Calculating the dimension of the tensor product of modules
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
This is a recent qual problem that I am struggling with.
Put $M=mathbbC[x]/(x^2+x)$ and $N=mathbbC[x]/(x-1)$.
$(a)$ What is the dimension of $M otimes_mathbbC[x]N$ as a vector space over $mathbbC$?
$(b)$ What is the dimension of $M otimes_mathbbCN$ as a vector space over $mathbbC$?
$(c)$ What is the dimension of $M otimes_mathbbC[x^2]N$ as a vector space over $mathbbC$?
I solved the part $(a)$ and $(b)$. The part $(b)$ is the easiest. Since $M$ (resp. $N$) is a free $mathbbC$-module with basis $1,x $ (resp. $1$), the dimension is $2$. To get the part $(a)$, I used the direct computations:
$$
1 otimes 1 = (1+x^2+x) otimes 1 = 1 otimes (1+x^2+x) = 1 otimes 3 = 3(1 otimes 1).
$$
Thus, we get $1 otimes 1 = 0$, and the dimension is $0$. For another proof of the part $(a)$, we might be able to use that the tensor product is a right-exact functor because
$$
mathbbC[x]/(x^2+x,x-1) = 0.
$$
I don't know how to solve the part $(c)$. First, I showed
$$
M otimes_mathbbC[x^2] N = c(1 otimes_mathbbC[x^2] 1): c in mathbbC .
$$
via direct computations. However, this does not guarantee that the dimension is $1$. We need to check that $1 otimes_mathbbC[x^2] 1 neq 0$. Do you have any ideas? I know the property that the tensor product is a right-exact functor, but I have no idea how to apply it here.
Any comments are welcomed. Thank you in advance.
abstract-algebra modules tensor-products
asked Aug 19 at 5:52
C. Oh
908
add a comment |Â
up vote
3
down vote
favorite
This is a recent qual problem that I am struggling with.
Put $M=mathbbC[x]/(x^2+x)$ and $N=mathbbC[x]/(x-1)$.
$(a)$ What is the dimension of $M otimes_mathbbC[x]N$ as a vector space over $mathbbC$?
$(b)$ What is the dimension of $M otimes_mathbbCN$ as a vector space over $mathbbC$?
$(c)$ What is the dimension of $M otimes_mathbbC[x^2]N$ as a vector space over $mathbbC$?
I solved the part $(a)$ and $(b)$. The part $(b)$ is the easiest. Since $M$ (resp. $N$) is a free $mathbbC$-module with basis $1,x $ (resp. $1$), the dimension is $2$. To get the part $(a)$, I used the direct computations:
$$
1 otimes 1 = (1+x^2+x) otimes 1 = 1 otimes (1+x^2+x) = 1 otimes 3 = 3(1 otimes 1).
$$
Thus, we get $1 otimes 1 = 0$, and the dimension is $0$. For another proof of the part $(a)$, we might be able to use that the tensor product is a right-exact functor because
$$
mathbbC[x]/(x^2+x,x-1) = 0.
$$
I don't know how to solve the part $(c)$. First, I showed
$$
M otimes_mathbbC[x^2] N = c(1 otimes_mathbbC[x^2] 1): c in mathbbC .
$$
via direct computations. However, this does not guarantee that the dimension is $1$. We need to check that $1 otimes_mathbbC[x^2] 1 neq 0$. Do you have any ideas? I know the property that the tensor product is a right-exact functor, but I have no idea how to apply it here.
Any comments are welcomed. Thank you in advance.
abstract-algebra modules tensor-products
asked Aug 19 at 5:52
C. Oh
908
2
For part $(a)$, the simplest argument is that for any commutative ring $R$, $;R/Iotimes_RR/Jsimeq R/I+J$, and here, $I$ and $J$ are generated by coprime lements in the P.I.D. $mathbf C[x]$.
– Bernard
Aug 19 at 9:12
@Bernard Very short and clear proof! Thank you for another solution of part $(a)$.
– C. Oh
Aug 19 at 9:20
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is a recent qual problem that I am struggling with.
Put $M=mathbbC[x]/(x^2+x)$ and $N=mathbbC[x]/(x-1)$.
$(a)$ What is the dimension of $M otimes_mathbbC[x]N$ as a vector space over $mathbbC$?
$(b)$ What is the dimension of $M otimes_mathbbCN$ as a vector space over $mathbbC$?
$(c)$ What is the dimension of $M otimes_mathbbC[x^2]N$ as a vector space over $mathbbC$?
I solved the part $(a)$ and $(b)$. The part $(b)$ is the easiest. Since $M$ (resp. $N$) is a free $mathbbC$-module with basis $1,x $ (resp. $1$), the dimension is $2$. To get the part $(a)$, I used the direct computations:
$$
1 otimes 1 = (1+x^2+x) otimes 1 = 1 otimes (1+x^2+x) = 1 otimes 3 = 3(1 otimes 1).
$$
Thus, we get $1 otimes 1 = 0$, and the dimension is $0$. For another proof of the part $(a)$, we might be able to use that the tensor product is a right-exact functor because
$$
mathbbC[x]/(x^2+x,x-1) = 0.
$$
I don't know how to solve the part $(c)$. First, I showed
$$
M otimes_mathbbC[x^2] N = c(1 otimes_mathbbC[x^2] 1): c in mathbbC .
$$
via direct computations. However, this does not guarantee that the dimension is $1$. We need to check that $1 otimes_mathbbC[x^2] 1 neq 0$. Do you have any ideas? I know the property that the tensor product is a right-exact functor, but I have no idea how to apply it here.
Any comments are welcomed. Thank you in advance.
abstract-algebra modules tensor-products
asked Aug 19 at 5:52
C. Oh
908
This is a recent qual problem that I am struggling with.
Put $M=mathbbC[x]/(x^2+x)$ and $N=mathbbC[x]/(x-1)$.
$(a)$ What is the dimension of $M otimes_mathbbC[x]N$ as a vector space over $mathbbC$?
$(b)$ What is the dimension of $M otimes_mathbbCN$ as a vector space over $mathbbC$?
$(c)$ What is the dimension of $M otimes_mathbbC[x^2]N$ as a vector space over $mathbbC$?
I solved the part $(a)$ and $(b)$. The part $(b)$ is the easiest. Since $M$ (resp. $N$) is a free $mathbbC$-module with basis $1,x $ (resp. $1$), the dimension is $2$. To get the part $(a)$, I used the direct computations:
$$
1 otimes 1 = (1+x^2+x) otimes 1 = 1 otimes (1+x^2+x) = 1 otimes 3 = 3(1 otimes 1).
$$
Thus, we get $1 otimes 1 = 0$, and the dimension is $0$. For another proof of the part $(a)$, we might be able to use that the tensor product is a right-exact functor because
$$
mathbbC[x]/(x^2+x,x-1) = 0.
$$
I don't know how to solve the part $(c)$. First, I showed
$$
M otimes_mathbbC[x^2] N = c(1 otimes_mathbbC[x^2] 1): c in mathbbC .
$$
via direct computations. However, this does not guarantee that the dimension is $1$. We need to check that $1 otimes_mathbbC[x^2] 1 neq 0$. Do you have any ideas? I know the property that the tensor product is a right-exact functor, but I have no idea how to apply it here.
Any comments are welcomed. Thank you in advance.
abstract-algebra modules tensor-products
asked Aug 19 at 5:52
C. Oh
908
edited Aug 19 at 10:44
asked Aug 19 at 5:52
C. Oh
908
asked Aug 19 at 5:52
C. Oh
908
asked Aug 19 at 5:52
C. Oh
908
908
2
For part $(a)$, the simplest argument is that for any commutative ring $R$, $;R/Iotimes_RR/Jsimeq R/I+J$, and here, $I$ and $J$ are generated by coprime lements in the P.I.D. $mathbf C[x]$.
– Bernard
Aug 19 at 9:12
@Bernard Very short and clear proof! Thank you for another solution of part $(a)$.
– C. Oh
Aug 19 at 9:20
add a comment |Â
2
For part $(a)$, the simplest argument is that for any commutative ring $R$, $;R/Iotimes_RR/Jsimeq R/I+J$, and here, $I$ and $J$ are generated by coprime lements in the P.I.D. $mathbf C[x]$.
– Bernard
Aug 19 at 9:12
@Bernard Very short and clear proof! Thank you for another solution of part $(a)$.
– C. Oh
Aug 19 at 9:20
2
2
For part $(a)$, the simplest argument is that for any commutative ring $R$, $;R/Iotimes_RR/Jsimeq R/I+J$, and here, $I$ and $J$ are generated by coprime lements in the P.I.D. $mathbf C[x]$.
– Bernard
Aug 19 at 9:12
For part $(a)$, the simplest argument is that for any commutative ring $R$, $;R/Iotimes_RR/Jsimeq R/I+J$, and here, $I$ and $J$ are generated by coprime lements in the P.I.D. $mathbf C[x]$.
– Bernard
Aug 19 at 9:12
@Bernard Very short and clear proof! Thank you for another solution of part $(a)$.
– C. Oh
Aug 19 at 9:20
@Bernard Very short and clear proof! Thank you for another solution of part $(a)$.
– C. Oh
Aug 19 at 9:20
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
We can define a bilinear map $beta:Mtimes NtoBbb C, (a+bx, ,c) mapsto (b-a)c$.
Then we only have to check that it's also $Bbb C[x^2]$-bilinear, where $x^2$ acts on $N=Bbb C[x]/(x-1)congBbb C $ as $1$:
$$((a+bx)x^2,, c) = (-ax+bx, , c) mapsto (b-a)c\
(a+bx, , x^2c) =(a+bx, c) mapsto (b-a)c$$
answered Aug 20 at 0:44
Berci
56.6k23570
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We can define a bilinear map $beta:Mtimes NtoBbb C, (a+bx, ,c) mapsto (b-a)c$.
Then we only have to check that it's also $Bbb C[x^2]$-bilinear, where $x^2$ acts on $N=Bbb C[x]/(x-1)congBbb C $ as $1$:
$$((a+bx)x^2,, c) = (-ax+bx, , c) mapsto (b-a)c\
(a+bx, , x^2c) =(a+bx, c) mapsto (b-a)c$$
answered Aug 20 at 0:44
Berci
56.6k23570
add a comment |Â
up vote
1
down vote
accepted
We can define a bilinear map $beta:Mtimes NtoBbb C, (a+bx, ,c) mapsto (b-a)c$.
Then we only have to check that it's also $Bbb C[x^2]$-bilinear, where $x^2$ acts on $N=Bbb C[x]/(x-1)congBbb C $ as $1$:
$$((a+bx)x^2,, c) = (-ax+bx, , c) mapsto (b-a)c\
(a+bx, , x^2c) =(a+bx, c) mapsto (b-a)c$$
answered Aug 20 at 0:44
Berci
56.6k23570
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We can define a bilinear map $beta:Mtimes NtoBbb C, (a+bx, ,c) mapsto (b-a)c$.
Then we only have to check that it's also $Bbb C[x^2]$-bilinear, where $x^2$ acts on $N=Bbb C[x]/(x-1)congBbb C $ as $1$:
$$((a+bx)x^2,, c) = (-ax+bx, , c) mapsto (b-a)c\
(a+bx, , x^2c) =(a+bx, c) mapsto (b-a)c$$
answered Aug 20 at 0:44
Berci
56.6k23570
We can define a bilinear map $beta:Mtimes NtoBbb C, (a+bx, ,c) mapsto (b-a)c$.
Then we only have to check that it's also $Bbb C[x^2]$-bilinear, where $x^2$ acts on $N=Bbb C[x]/(x-1)congBbb C $ as $1$:
$$((a+bx)x^2,, c) = (-ax+bx, , c) mapsto (b-a)c\
(a+bx, , x^2c) =(a+bx, c) mapsto (b-a)c$$
answered Aug 20 at 0:44
Berci
56.6k23570
answered Aug 20 at 0:44
Berci
56.6k23570
answered Aug 20 at 0:44
Berci
56.6k23570
answered Aug 20 at 0:44
Berci
56.6k23570
56.6k23570
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2887392%2fcalculating-the-dimension-of-the-tensor-product-of-modules%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
For part $(a)$, the simplest argument is that for any commutative ring $R$, $;R/Iotimes_RR/Jsimeq R/I+J$, and here, $I$ and $J$ are generated by coprime lements in the P.I.D. $mathbf C[x]$.
– Bernard
Aug 19 at 9:12
@Bernard Very short and clear proof! Thank you for another solution of part $(a)$.
– C. Oh
Aug 19 at 9:20