Centre of algebra under field extension
$begingroup$
Suppose $F$ is a field extension of a field $k$ and $R$ is an algebra over $k$. Is it true that $Z(R otimes_kF)=Z(R)otimes_k F$, where $Z(R)$ is the centre of $R$?
abstract-algebra ring-theory extension-field
asked Nov 28 '18 at 12:55
kissanpentukissanpentu
301112
$endgroup$
add a comment |
$begingroup$
Suppose $F$ is a field extension of a field $k$ and $R$ is an algebra over $k$. Is it true that $Z(R otimes_kF)=Z(R)otimes_k F$, where $Z(R)$ is the centre of $R$?
abstract-algebra ring-theory extension-field
asked Nov 28 '18 at 12:55
kissanpentukissanpentu
301112
$endgroup$
$begingroup$
Yes. As proved here, if $R, S$ are two associative unital algebras over a field $k$ (not necessarily finite dimensional), then $$Z(R otimes_k S) cong Z(R) otimes_k Z(S).$$
$endgroup$
– Watson
Nov 28 '18 at 13:18
add a comment |
$begingroup$
Suppose $F$ is a field extension of a field $k$ and $R$ is an algebra over $k$. Is it true that $Z(R otimes_kF)=Z(R)otimes_k F$, where $Z(R)$ is the centre of $R$?
abstract-algebra ring-theory extension-field
asked Nov 28 '18 at 12:55
kissanpentukissanpentu
301112
$endgroup$
Suppose $F$ is a field extension of a field $k$ and $R$ is an algebra over $k$. Is it true that $Z(R otimes_kF)=Z(R)otimes_k F$, where $Z(R)$ is the centre of $R$?
abstract-algebra ring-theory extension-field
abstract-algebra ring-theory extension-field
asked Nov 28 '18 at 12:55
kissanpentukissanpentu
301112
asked Nov 28 '18 at 12:55
kissanpentukissanpentu
301112
asked Nov 28 '18 at 12:55
kissanpentukissanpentu
301112
asked Nov 28 '18 at 12:55
kissanpentukissanpentu
301112
asked Nov 28 '18 at 12:55
kissanpentukissanpentu
301112
301112
$begingroup$
Yes. As proved here, if $R, S$ are two associative unital algebras over a field $k$ (not necessarily finite dimensional), then $$Z(R otimes_k S) cong Z(R) otimes_k Z(S).$$
$endgroup$
– Watson
Nov 28 '18 at 13:18
add a comment |
$begingroup$
Yes. As proved here, if $R, S$ are two associative unital algebras over a field $k$ (not necessarily finite dimensional), then $$Z(R otimes_k S) cong Z(R) otimes_k Z(S).$$
$endgroup$
– Watson
Nov 28 '18 at 13:18
$begingroup$
Yes. As proved here, if $R, S$ are two associative unital algebras over a field $k$ (not necessarily finite dimensional), then $$Z(R otimes_k S) cong Z(R) otimes_k Z(S).$$
$endgroup$
– Watson
Nov 28 '18 at 13:18
$begingroup$
Yes. As proved here, if $R, S$ are two associative unital algebras over a field $k$ (not necessarily finite dimensional), then $$Z(R otimes_k S) cong Z(R) otimes_k Z(S).$$
$endgroup$
– Watson
Nov 28 '18 at 13:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is correct.
Let $ a = sum r_i otimes x_i $ be an element in the center, $ r_i in R $ and $ x_i in F $. We may assume that the $ x_i $ are linearly independent over $ k $ and hence, are a part of a $ k $-basis of $ F $. Then the elements $ 1 otimes x_i $ are part of a basis of $ R otimes F $ as a free $ R $-module. We have $ a(r otimes 1) = (r otimes 1)a $ for all $ r in R $. This gives $$ sum (rr_i-r_ir) otimes x_i = 0 = sum (rr_i-r_ir)(1 otimes x_i) $$ where in the last sum, we are viewing $ R otimes F $ as a free $ R $-module. This means $ rr_i=r_ir $ for all $ i $ and $ r in R $ hence $ r_i in Z(R) $ giving the conclusion.
answered Nov 28 '18 at 13:17
hellHoundhellHound
48328
$endgroup$
$begingroup$
Brilliant! I have managed to obtain something similar, but for pure tensors. It turns out that the general case can also be elegant and simple. Thank you!
$endgroup$
– kissanpentu
Nov 28 '18 at 14:19
add a comment |
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1 Answer
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$begingroup$
This is correct.
Let $ a = sum r_i otimes x_i $ be an element in the center, $ r_i in R $ and $ x_i in F $. We may assume that the $ x_i $ are linearly independent over $ k $ and hence, are a part of a $ k $-basis of $ F $. Then the elements $ 1 otimes x_i $ are part of a basis of $ R otimes F $ as a free $ R $-module. We have $ a(r otimes 1) = (r otimes 1)a $ for all $ r in R $. This gives $$ sum (rr_i-r_ir) otimes x_i = 0 = sum (rr_i-r_ir)(1 otimes x_i) $$ where in the last sum, we are viewing $ R otimes F $ as a free $ R $-module. This means $ rr_i=r_ir $ for all $ i $ and $ r in R $ hence $ r_i in Z(R) $ giving the conclusion.
answered Nov 28 '18 at 13:17
hellHoundhellHound
48328
$endgroup$
$begingroup$
Brilliant! I have managed to obtain something similar, but for pure tensors. It turns out that the general case can also be elegant and simple. Thank you!
$endgroup$
– kissanpentu
Nov 28 '18 at 14:19
add a comment |
$begingroup$
This is correct.
Let $ a = sum r_i otimes x_i $ be an element in the center, $ r_i in R $ and $ x_i in F $. We may assume that the $ x_i $ are linearly independent over $ k $ and hence, are a part of a $ k $-basis of $ F $. Then the elements $ 1 otimes x_i $ are part of a basis of $ R otimes F $ as a free $ R $-module. We have $ a(r otimes 1) = (r otimes 1)a $ for all $ r in R $. This gives $$ sum (rr_i-r_ir) otimes x_i = 0 = sum (rr_i-r_ir)(1 otimes x_i) $$ where in the last sum, we are viewing $ R otimes F $ as a free $ R $-module. This means $ rr_i=r_ir $ for all $ i $ and $ r in R $ hence $ r_i in Z(R) $ giving the conclusion.
answered Nov 28 '18 at 13:17
hellHoundhellHound
48328
$endgroup$
$begingroup$
Brilliant! I have managed to obtain something similar, but for pure tensors. It turns out that the general case can also be elegant and simple. Thank you!
$endgroup$
– kissanpentu
Nov 28 '18 at 14:19
add a comment |
$begingroup$
This is correct.
Let $ a = sum r_i otimes x_i $ be an element in the center, $ r_i in R $ and $ x_i in F $. We may assume that the $ x_i $ are linearly independent over $ k $ and hence, are a part of a $ k $-basis of $ F $. Then the elements $ 1 otimes x_i $ are part of a basis of $ R otimes F $ as a free $ R $-module. We have $ a(r otimes 1) = (r otimes 1)a $ for all $ r in R $. This gives $$ sum (rr_i-r_ir) otimes x_i = 0 = sum (rr_i-r_ir)(1 otimes x_i) $$ where in the last sum, we are viewing $ R otimes F $ as a free $ R $-module. This means $ rr_i=r_ir $ for all $ i $ and $ r in R $ hence $ r_i in Z(R) $ giving the conclusion.
answered Nov 28 '18 at 13:17
hellHoundhellHound
48328
$endgroup$
This is correct.
Let $ a = sum r_i otimes x_i $ be an element in the center, $ r_i in R $ and $ x_i in F $. We may assume that the $ x_i $ are linearly independent over $ k $ and hence, are a part of a $ k $-basis of $ F $. Then the elements $ 1 otimes x_i $ are part of a basis of $ R otimes F $ as a free $ R $-module. We have $ a(r otimes 1) = (r otimes 1)a $ for all $ r in R $. This gives $$ sum (rr_i-r_ir) otimes x_i = 0 = sum (rr_i-r_ir)(1 otimes x_i) $$ where in the last sum, we are viewing $ R otimes F $ as a free $ R $-module. This means $ rr_i=r_ir $ for all $ i $ and $ r in R $ hence $ r_i in Z(R) $ giving the conclusion.
answered Nov 28 '18 at 13:17
hellHoundhellHound
48328
answered Nov 28 '18 at 13:17
hellHoundhellHound
48328
answered Nov 28 '18 at 13:17
hellHoundhellHound
48328
answered Nov 28 '18 at 13:17
hellHoundhellHound
48328
48328
$begingroup$
Brilliant! I have managed to obtain something similar, but for pure tensors. It turns out that the general case can also be elegant and simple. Thank you!
$endgroup$
– kissanpentu
Nov 28 '18 at 14:19
add a comment |
$begingroup$
Brilliant! I have managed to obtain something similar, but for pure tensors. It turns out that the general case can also be elegant and simple. Thank you!
$endgroup$
– kissanpentu
Nov 28 '18 at 14:19
$begingroup$
Brilliant! I have managed to obtain something similar, but for pure tensors. It turns out that the general case can also be elegant and simple. Thank you!
$endgroup$
– kissanpentu
Nov 28 '18 at 14:19
$begingroup$
Brilliant! I have managed to obtain something similar, but for pure tensors. It turns out that the general case can also be elegant and simple. Thank you!
$endgroup$
– kissanpentu
Nov 28 '18 at 14:19
add a comment |
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$begingroup$
Yes. As proved here, if $R, S$ are two associative unital algebras over a field $k$ (not necessarily finite dimensional), then $$Z(R otimes_k S) cong Z(R) otimes_k Z(S).$$
$endgroup$
– Watson
Nov 28 '18 at 13:18