x]/x^2$
– Daven
Nov 20 at 16:41
1
1
$Rotimes_R M cong M$ for any $R$-module $M$. In particular, $Rotimes_R R cong R$.
– Slade
Nov 20 at 16:42
$Rotimes_R M cong M$ for any $R$-module $M$. In particular, $Rotimes_R R cong R$.
– Slade
Nov 20 at 16:42
Okay thanks! I guess I'll have to find some other issue with my work...
– Daven
Nov 20 at 16:43
Okay thanks! I guess I'll have to find some other issue with my work...
– Daven
Nov 20 at 16:43
add a comment |
1 Answer
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1
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accepted
For any ring $R$ and any $R$-module $M$ we have $Motimes_RRcong M$ canonically. Taking $R=M=Bbb{C}[x]/(x^2)$ yields the desired result.
answered Nov 20 at 16:42
Servaes
22.2k33793
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
1
down vote
accepted
For any ring $R$ and any $R$-module $M$ we have $Motimes_RRcong M$ canonically. Taking $R=M=Bbb{C}[x]/(x^2)$ yields the desired result.
answered Nov 20 at 16:42
Servaes
22.2k33793
add a comment |
up vote
1
down vote
accepted
For any ring $R$ and any $R$-module $M$ we have $Motimes_RRcong M$ canonically. Taking $R=M=Bbb{C}[x]/(x^2)$ yields the desired result.
answered Nov 20 at 16:42
Servaes
22.2k33793
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For any ring $R$ and any $R$-module $M$ we have $Motimes_RRcong M$ canonically. Taking $R=M=Bbb{C}[x]/(x^2)$ yields the desired result.
answered Nov 20 at 16:42
Servaes
22.2k33793
For any ring $R$ and any $R$-module $M$ we have $Motimes_RRcong M$ canonically. Taking $R=M=Bbb{C}[x]/(x^2)$ yields the desired result.
answered Nov 20 at 16:42
Servaes
22.2k33793
answered Nov 20 at 16:42
Servaes
22.2k33793
answered Nov 20 at 16:42
Servaes
22.2k33793
answered Nov 20 at 16:42
Servaes
22.2k33793
22.2k33793
add a comment |
add a comment |
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1
Is the tensor product over $Bbb{C}$ or over $Bbb{C}[x]/(x^2)$ or something else?
– Servaes
Nov 20 at 16:37
The second; the tensor product is over $mathbb{C}[x]/x^2$
– Daven
Nov 20 at 16:41
1
$Rotimes_R M cong M$ for any $R$-module $M$. In particular, $Rotimes_R R cong R$.
– Slade
Nov 20 at 16:42
Okay thanks! I guess I'll have to find some other issue with my work...
– Daven
Nov 20 at 16:43