Mixing
The set of integral elements form a ring.
$begingroup$
Let $A subset B$ be two rings.
I know that an element $x in B$ is integral over $A$ iff $A[x]$ is contained in a finitely generated $A$-module $T subset B$.
I also know that if $b_1,...,b_n$ are integral over $A$ then $A[b_1,...,b_n]$ is a finitely generated $A$-module.
Why does these imply that the set of elements of $B$ that are integral over $A$ form a ring?
abstract-algebra modules integral-extensions
edited Jan 28 at 22:31
user26857
39.5k124283
asked Jan 28 at 18:49
roi_saumonroi_saumon
63438
$endgroup$
add a comment |
$begingroup$
Let $A subset B$ be two rings.
I know that an element $x in B$ is integral over $A$ iff $A[x]$ is contained in a finitely generated $A$-module $T subset B$.
I also know that if $b_1,...,b_n$ are integral over $A$ then $A[b_1,...,b_n]$ is a finitely generated $A$-module.
Why does these imply that the set of elements of $B$ that are integral over $A$ form a ring?
abstract-algebra modules integral-extensions
edited Jan 28 at 22:31
user26857
39.5k124283
asked Jan 28 at 18:49
roi_saumonroi_saumon
63438
$endgroup$
$begingroup$
See for example this question.
$endgroup$
– Dietrich Burde
Jan 28 at 19:11
add a comment |
$begingroup$
Let $A subset B$ be two rings.
I know that an element $x in B$ is integral over $A$ iff $A[x]$ is contained in a finitely generated $A$-module $T subset B$.
I also know that if $b_1,...,b_n$ are integral over $A$ then $A[b_1,...,b_n]$ is a finitely generated $A$-module.
Why does these imply that the set of elements of $B$ that are integral over $A$ form a ring?
abstract-algebra modules integral-extensions
edited Jan 28 at 22:31
user26857
39.5k124283
asked Jan 28 at 18:49
roi_saumonroi_saumon
63438
$endgroup$
Let $A subset B$ be two rings.
I know that an element $x in B$ is integral over $A$ iff $A[x]$ is contained in a finitely generated $A$-module $T subset B$.
I also know that if $b_1,...,b_n$ are integral over $A$ then $A[b_1,...,b_n]$ is a finitely generated $A$-module.
Why does these imply that the set of elements of $B$ that are integral over $A$ form a ring?
abstract-algebra modules integral-extensions
abstract-algebra modules integral-extensions
edited Jan 28 at 22:31
user26857
39.5k124283
asked Jan 28 at 18:49
roi_saumonroi_saumon
63438
edited Jan 28 at 22:31
user26857
39.5k124283
asked Jan 28 at 18:49
roi_saumonroi_saumon
63438
edited Jan 28 at 22:31
user26857
39.5k124283
edited Jan 28 at 22:31
user26857
39.5k124283
edited Jan 28 at 22:31
user26857
39.5k124283
39.5k124283
asked Jan 28 at 18:49
roi_saumonroi_saumon
63438
asked Jan 28 at 18:49
roi_saumonroi_saumon
63438
asked Jan 28 at 18:49
roi_saumonroi_saumon
63438
63438
$begingroup$
See for example this question.
$endgroup$
– Dietrich Burde
Jan 28 at 19:11
add a comment |
$begingroup$
See for example this question.
$endgroup$
– Dietrich Burde
Jan 28 at 19:11
$begingroup$
See for example this question.
$endgroup$
– Dietrich Burde
Jan 28 at 19:11
$begingroup$
See for example this question.
$endgroup$
– Dietrich Burde
Jan 28 at 19:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $a,b$ be two elements of $B$ integral over A. Your second assertion implies that $ain{}M, bin{}N$ which are both finitelly generated submodules of $B$. Then $MN={sum{}a_ib_j, a_iin{}M,b_jin{}N}$ is a finitelly generated submodule of $B$. Notice that $MN$ contains both $a+b$ and $ab$ and the conclusion follows...
Note that you are taking the $x$-subalgebra generated by $x$ (I mean $A[x]$) but you really need a finitely generated submodule of $A$ (let's say $M$) such that $xMsubset{}M$. Since Dedekind's proof uses this inclusion to work with Cramer's rule it is this essential property that's needed.
answered Jan 28 at 18:58
Μάρκος ΚαραμέρηςΜάρκος Καραμέρης
47610
$endgroup$
$begingroup$
Thanks for the answer. Shouldn't it be simpler than that though?
$endgroup$
– roi_saumon
Jan 29 at 0:27
$begingroup$
@roi_saumon: How much easier could it possibly get?
$endgroup$
– RghtHndSd
Jan 29 at 2:30
$begingroup$
I think it is pretty straightforward to prove that $MN$ is a finitely generated submodule, so not sure what would be an even shorter proof
$endgroup$
– Μάρκος Καραμέρης
Jan 29 at 6:36
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Let $a,b$ be two elements of $B$ integral over A. Your second assertion implies that $ain{}M, bin{}N$ which are both finitelly generated submodules of $B$. Then $MN={sum{}a_ib_j, a_iin{}M,b_jin{}N}$ is a finitelly generated submodule of $B$. Notice that $MN$ contains both $a+b$ and $ab$ and the conclusion follows...
Note that you are taking the $x$-subalgebra generated by $x$ (I mean $A[x]$) but you really need a finitely generated submodule of $A$ (let's say $M$) such that $xMsubset{}M$. Since Dedekind's proof uses this inclusion to work with Cramer's rule it is this essential property that's needed.
answered Jan 28 at 18:58
Μάρκος ΚαραμέρηςΜάρκος Καραμέρης
47610
$endgroup$
$begingroup$
Thanks for the answer. Shouldn't it be simpler than that though?
$endgroup$
– roi_saumon
Jan 29 at 0:27
$begingroup$
@roi_saumon: How much easier could it possibly get?
$endgroup$
– RghtHndSd
Jan 29 at 2:30
$begingroup$
I think it is pretty straightforward to prove that $MN$ is a finitely generated submodule, so not sure what would be an even shorter proof
$endgroup$
– Μάρκος Καραμέρης
Jan 29 at 6:36
add a comment |
$begingroup$
Let $a,b$ be two elements of $B$ integral over A. Your second assertion implies that $ain{}M, bin{}N$ which are both finitelly generated submodules of $B$. Then $MN={sum{}a_ib_j, a_iin{}M,b_jin{}N}$ is a finitelly generated submodule of $B$. Notice that $MN$ contains both $a+b$ and $ab$ and the conclusion follows...
Note that you are taking the $x$-subalgebra generated by $x$ (I mean $A[x]$) but you really need a finitely generated submodule of $A$ (let's say $M$) such that $xMsubset{}M$. Since Dedekind's proof uses this inclusion to work with Cramer's rule it is this essential property that's needed.
answered Jan 28 at 18:58
Μάρκος ΚαραμέρηςΜάρκος Καραμέρης
47610
$endgroup$
$begingroup$
Thanks for the answer. Shouldn't it be simpler than that though?
$endgroup$
– roi_saumon
Jan 29 at 0:27
$begingroup$
@roi_saumon: How much easier could it possibly get?
$endgroup$
– RghtHndSd
Jan 29 at 2:30
$begingroup$
I think it is pretty straightforward to prove that $MN$ is a finitely generated submodule, so not sure what would be an even shorter proof
$endgroup$
– Μάρκος Καραμέρης
Jan 29 at 6:36
add a comment |
$begingroup$
Let $a,b$ be two elements of $B$ integral over A. Your second assertion implies that $ain{}M, bin{}N$ which are both finitelly generated submodules of $B$. Then $MN={sum{}a_ib_j, a_iin{}M,b_jin{}N}$ is a finitelly generated submodule of $B$. Notice that $MN$ contains both $a+b$ and $ab$ and the conclusion follows...
Note that you are taking the $x$-subalgebra generated by $x$ (I mean $A[x]$) but you really need a finitely generated submodule of $A$ (let's say $M$) such that $xMsubset{}M$. Since Dedekind's proof uses this inclusion to work with Cramer's rule it is this essential property that's needed.
answered Jan 28 at 18:58
Μάρκος ΚαραμέρηςΜάρκος Καραμέρης
47610
$endgroup$
Let $a,b$ be two elements of $B$ integral over A. Your second assertion implies that $ain{}M, bin{}N$ which are both finitelly generated submodules of $B$. Then $MN={sum{}a_ib_j, a_iin{}M,b_jin{}N}$ is a finitelly generated submodule of $B$. Notice that $MN$ contains both $a+b$ and $ab$ and the conclusion follows...
Note that you are taking the $x$-subalgebra generated by $x$ (I mean $A[x]$) but you really need a finitely generated submodule of $A$ (let's say $M$) such that $xMsubset{}M$. Since Dedekind's proof uses this inclusion to work with Cramer's rule it is this essential property that's needed.
answered Jan 28 at 18:58
Μάρκος ΚαραμέρηςΜάρκος Καραμέρης
47610
edited Jan 28 at 19:07
answered Jan 28 at 18:58
Μάρκος ΚαραμέρηςΜάρκος Καραμέρης
47610
answered Jan 28 at 18:58
Μάρκος ΚαραμέρηςΜάρκος Καραμέρης
47610
answered Jan 28 at 18:58
Μάρκος ΚαραμέρηςΜάρκος Καραμέρης
47610
47610
$begingroup$
Thanks for the answer. Shouldn't it be simpler than that though?
$endgroup$
– roi_saumon
Jan 29 at 0:27
$begingroup$
@roi_saumon: How much easier could it possibly get?
$endgroup$
– RghtHndSd
Jan 29 at 2:30
$begingroup$
I think it is pretty straightforward to prove that $MN$ is a finitely generated submodule, so not sure what would be an even shorter proof
$endgroup$
– Μάρκος Καραμέρης
Jan 29 at 6:36
add a comment |
$begingroup$
Thanks for the answer. Shouldn't it be simpler than that though?
$endgroup$
– roi_saumon
Jan 29 at 0:27
$begingroup$
@roi_saumon: How much easier could it possibly get?
$endgroup$
– RghtHndSd
Jan 29 at 2:30
$begingroup$
I think it is pretty straightforward to prove that $MN$ is a finitely generated submodule, so not sure what would be an even shorter proof
$endgroup$
– Μάρκος Καραμέρης
Jan 29 at 6:36
$begingroup$
Thanks for the answer. Shouldn't it be simpler than that though?
$endgroup$
– roi_saumon
Jan 29 at 0:27
$begingroup$
Thanks for the answer. Shouldn't it be simpler than that though?
$endgroup$
– roi_saumon
Jan 29 at 0:27
$begingroup$
@roi_saumon: How much easier could it possibly get?
$endgroup$
– RghtHndSd
Jan 29 at 2:30
$begingroup$
@roi_saumon: How much easier could it possibly get?
$endgroup$
– RghtHndSd
Jan 29 at 2:30
$begingroup$
I think it is pretty straightforward to prove that $MN$ is a finitely generated submodule, so not sure what would be an even shorter proof
$endgroup$
– Μάρκος Καραμέρης
Jan 29 at 6:36
$begingroup$
I think it is pretty straightforward to prove that $MN$ is a finitely generated submodule, so not sure what would be an even shorter proof
$endgroup$
– Μάρκος Καραμέρης
Jan 29 at 6:36
add a comment |
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$begingroup$
See for example this question.
$endgroup$
– Dietrich Burde
Jan 28 at 19:11