Krull Dimension of $K otimes_k L$
$begingroup$
Let $k$ be a field and $K,L$ are arbitrary extensions of $k$.
What do we know about $dim(K otimes_k L)$ (as Krull dimension). How does it depend on transcendence degrees $trdeg_k(K), trdeg_k(L)$?
Futhermore: If we cosider by $k$ instead of a field a ring and $K,L$ be it's ring extensions. How does $dim(K otimes_k L)$ depend on $dim(K), dim(L)$?
ring-theory commutative-algebra tensor-products krull-dimension
asked Dec 30 '18 at 0:04
KarlPeterKarlPeter
3201315
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field and $K,L$ are arbitrary extensions of $k$.
What do we know about $dim(K otimes_k L)$ (as Krull dimension). How does it depend on transcendence degrees $trdeg_k(K), trdeg_k(L)$?
Futhermore: If we cosider by $k$ instead of a field a ring and $K,L$ be it's ring extensions. How does $dim(K otimes_k L)$ depend on $dim(K), dim(L)$?
ring-theory commutative-algebra tensor-products krull-dimension
asked Dec 30 '18 at 0:04
KarlPeterKarlPeter
3201315
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field and $K,L$ are arbitrary extensions of $k$.
What do we know about $dim(K otimes_k L)$ (as Krull dimension). How does it depend on transcendence degrees $trdeg_k(K), trdeg_k(L)$?
Futhermore: If we cosider by $k$ instead of a field a ring and $K,L$ be it's ring extensions. How does $dim(K otimes_k L)$ depend on $dim(K), dim(L)$?
ring-theory commutative-algebra tensor-products krull-dimension
asked Dec 30 '18 at 0:04
KarlPeterKarlPeter
3201315
$endgroup$
Let $k$ be a field and $K,L$ are arbitrary extensions of $k$.
What do we know about $dim(K otimes_k L)$ (as Krull dimension). How does it depend on transcendence degrees $trdeg_k(K), trdeg_k(L)$?
Futhermore: If we cosider by $k$ instead of a field a ring and $K,L$ be it's ring extensions. How does $dim(K otimes_k L)$ depend on $dim(K), dim(L)$?
ring-theory commutative-algebra tensor-products krull-dimension
ring-theory commutative-algebra tensor-products krull-dimension
asked Dec 30 '18 at 0:04
KarlPeterKarlPeter
3201315
asked Dec 30 '18 at 0:04
KarlPeterKarlPeter
3201315
asked Dec 30 '18 at 0:04
KarlPeterKarlPeter
3201315
asked Dec 30 '18 at 0:04
KarlPeterKarlPeter
3201315
asked Dec 30 '18 at 0:04
KarlPeterKarlPeter
3201315
3201315
add a comment |
add a comment |
1 Answer
1
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In the case where $k$ is a field we have
$$ dim (K otimes_k L) = min(operatorname{trdeg}_k K, operatorname{trdeg}_k L) $$
There is a Proof in Görtz's and Wedhorn's Algebraic Geometry 1 (Lemma B.97).
No such thing is possible in the second case: Suppose $k$ is a ring that "happens to be" a field and $K = L = k(x_1, dots, x_n)$ are the ring extensions of $k$. We have $dim K = dim L = 0$ since they are fields but $dim(K otimes_k L) = n$. So $dim(K otimes_k L)$ does not only depend on $dim K$ and $dim L$.
answered Dec 30 '18 at 0:45
0x5390x539
1,415518
$endgroup$
$begingroup$
Thank you. Do you know anything about results if we consider the same story with rings instead of fields?
$endgroup$
– KarlPeter
Dec 30 '18 at 0:55
$begingroup$
looks that in case of rings it has something to do with transcendence degrees of quotient fields of the rings
$endgroup$
– KarlPeter
Dec 30 '18 at 2:20
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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$begingroup$
In the case where $k$ is a field we have
$$ dim (K otimes_k L) = min(operatorname{trdeg}_k K, operatorname{trdeg}_k L) $$
There is a Proof in Görtz's and Wedhorn's Algebraic Geometry 1 (Lemma B.97).
No such thing is possible in the second case: Suppose $k$ is a ring that "happens to be" a field and $K = L = k(x_1, dots, x_n)$ are the ring extensions of $k$. We have $dim K = dim L = 0$ since they are fields but $dim(K otimes_k L) = n$. So $dim(K otimes_k L)$ does not only depend on $dim K$ and $dim L$.
answered Dec 30 '18 at 0:45
0x5390x539
1,415518
$endgroup$
$begingroup$
Thank you. Do you know anything about results if we consider the same story with rings instead of fields?
$endgroup$
– KarlPeter
Dec 30 '18 at 0:55
$begingroup$
looks that in case of rings it has something to do with transcendence degrees of quotient fields of the rings
$endgroup$
– KarlPeter
Dec 30 '18 at 2:20
add a comment |
$begingroup$
In the case where $k$ is a field we have
$$ dim (K otimes_k L) = min(operatorname{trdeg}_k K, operatorname{trdeg}_k L) $$
There is a Proof in Görtz's and Wedhorn's Algebraic Geometry 1 (Lemma B.97).
No such thing is possible in the second case: Suppose $k$ is a ring that "happens to be" a field and $K = L = k(x_1, dots, x_n)$ are the ring extensions of $k$. We have $dim K = dim L = 0$ since they are fields but $dim(K otimes_k L) = n$. So $dim(K otimes_k L)$ does not only depend on $dim K$ and $dim L$.
answered Dec 30 '18 at 0:45
0x5390x539
1,415518
$endgroup$
$begingroup$
Thank you. Do you know anything about results if we consider the same story with rings instead of fields?
$endgroup$
– KarlPeter
Dec 30 '18 at 0:55
$begingroup$
looks that in case of rings it has something to do with transcendence degrees of quotient fields of the rings
$endgroup$
– KarlPeter
Dec 30 '18 at 2:20
add a comment |
$begingroup$
In the case where $k$ is a field we have
$$ dim (K otimes_k L) = min(operatorname{trdeg}_k K, operatorname{trdeg}_k L) $$
There is a Proof in Görtz's and Wedhorn's Algebraic Geometry 1 (Lemma B.97).
No such thing is possible in the second case: Suppose $k$ is a ring that "happens to be" a field and $K = L = k(x_1, dots, x_n)$ are the ring extensions of $k$. We have $dim K = dim L = 0$ since they are fields but $dim(K otimes_k L) = n$. So $dim(K otimes_k L)$ does not only depend on $dim K$ and $dim L$.
answered Dec 30 '18 at 0:45
0x5390x539
1,415518
$endgroup$
In the case where $k$ is a field we have
$$ dim (K otimes_k L) = min(operatorname{trdeg}_k K, operatorname{trdeg}_k L) $$
There is a Proof in Görtz's and Wedhorn's Algebraic Geometry 1 (Lemma B.97).
No such thing is possible in the second case: Suppose $k$ is a ring that "happens to be" a field and $K = L = k(x_1, dots, x_n)$ are the ring extensions of $k$. We have $dim K = dim L = 0$ since they are fields but $dim(K otimes_k L) = n$. So $dim(K otimes_k L)$ does not only depend on $dim K$ and $dim L$.
answered Dec 30 '18 at 0:45
0x5390x539
1,415518
edited Dec 30 '18 at 1:07
answered Dec 30 '18 at 0:45
0x5390x539
1,415518
answered Dec 30 '18 at 0:45
0x5390x539
1,415518
answered Dec 30 '18 at 0:45
0x5390x539
1,415518
1,415518
$begingroup$
Thank you. Do you know anything about results if we consider the same story with rings instead of fields?
$endgroup$
– KarlPeter
Dec 30 '18 at 0:55
$begingroup$
looks that in case of rings it has something to do with transcendence degrees of quotient fields of the rings
$endgroup$
– KarlPeter
Dec 30 '18 at 2:20
add a comment |
$begingroup$
Thank you. Do you know anything about results if we consider the same story with rings instead of fields?
$endgroup$
– KarlPeter
Dec 30 '18 at 0:55
$begingroup$
looks that in case of rings it has something to do with transcendence degrees of quotient fields of the rings
$endgroup$
– KarlPeter
Dec 30 '18 at 2:20
$begingroup$
Thank you. Do you know anything about results if we consider the same story with rings instead of fields?
$endgroup$
– KarlPeter
Dec 30 '18 at 0:55
$begingroup$
Thank you. Do you know anything about results if we consider the same story with rings instead of fields?
$endgroup$
– KarlPeter
Dec 30 '18 at 0:55
$begingroup$
looks that in case of rings it has something to do with transcendence degrees of quotient fields of the rings
$endgroup$
– KarlPeter
Dec 30 '18 at 2:20
$begingroup$
looks that in case of rings it has something to do with transcendence degrees of quotient fields of the rings
$endgroup$
– KarlPeter
Dec 30 '18 at 2:20
add a comment |
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