Condition for an A-module to be an A-algebra
Let B be an A-algebra where A is a noetherian ring
Prove that E the set of integral elements over A in B form a subring of B.
What I did:
I tried to prove that E is a subalgebra of B.
I proved by induction that E is a finitely-generated sub-module of B. Does it make it a sub-algebra of B ? on which conditions?
abstract-algebra
asked Dec 22 '18 at 14:12
PerelMan
519211
add a comment |
Let B be an A-algebra where A is a noetherian ring
Prove that E the set of integral elements over A in B form a subring of B.
What I did:
I tried to prove that E is a subalgebra of B.
I proved by induction that E is a finitely-generated sub-module of B. Does it make it a sub-algebra of B ? on which conditions?
abstract-algebra
asked Dec 22 '18 at 14:12
PerelMan
519211
add a comment |
Let B be an A-algebra where A is a noetherian ring
Prove that E the set of integral elements over A in B form a subring of B.
What I did:
I tried to prove that E is a subalgebra of B.
I proved by induction that E is a finitely-generated sub-module of B. Does it make it a sub-algebra of B ? on which conditions?
abstract-algebra
asked Dec 22 '18 at 14:12
PerelMan
519211
Let B be an A-algebra where A is a noetherian ring
Prove that E the set of integral elements over A in B form a subring of B.
What I did:
I tried to prove that E is a subalgebra of B.
I proved by induction that E is a finitely-generated sub-module of B. Does it make it a sub-algebra of B ? on which conditions?
abstract-algebra
abstract-algebra
asked Dec 22 '18 at 14:12
PerelMan
519211
asked Dec 22 '18 at 14:12
PerelMan
519211
asked Dec 22 '18 at 14:12
PerelMan
519211
asked Dec 22 '18 at 14:12
PerelMan
519211
asked Dec 22 '18 at 14:12
PerelMan
519211
519211
add a comment |
add a comment |
2 Answers
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I'm not sure how you proved that $E$ is a finitely generated $B$-submodule, which it isn't generally; it is generally not a finitely generated $A$-module either.
The condition for an element $bin B$ to be integral over $A$ is that $A[b]$ is a finitely generated $A$-module.
If $b,cin E$, then $c$ is integral over $A[b]$, so $A[b][c]$ is a finitely generated $A[b]$-module; hence also a finitely generated $A$-module (prove it). Since $b+c$ and $bc$ belong to $A[b][c]$, we see that both $A[b+c]$ and $A[bc]$ are finitely generated $A$-modules. Try and see where $A$ being Noetherian is used.
answered Dec 22 '18 at 15:10
egreg
178k1484201
If $A[b+c]$ and $A[bc]$ are finitely generated A-modules, then ab and a+b are integral over A since A is noetherian. Thank you both for your help.
– PerelMan
Dec 22 '18 at 16:16
add a comment |
Not necessarily. Consider for example $B=A[X]$. The submodule generated by $1$ and $X$ is not closed under multiplication and so is not a subalgebra.
answered Dec 22 '18 at 14:47
Henning Makholm
238k16303539
add a comment |
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2 Answers
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2 Answers
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I'm not sure how you proved that $E$ is a finitely generated $B$-submodule, which it isn't generally; it is generally not a finitely generated $A$-module either.
The condition for an element $bin B$ to be integral over $A$ is that $A[b]$ is a finitely generated $A$-module.
If $b,cin E$, then $c$ is integral over $A[b]$, so $A[b][c]$ is a finitely generated $A[b]$-module; hence also a finitely generated $A$-module (prove it). Since $b+c$ and $bc$ belong to $A[b][c]$, we see that both $A[b+c]$ and $A[bc]$ are finitely generated $A$-modules. Try and see where $A$ being Noetherian is used.
answered Dec 22 '18 at 15:10
egreg
178k1484201
If $A[b+c]$ and $A[bc]$ are finitely generated A-modules, then ab and a+b are integral over A since A is noetherian. Thank you both for your help.
– PerelMan
Dec 22 '18 at 16:16
add a comment |
I'm not sure how you proved that $E$ is a finitely generated $B$-submodule, which it isn't generally; it is generally not a finitely generated $A$-module either.
The condition for an element $bin B$ to be integral over $A$ is that $A[b]$ is a finitely generated $A$-module.
If $b,cin E$, then $c$ is integral over $A[b]$, so $A[b][c]$ is a finitely generated $A[b]$-module; hence also a finitely generated $A$-module (prove it). Since $b+c$ and $bc$ belong to $A[b][c]$, we see that both $A[b+c]$ and $A[bc]$ are finitely generated $A$-modules. Try and see where $A$ being Noetherian is used.
answered Dec 22 '18 at 15:10
egreg
178k1484201
If $A[b+c]$ and $A[bc]$ are finitely generated A-modules, then ab and a+b are integral over A since A is noetherian. Thank you both for your help.
– PerelMan
Dec 22 '18 at 16:16
add a comment |
I'm not sure how you proved that $E$ is a finitely generated $B$-submodule, which it isn't generally; it is generally not a finitely generated $A$-module either.
The condition for an element $bin B$ to be integral over $A$ is that $A[b]$ is a finitely generated $A$-module.
If $b,cin E$, then $c$ is integral over $A[b]$, so $A[b][c]$ is a finitely generated $A[b]$-module; hence also a finitely generated $A$-module (prove it). Since $b+c$ and $bc$ belong to $A[b][c]$, we see that both $A[b+c]$ and $A[bc]$ are finitely generated $A$-modules. Try and see where $A$ being Noetherian is used.
answered Dec 22 '18 at 15:10
egreg
178k1484201
I'm not sure how you proved that $E$ is a finitely generated $B$-submodule, which it isn't generally; it is generally not a finitely generated $A$-module either.
The condition for an element $bin B$ to be integral over $A$ is that $A[b]$ is a finitely generated $A$-module.
If $b,cin E$, then $c$ is integral over $A[b]$, so $A[b][c]$ is a finitely generated $A[b]$-module; hence also a finitely generated $A$-module (prove it). Since $b+c$ and $bc$ belong to $A[b][c]$, we see that both $A[b+c]$ and $A[bc]$ are finitely generated $A$-modules. Try and see where $A$ being Noetherian is used.
answered Dec 22 '18 at 15:10
egreg
178k1484201
answered Dec 22 '18 at 15:10
egreg
178k1484201
answered Dec 22 '18 at 15:10
egreg
178k1484201
answered Dec 22 '18 at 15:10
egreg
178k1484201
178k1484201
If $A[b+c]$ and $A[bc]$ are finitely generated A-modules, then ab and a+b are integral over A since A is noetherian. Thank you both for your help.
– PerelMan
Dec 22 '18 at 16:16
add a comment |
If $A[b+c]$ and $A[bc]$ are finitely generated A-modules, then ab and a+b are integral over A since A is noetherian. Thank you both for your help.
– PerelMan
Dec 22 '18 at 16:16
If $A[b+c]$ and $A[bc]$ are finitely generated A-modules, then ab and a+b are integral over A since A is noetherian. Thank you both for your help.
– PerelMan
Dec 22 '18 at 16:16
If $A[b+c]$ and $A[bc]$ are finitely generated A-modules, then ab and a+b are integral over A since A is noetherian. Thank you both for your help.
– PerelMan
Dec 22 '18 at 16:16
add a comment |
Not necessarily. Consider for example $B=A[X]$. The submodule generated by $1$ and $X$ is not closed under multiplication and so is not a subalgebra.
answered Dec 22 '18 at 14:47
Henning Makholm
238k16303539
add a comment |
Not necessarily. Consider for example $B=A[X]$. The submodule generated by $1$ and $X$ is not closed under multiplication and so is not a subalgebra.
answered Dec 22 '18 at 14:47
Henning Makholm
238k16303539
add a comment |
Not necessarily. Consider for example $B=A[X]$. The submodule generated by $1$ and $X$ is not closed under multiplication and so is not a subalgebra.
answered Dec 22 '18 at 14:47
Henning Makholm
238k16303539
Not necessarily. Consider for example $B=A[X]$. The submodule generated by $1$ and $X$ is not closed under multiplication and so is not a subalgebra.
answered Dec 22 '18 at 14:47
Henning Makholm
238k16303539
answered Dec 22 '18 at 14:47
Henning Makholm
238k16303539
answered Dec 22 '18 at 14:47
Henning Makholm
238k16303539
answered Dec 22 '18 at 14:47
Henning Makholm
238k16303539
238k16303539
add a comment |
add a comment |
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