Why is the decomposition of $operatorname{Tor}(M)$ into cyclic modules a “subset” of the decomposition of...
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In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then
$$M cong R^r oplus R/(a_1) oplus cdots oplus R/(a_n)$$
and that
$$operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_n)$$
I don't understand how the second part follows from the first. First, I don't see why $operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.
Secondly, if we assume that $operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get
$$operatorname{Tor}(M) cong R^k oplus R/(b_1) oplus cdots oplus R/(b_m)$$
I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) oplus cdots oplus R/(b_m)$ part has to be the same as the $R/(a_1) oplus cdots oplus R/(a_n)$ part.
abstract-algebra modules finitely-generated
edited Nov 18 at 15:14
Bernard
117k637109
asked Nov 18 at 14:32
Ovi
12.1k938108
add a comment |
up vote
1
down vote
favorite
In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then
$$M cong R^r oplus R/(a_1) oplus cdots oplus R/(a_n)$$
and that
$$operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_n)$$
I don't understand how the second part follows from the first. First, I don't see why $operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.
Secondly, if we assume that $operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get
$$operatorname{Tor}(M) cong R^k oplus R/(b_1) oplus cdots oplus R/(b_m)$$
I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) oplus cdots oplus R/(b_m)$ part has to be the same as the $R/(a_1) oplus cdots oplus R/(a_n)$ part.
abstract-algebra modules finitely-generated
edited Nov 18 at 15:14
Bernard
117k637109
asked Nov 18 at 14:32
Ovi
12.1k938108
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then
$$M cong R^r oplus R/(a_1) oplus cdots oplus R/(a_n)$$
and that
$$operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_n)$$
I don't understand how the second part follows from the first. First, I don't see why $operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.
Secondly, if we assume that $operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get
$$operatorname{Tor}(M) cong R^k oplus R/(b_1) oplus cdots oplus R/(b_m)$$
I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) oplus cdots oplus R/(b_m)$ part has to be the same as the $R/(a_1) oplus cdots oplus R/(a_n)$ part.
abstract-algebra modules finitely-generated
edited Nov 18 at 15:14
Bernard
117k637109
asked Nov 18 at 14:32
Ovi
12.1k938108
In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then
$$M cong R^r oplus R/(a_1) oplus cdots oplus R/(a_n)$$
and that
$$operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_n)$$
I don't understand how the second part follows from the first. First, I don't see why $operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.
Secondly, if we assume that $operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get
$$operatorname{Tor}(M) cong R^k oplus R/(b_1) oplus cdots oplus R/(b_m)$$
I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) oplus cdots oplus R/(b_m)$ part has to be the same as the $R/(a_1) oplus cdots oplus R/(a_n)$ part.
abstract-algebra modules finitely-generated
abstract-algebra modules finitely-generated
edited Nov 18 at 15:14
Bernard
117k637109
asked Nov 18 at 14:32
Ovi
12.1k938108
edited Nov 18 at 15:14
Bernard
117k637109
asked Nov 18 at 14:32
Ovi
12.1k938108
edited Nov 18 at 15:14
Bernard
117k637109
edited Nov 18 at 15:14
Bernard
117k637109
edited Nov 18 at 15:14
Bernard
117k637109
117k637109
asked Nov 18 at 14:32
Ovi
12.1k938108
asked Nov 18 at 14:32
Ovi
12.1k938108
asked Nov 18 at 14:32
Ovi
12.1k938108
12.1k938108
add a comment |
add a comment |
2 Answers
2
active
oldest
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up vote
1
down vote
accepted
Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.
answered Nov 25 at 4:20
Ovi
12.1k938108
add a comment |
up vote
1
down vote
- A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.
- There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.
answered Nov 18 at 14:40
Bernard
117k637109
Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
– Ovi
Nov 18 at 14:46
But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
– Bernard
Nov 18 at 15:26
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.
answered Nov 25 at 4:20
Ovi
12.1k938108
add a comment |
up vote
1
down vote
accepted
Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.
answered Nov 25 at 4:20
Ovi
12.1k938108
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.
answered Nov 25 at 4:20
Ovi
12.1k938108
Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.
answered Nov 25 at 4:20
Ovi
12.1k938108
answered Nov 25 at 4:20
Ovi
12.1k938108
answered Nov 25 at 4:20
Ovi
12.1k938108
answered Nov 25 at 4:20
Ovi
12.1k938108
12.1k938108
add a comment |
add a comment |
up vote
1
down vote
- A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.
- There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.
answered Nov 18 at 14:40
Bernard
117k637109
Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
– Ovi
Nov 18 at 14:46
But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
– Bernard
Nov 18 at 15:26
add a comment |
up vote
1
down vote
- A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.
- There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.
answered Nov 18 at 14:40
Bernard
117k637109
Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
– Ovi
Nov 18 at 14:46
But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
– Bernard
Nov 18 at 15:26
add a comment |
up vote
1
down vote
up vote
1
down vote
- A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.
- There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.
answered Nov 18 at 14:40
Bernard
117k637109
- A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.
- There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.
answered Nov 18 at 14:40
Bernard
117k637109
edited Nov 25 at 11:07
answered Nov 18 at 14:40
Bernard
117k637109
answered Nov 18 at 14:40
Bernard
117k637109
answered Nov 18 at 14:40
Bernard
117k637109
117k637109
Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
– Ovi
Nov 18 at 14:46
But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
– Bernard
Nov 18 at 15:26
add a comment |
Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
– Ovi
Nov 18 at 14:46
But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
– Bernard
Nov 18 at 15:26
Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
– Ovi
Nov 18 at 14:46
Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
– Ovi
Nov 18 at 14:46
But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
– Bernard
Nov 18 at 15:26
But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
– Bernard
Nov 18 at 15:26
add a comment |
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