Tensor Product of Fields is a Field Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Tensor product of fields and homomorphismstensor product of a number field with $ mathbb R $Explicit examples of tensor productsTensor product of algebraic number field with p-adic fieldTensor product of modules over an arbitrary algebra is not always defined?Showing that multilinear maps factor through tensor productsTensor product of $K$-skew fields is nilpotent iff every $D_i$ is nilpotent?Number field tensor $mathbbQ$ isomorphisms.Tensor Product of Hilbert Spaces: how to prove completenessComposite of certain Finite Subextensions $L|K$ Galois
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Tensor Product of Fields is a Field
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Tensor product of fields and homomorphismstensor product of a number field with $ mathbb R $Explicit examples of tensor productsTensor product of algebraic number field with p-adic fieldTensor product of modules over an arbitrary algebra is not always defined?Showing that multilinear maps factor through tensor productsTensor product of $K$-skew fields is nilpotent iff every $D_i$ is nilpotent?Number field tensor $mathbbQ$ isomorphisms.Tensor Product of Hilbert Spaces: how to prove completenessComposite of certain Finite Subextensions $L|K$ Galois
$begingroup$
I have two questions about a construction introduced in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 122):
We fix an integral proper normal curve $X$ over a field $k$. We consider it's function field $K$ which is a finite extension of $k(t)$ and take an arbitrary field extension $L vert k$.
The point of interest is the resulting tensor product $K otimes L$. We know that $K otimes L$ is finite dimensional $L(t)$-algebra.
Following two questions:
Assume $K otimes L$ is a finite direct product of fields $L_i$. Why these fields are finitely generated (as $L$-modules)?
Assume non $k$ is algebraically closed. Why is $K otimes L$ then a field?
1.
abstract-algebra extension-field tensor-products
asked Apr 1 at 1:18
KarlPeterKarlPeter
7091416
$endgroup$
|
show 8 more comments
$begingroup$
I have two questions about a construction introduced in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 122):
We fix an integral proper normal curve $X$ over a field $k$. We consider it's function field $K$ which is a finite extension of $k(t)$ and take an arbitrary field extension $L vert k$.
The point of interest is the resulting tensor product $K otimes L$. We know that $K otimes L$ is finite dimensional $L(t)$-algebra.
Following two questions:
Assume $K otimes L$ is a finite direct product of fields $L_i$. Why these fields are finitely generated (as $L$-modules)?
Assume non $k$ is algebraically closed. Why is $K otimes L$ then a field?
1.
abstract-algebra extension-field tensor-products
asked Apr 1 at 1:18
KarlPeterKarlPeter
7091416
$endgroup$
1
$begingroup$
If $L/k$ is separable then the primitive element gives $L = k[x]/(f)$ and $K otimes_k L = K[x]/(f) = K[x]/(prod_j f_j) = prod_j K[x]/(f_j)$ since $f$ separable implies the $(f_j)$ are comaximal. If $k$ is algebraically closed and $L/k$ is a tower of purely transcendental and algebraic extensions then rename the transcendental elements so they are not in $overlineK$ thus the algebraic extensions keep the same degree with $k$ or $K$ as the basefield
$endgroup$
– reuns
Apr 1 at 1:42
$begingroup$
I'm quite not sure if it is possible to reduce the problem 1. to the case when $L vert k$ is separable. Or why do you implicitely assumed that?
$endgroup$
– KarlPeter
Apr 1 at 1:54
1
$begingroup$
If $K$ is a finite extension of $k(t)$ then $K = k(t)[Y_1,ldots,Y_m]/I$, finitely generated as field, not as an algebra or module, and if $K otimes_k L = L(t)[Y_1,ldots,Y_m]/I= prod_j L_j$ then each $L_j$ is a quotient of $L(t)[Y_1,ldots,Y_m]/I$, finitely generated as a field over $L$.
$endgroup$
– reuns
Apr 1 at 1:56
$begingroup$
@reuns: yes, so the 1. question is solved. Concerning the second one I'm a bit confused. Assume $k$ is algebraically closed. $K$ is finite extension of $k(t)$ and therefore has transcendence degree $1$ over $k$. On the other hand $L vert k$ might be an arbitrary extension.
$endgroup$
– KarlPeter
Apr 1 at 12:52
$begingroup$
Then Grothendieck-Sharp tells as in mathoverflow.net/questions/82083/… that $dim_mathrmKrull(Kotimes_k L) = min(operatornametrdeg_k(K),operatornametrdeg_k(L))$. Then if we take as $L$ for example a transcendent extension $k(t)$ of $k$ we obtain $dim_mathrmKrull(Kotimes_k L)=1$ so the tensor product can't be a field. But this contradicts the statement 2. The author didn't explicitely siad that $L vert k$ should be a finite extension
$endgroup$
– KarlPeter
Apr 1 at 12:52
|
show 8 more comments
$begingroup$
I have two questions about a construction introduced in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 122):
We fix an integral proper normal curve $X$ over a field $k$. We consider it's function field $K$ which is a finite extension of $k(t)$ and take an arbitrary field extension $L vert k$.
The point of interest is the resulting tensor product $K otimes L$. We know that $K otimes L$ is finite dimensional $L(t)$-algebra.
Following two questions:
Assume $K otimes L$ is a finite direct product of fields $L_i$. Why these fields are finitely generated (as $L$-modules)?
Assume non $k$ is algebraically closed. Why is $K otimes L$ then a field?
1.
abstract-algebra extension-field tensor-products
asked Apr 1 at 1:18
KarlPeterKarlPeter
7091416
$endgroup$
I have two questions about a construction introduced in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 122):
We fix an integral proper normal curve $X$ over a field $k$. We consider it's function field $K$ which is a finite extension of $k(t)$ and take an arbitrary field extension $L vert k$.
The point of interest is the resulting tensor product $K otimes L$. We know that $K otimes L$ is finite dimensional $L(t)$-algebra.
Following two questions:
Assume $K otimes L$ is a finite direct product of fields $L_i$. Why these fields are finitely generated (as $L$-modules)?
Assume non $k$ is algebraically closed. Why is $K otimes L$ then a field?
1.
abstract-algebra extension-field tensor-products
abstract-algebra extension-field tensor-products
asked Apr 1 at 1:18
KarlPeterKarlPeter
7091416
asked Apr 1 at 1:18
KarlPeterKarlPeter
7091416
asked Apr 1 at 1:18
KarlPeterKarlPeter
7091416
asked Apr 1 at 1:18
KarlPeterKarlPeter
7091416
asked Apr 1 at 1:18
KarlPeterKarlPeter
7091416
7091416
1
$begingroup$
If $L/k$ is separable then the primitive element gives $L = k[x]/(f)$ and $K otimes_k L = K[x]/(f) = K[x]/(prod_j f_j) = prod_j K[x]/(f_j)$ since $f$ separable implies the $(f_j)$ are comaximal. If $k$ is algebraically closed and $L/k$ is a tower of purely transcendental and algebraic extensions then rename the transcendental elements so they are not in $overlineK$ thus the algebraic extensions keep the same degree with $k$ or $K$ as the basefield
$endgroup$
– reuns
Apr 1 at 1:42
$begingroup$
I'm quite not sure if it is possible to reduce the problem 1. to the case when $L vert k$ is separable. Or why do you implicitely assumed that?
$endgroup$
– KarlPeter
Apr 1 at 1:54
1
$begingroup$
If $K$ is a finite extension of $k(t)$ then $K = k(t)[Y_1,ldots,Y_m]/I$, finitely generated as field, not as an algebra or module, and if $K otimes_k L = L(t)[Y_1,ldots,Y_m]/I= prod_j L_j$ then each $L_j$ is a quotient of $L(t)[Y_1,ldots,Y_m]/I$, finitely generated as a field over $L$.
$endgroup$
– reuns
Apr 1 at 1:56
$begingroup$
@reuns: yes, so the 1. question is solved. Concerning the second one I'm a bit confused. Assume $k$ is algebraically closed. $K$ is finite extension of $k(t)$ and therefore has transcendence degree $1$ over $k$. On the other hand $L vert k$ might be an arbitrary extension.
$endgroup$
– KarlPeter
Apr 1 at 12:52
$begingroup$
Then Grothendieck-Sharp tells as in mathoverflow.net/questions/82083/… that $dim_mathrmKrull(Kotimes_k L) = min(operatornametrdeg_k(K),operatornametrdeg_k(L))$. Then if we take as $L$ for example a transcendent extension $k(t)$ of $k$ we obtain $dim_mathrmKrull(Kotimes_k L)=1$ so the tensor product can't be a field. But this contradicts the statement 2. The author didn't explicitely siad that $L vert k$ should be a finite extension
$endgroup$
– KarlPeter
Apr 1 at 12:52
|
show 8 more comments
1
$begingroup$
If $L/k$ is separable then the primitive element gives $L = k[x]/(f)$ and $K otimes_k L = K[x]/(f) = K[x]/(prod_j f_j) = prod_j K[x]/(f_j)$ since $f$ separable implies the $(f_j)$ are comaximal. If $k$ is algebraically closed and $L/k$ is a tower of purely transcendental and algebraic extensions then rename the transcendental elements so they are not in $overlineK$ thus the algebraic extensions keep the same degree with $k$ or $K$ as the basefield
$endgroup$
– reuns
Apr 1 at 1:42
$begingroup$
I'm quite not sure if it is possible to reduce the problem 1. to the case when $L vert k$ is separable. Or why do you implicitely assumed that?
$endgroup$
– KarlPeter
Apr 1 at 1:54
1
$begingroup$
If $K$ is a finite extension of $k(t)$ then $K = k(t)[Y_1,ldots,Y_m]/I$, finitely generated as field, not as an algebra or module, and if $K otimes_k L = L(t)[Y_1,ldots,Y_m]/I= prod_j L_j$ then each $L_j$ is a quotient of $L(t)[Y_1,ldots,Y_m]/I$, finitely generated as a field over $L$.
$endgroup$
– reuns
Apr 1 at 1:56
$begingroup$
@reuns: yes, so the 1. question is solved. Concerning the second one I'm a bit confused. Assume $k$ is algebraically closed. $K$ is finite extension of $k(t)$ and therefore has transcendence degree $1$ over $k$. On the other hand $L vert k$ might be an arbitrary extension.
$endgroup$
– KarlPeter
Apr 1 at 12:52
$begingroup$
Then Grothendieck-Sharp tells as in mathoverflow.net/questions/82083/… that $dim_mathrmKrull(Kotimes_k L) = min(operatornametrdeg_k(K),operatornametrdeg_k(L))$. Then if we take as $L$ for example a transcendent extension $k(t)$ of $k$ we obtain $dim_mathrmKrull(Kotimes_k L)=1$ so the tensor product can't be a field. But this contradicts the statement 2. The author didn't explicitely siad that $L vert k$ should be a finite extension
$endgroup$
– KarlPeter
Apr 1 at 12:52
1
1
$begingroup$
If $L/k$ is separable then the primitive element gives $L = k[x]/(f)$ and $K otimes_k L = K[x]/(f) = K[x]/(prod_j f_j) = prod_j K[x]/(f_j)$ since $f$ separable implies the $(f_j)$ are comaximal. If $k$ is algebraically closed and $L/k$ is a tower of purely transcendental and algebraic extensions then rename the transcendental elements so they are not in $overlineK$ thus the algebraic extensions keep the same degree with $k$ or $K$ as the basefield
$endgroup$
– reuns
Apr 1 at 1:42
$begingroup$
If $L/k$ is separable then the primitive element gives $L = k[x]/(f)$ and $K otimes_k L = K[x]/(f) = K[x]/(prod_j f_j) = prod_j K[x]/(f_j)$ since $f$ separable implies the $(f_j)$ are comaximal. If $k$ is algebraically closed and $L/k$ is a tower of purely transcendental and algebraic extensions then rename the transcendental elements so they are not in $overlineK$ thus the algebraic extensions keep the same degree with $k$ or $K$ as the basefield
$endgroup$
– reuns
Apr 1 at 1:42
$begingroup$
I'm quite not sure if it is possible to reduce the problem 1. to the case when $L vert k$ is separable. Or why do you implicitely assumed that?
$endgroup$
– KarlPeter
Apr 1 at 1:54
$begingroup$
I'm quite not sure if it is possible to reduce the problem 1. to the case when $L vert k$ is separable. Or why do you implicitely assumed that?
$endgroup$
– KarlPeter
Apr 1 at 1:54
1
1
$begingroup$
If $K$ is a finite extension of $k(t)$ then $K = k(t)[Y_1,ldots,Y_m]/I$, finitely generated as field, not as an algebra or module, and if $K otimes_k L = L(t)[Y_1,ldots,Y_m]/I= prod_j L_j$ then each $L_j$ is a quotient of $L(t)[Y_1,ldots,Y_m]/I$, finitely generated as a field over $L$.
$endgroup$
– reuns
Apr 1 at 1:56
$begingroup$
If $K$ is a finite extension of $k(t)$ then $K = k(t)[Y_1,ldots,Y_m]/I$, finitely generated as field, not as an algebra or module, and if $K otimes_k L = L(t)[Y_1,ldots,Y_m]/I= prod_j L_j$ then each $L_j$ is a quotient of $L(t)[Y_1,ldots,Y_m]/I$, finitely generated as a field over $L$.
$endgroup$
– reuns
Apr 1 at 1:56
$begingroup$
@reuns: yes, so the 1. question is solved. Concerning the second one I'm a bit confused. Assume $k$ is algebraically closed. $K$ is finite extension of $k(t)$ and therefore has transcendence degree $1$ over $k$. On the other hand $L vert k$ might be an arbitrary extension.
$endgroup$
– KarlPeter
Apr 1 at 12:52
$begingroup$
@reuns: yes, so the 1. question is solved. Concerning the second one I'm a bit confused. Assume $k$ is algebraically closed. $K$ is finite extension of $k(t)$ and therefore has transcendence degree $1$ over $k$. On the other hand $L vert k$ might be an arbitrary extension.
$endgroup$
– KarlPeter
Apr 1 at 12:52
$begingroup$
Then Grothendieck-Sharp tells as in mathoverflow.net/questions/82083/… that $dim_mathrmKrull(Kotimes_k L) = min(operatornametrdeg_k(K),operatornametrdeg_k(L))$. Then if we take as $L$ for example a transcendent extension $k(t)$ of $k$ we obtain $dim_mathrmKrull(Kotimes_k L)=1$ so the tensor product can't be a field. But this contradicts the statement 2. The author didn't explicitely siad that $L vert k$ should be a finite extension
$endgroup$
– KarlPeter
Apr 1 at 12:52
$begingroup$
Then Grothendieck-Sharp tells as in mathoverflow.net/questions/82083/… that $dim_mathrmKrull(Kotimes_k L) = min(operatornametrdeg_k(K),operatornametrdeg_k(L))$. Then if we take as $L$ for example a transcendent extension $k(t)$ of $k$ we obtain $dim_mathrmKrull(Kotimes_k L)=1$ so the tensor product can't be a field. But this contradicts the statement 2. The author didn't explicitely siad that $L vert k$ should be a finite extension
$endgroup$
– KarlPeter
Apr 1 at 12:52
|
show 8 more comments
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1
$begingroup$
If $L/k$ is separable then the primitive element gives $L = k[x]/(f)$ and $K otimes_k L = K[x]/(f) = K[x]/(prod_j f_j) = prod_j K[x]/(f_j)$ since $f$ separable implies the $(f_j)$ are comaximal. If $k$ is algebraically closed and $L/k$ is a tower of purely transcendental and algebraic extensions then rename the transcendental elements so they are not in $overlineK$ thus the algebraic extensions keep the same degree with $k$ or $K$ as the basefield
$endgroup$
– reuns
Apr 1 at 1:42
$begingroup$
I'm quite not sure if it is possible to reduce the problem 1. to the case when $L vert k$ is separable. Or why do you implicitely assumed that?
$endgroup$
– KarlPeter
Apr 1 at 1:54
1
$begingroup$
If $K$ is a finite extension of $k(t)$ then $K = k(t)[Y_1,ldots,Y_m]/I$, finitely generated as field, not as an algebra or module, and if $K otimes_k L = L(t)[Y_1,ldots,Y_m]/I= prod_j L_j$ then each $L_j$ is a quotient of $L(t)[Y_1,ldots,Y_m]/I$, finitely generated as a field over $L$.
$endgroup$
– reuns
Apr 1 at 1:56
$begingroup$
@reuns: yes, so the 1. question is solved. Concerning the second one I'm a bit confused. Assume $k$ is algebraically closed. $K$ is finite extension of $k(t)$ and therefore has transcendence degree $1$ over $k$. On the other hand $L vert k$ might be an arbitrary extension.
$endgroup$
– KarlPeter
Apr 1 at 12:52
$begingroup$
Then Grothendieck-Sharp tells as in mathoverflow.net/questions/82083/… that $dim_mathrmKrull(Kotimes_k L) = min(operatornametrdeg_k(K),operatornametrdeg_k(L))$. Then if we take as $L$ for example a transcendent extension $k(t)$ of $k$ we obtain $dim_mathrmKrull(Kotimes_k L)=1$ so the tensor product can't be a field. But this contradicts the statement 2. The author didn't explicitely siad that $L vert k$ should be a finite extension
$endgroup$
– KarlPeter
Apr 1 at 12:52