Principal ideal and free module
$begingroup$
Let $R$ be a commutative ring and $I$ be an ideal of $R$.
Is it true that $I$ is a principal ideal if and only if $I$ is a free $R$-module?
abstract-algebra commutative-algebra ring-theory
edited Jun 18 '13 at 14:25
Gil
633514
asked Jun 18 '13 at 14:08
thatha
18616
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring and $I$ be an ideal of $R$.
Is it true that $I$ is a principal ideal if and only if $I$ is a free $R$-module?
abstract-algebra commutative-algebra ring-theory
edited Jun 18 '13 at 14:25
Gil
633514
asked Jun 18 '13 at 14:08
thatha
18616
$endgroup$
1
$begingroup$
I'm not sure about the context the question came from. I get the idea that it might have been intended to be a fairly simple question, but even then, it really ought to explicitly exclude the zero ideal.
$endgroup$
– rschwieb
Jun 18 '13 at 14:12
add a comment |
$begingroup$
Let $R$ be a commutative ring and $I$ be an ideal of $R$.
Is it true that $I$ is a principal ideal if and only if $I$ is a free $R$-module?
abstract-algebra commutative-algebra ring-theory
edited Jun 18 '13 at 14:25
Gil
633514
asked Jun 18 '13 at 14:08
thatha
18616
$endgroup$
Let $R$ be a commutative ring and $I$ be an ideal of $R$.
Is it true that $I$ is a principal ideal if and only if $I$ is a free $R$-module?
abstract-algebra commutative-algebra ring-theory
abstract-algebra commutative-algebra ring-theory
edited Jun 18 '13 at 14:25
Gil
633514
asked Jun 18 '13 at 14:08
thatha
18616
edited Jun 18 '13 at 14:25
Gil
633514
asked Jun 18 '13 at 14:08
thatha
18616
edited Jun 18 '13 at 14:25
Gil
633514
edited Jun 18 '13 at 14:25
Gil
633514
edited Jun 18 '13 at 14:25
Gil
633514
633514
asked Jun 18 '13 at 14:08
thatha
18616
asked Jun 18 '13 at 14:08
thatha
18616
asked Jun 18 '13 at 14:08
thatha
18616
18616
1
$begingroup$
I'm not sure about the context the question came from. I get the idea that it might have been intended to be a fairly simple question, but even then, it really ought to explicitly exclude the zero ideal.
$endgroup$
– rschwieb
Jun 18 '13 at 14:12
add a comment |
1
$begingroup$
I'm not sure about the context the question came from. I get the idea that it might have been intended to be a fairly simple question, but even then, it really ought to explicitly exclude the zero ideal.
$endgroup$
– rschwieb
Jun 18 '13 at 14:12
1
1
$begingroup$
I'm not sure about the context the question came from. I get the idea that it might have been intended to be a fairly simple question, but even then, it really ought to explicitly exclude the zero ideal.
$endgroup$
– rschwieb
Jun 18 '13 at 14:12
$begingroup$
I'm not sure about the context the question came from. I get the idea that it might have been intended to be a fairly simple question, but even then, it really ought to explicitly exclude the zero ideal.
$endgroup$
– rschwieb
Jun 18 '13 at 14:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Definitely not. Any proper principal ideal in a finite commutative ring is a counterexample.
On the other hand, a commutative ring is a principal ideal domain if and only if all of its nonzero ideals are free modules with unique rank. This is a result on "free ideal rings" (FIRs) studied by P.M. Cohn.
As you mentioned, it is easy to show that every principal ideal of a domain is isomorphic to $R$. (One way is to notice that $xRcong R/ann(x)$, and $ann(x)=0$ if $x$ is nonzero.)
Now suppose $Jneq 0$ is a free principal ideal of a commutative domain. Then $Jcong oplus_{iin I} R$ for some copies of $R$. In particular, $J$ this says (through the isomorphism) that $J$ has submodules corresponding to the copies of $R$, and so they are also ideals of $R$.
Suppose for a moment more than one copy of $R$ is used. Since the sum is direct, these copies have pairwise intersection zero. However, nonzero ideals of a domain always have nonzero intersection! To avoid this contradiction the sum can only have one term, hence $Jcong R$. Being isomorphic to a cyclic module, $J$ is itself cyclic (so it is a principal ideal).
answered Jun 18 '13 at 14:10
rschwiebrschwieb
106k12102250
$endgroup$
$begingroup$
Thank you very much.I could prove that if R is a domain,any principal ideal is isomorhphic to R.But I couldn't prove that if I is a free R module ,then I is a principai ideal .Please give me an advice.....
$endgroup$
– tha
Jun 18 '13 at 14:45
1
$begingroup$
@smith OK, I added some more to the solution for your comment's question.
$endgroup$
– rschwieb
Jun 18 '13 at 16:48
add a comment |
$begingroup$
It is true that, if $R$ is a commutative ring and $I$ is an ideal of $R$, then $I$ is free iff $I$ is principal and generated by a non-zerodivisor.
Proof: Say $I$ is free. By way of contradiction, suppose $I$ has an $R$-basis containing more than one element. Let $e_1$ and $e_2$ be distinct elements of this basis. Then we have $e_2e_1-e_1e_2=0$, which is impossible, since the $e_i$ are linearly independent (this is where we use commutativity). Thus, $I$ has finite rank, and its rank is $1$. Say $I$ is generated over $R$ by $e in R$; notice that $e$ must be a non-zerodivisor, or else ${e}$ would not be linearly independent.
Conversely, suppose $I$ is a principal ideal generated by a non-zerodivisor $e$. Then the map $R to I$ given by $r mapsto re$ is an isomorphism of $R$-modules.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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oldest
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$begingroup$
Definitely not. Any proper principal ideal in a finite commutative ring is a counterexample.
On the other hand, a commutative ring is a principal ideal domain if and only if all of its nonzero ideals are free modules with unique rank. This is a result on "free ideal rings" (FIRs) studied by P.M. Cohn.
As you mentioned, it is easy to show that every principal ideal of a domain is isomorphic to $R$. (One way is to notice that $xRcong R/ann(x)$, and $ann(x)=0$ if $x$ is nonzero.)
Now suppose $Jneq 0$ is a free principal ideal of a commutative domain. Then $Jcong oplus_{iin I} R$ for some copies of $R$. In particular, $J$ this says (through the isomorphism) that $J$ has submodules corresponding to the copies of $R$, and so they are also ideals of $R$.
Suppose for a moment more than one copy of $R$ is used. Since the sum is direct, these copies have pairwise intersection zero. However, nonzero ideals of a domain always have nonzero intersection! To avoid this contradiction the sum can only have one term, hence $Jcong R$. Being isomorphic to a cyclic module, $J$ is itself cyclic (so it is a principal ideal).
answered Jun 18 '13 at 14:10
rschwiebrschwieb
106k12102250
$endgroup$
$begingroup$
Thank you very much.I could prove that if R is a domain,any principal ideal is isomorhphic to R.But I couldn't prove that if I is a free R module ,then I is a principai ideal .Please give me an advice.....
$endgroup$
– tha
Jun 18 '13 at 14:45
1
$begingroup$
@smith OK, I added some more to the solution for your comment's question.
$endgroup$
– rschwieb
Jun 18 '13 at 16:48
add a comment |
$begingroup$
Definitely not. Any proper principal ideal in a finite commutative ring is a counterexample.
On the other hand, a commutative ring is a principal ideal domain if and only if all of its nonzero ideals are free modules with unique rank. This is a result on "free ideal rings" (FIRs) studied by P.M. Cohn.
As you mentioned, it is easy to show that every principal ideal of a domain is isomorphic to $R$. (One way is to notice that $xRcong R/ann(x)$, and $ann(x)=0$ if $x$ is nonzero.)
Now suppose $Jneq 0$ is a free principal ideal of a commutative domain. Then $Jcong oplus_{iin I} R$ for some copies of $R$. In particular, $J$ this says (through the isomorphism) that $J$ has submodules corresponding to the copies of $R$, and so they are also ideals of $R$.
Suppose for a moment more than one copy of $R$ is used. Since the sum is direct, these copies have pairwise intersection zero. However, nonzero ideals of a domain always have nonzero intersection! To avoid this contradiction the sum can only have one term, hence $Jcong R$. Being isomorphic to a cyclic module, $J$ is itself cyclic (so it is a principal ideal).
answered Jun 18 '13 at 14:10
rschwiebrschwieb
106k12102250
$endgroup$
$begingroup$
Thank you very much.I could prove that if R is a domain,any principal ideal is isomorhphic to R.But I couldn't prove that if I is a free R module ,then I is a principai ideal .Please give me an advice.....
$endgroup$
– tha
Jun 18 '13 at 14:45
1
$begingroup$
@smith OK, I added some more to the solution for your comment's question.
$endgroup$
– rschwieb
Jun 18 '13 at 16:48
add a comment |
$begingroup$
Definitely not. Any proper principal ideal in a finite commutative ring is a counterexample.
On the other hand, a commutative ring is a principal ideal domain if and only if all of its nonzero ideals are free modules with unique rank. This is a result on "free ideal rings" (FIRs) studied by P.M. Cohn.
As you mentioned, it is easy to show that every principal ideal of a domain is isomorphic to $R$. (One way is to notice that $xRcong R/ann(x)$, and $ann(x)=0$ if $x$ is nonzero.)
Now suppose $Jneq 0$ is a free principal ideal of a commutative domain. Then $Jcong oplus_{iin I} R$ for some copies of $R$. In particular, $J$ this says (through the isomorphism) that $J$ has submodules corresponding to the copies of $R$, and so they are also ideals of $R$.
Suppose for a moment more than one copy of $R$ is used. Since the sum is direct, these copies have pairwise intersection zero. However, nonzero ideals of a domain always have nonzero intersection! To avoid this contradiction the sum can only have one term, hence $Jcong R$. Being isomorphic to a cyclic module, $J$ is itself cyclic (so it is a principal ideal).
answered Jun 18 '13 at 14:10
rschwiebrschwieb
106k12102250
$endgroup$
Definitely not. Any proper principal ideal in a finite commutative ring is a counterexample.
On the other hand, a commutative ring is a principal ideal domain if and only if all of its nonzero ideals are free modules with unique rank. This is a result on "free ideal rings" (FIRs) studied by P.M. Cohn.
As you mentioned, it is easy to show that every principal ideal of a domain is isomorphic to $R$. (One way is to notice that $xRcong R/ann(x)$, and $ann(x)=0$ if $x$ is nonzero.)
Now suppose $Jneq 0$ is a free principal ideal of a commutative domain. Then $Jcong oplus_{iin I} R$ for some copies of $R$. In particular, $J$ this says (through the isomorphism) that $J$ has submodules corresponding to the copies of $R$, and so they are also ideals of $R$.
Suppose for a moment more than one copy of $R$ is used. Since the sum is direct, these copies have pairwise intersection zero. However, nonzero ideals of a domain always have nonzero intersection! To avoid this contradiction the sum can only have one term, hence $Jcong R$. Being isomorphic to a cyclic module, $J$ is itself cyclic (so it is a principal ideal).
answered Jun 18 '13 at 14:10
rschwiebrschwieb
106k12102250
edited Jun 18 '13 at 16:52
answered Jun 18 '13 at 14:10
rschwiebrschwieb
106k12102250
answered Jun 18 '13 at 14:10
rschwiebrschwieb
106k12102250
answered Jun 18 '13 at 14:10
rschwiebrschwieb
106k12102250
106k12102250
$begingroup$
Thank you very much.I could prove that if R is a domain,any principal ideal is isomorhphic to R.But I couldn't prove that if I is a free R module ,then I is a principai ideal .Please give me an advice.....
$endgroup$
– tha
Jun 18 '13 at 14:45
1
$begingroup$
@smith OK, I added some more to the solution for your comment's question.
$endgroup$
– rschwieb
Jun 18 '13 at 16:48
add a comment |
$begingroup$
Thank you very much.I could prove that if R is a domain,any principal ideal is isomorhphic to R.But I couldn't prove that if I is a free R module ,then I is a principai ideal .Please give me an advice.....
$endgroup$
– tha
Jun 18 '13 at 14:45
1
$begingroup$
@smith OK, I added some more to the solution for your comment's question.
$endgroup$
– rschwieb
Jun 18 '13 at 16:48
$begingroup$
Thank you very much.I could prove that if R is a domain,any principal ideal is isomorhphic to R.But I couldn't prove that if I is a free R module ,then I is a principai ideal .Please give me an advice.....
$endgroup$
– tha
Jun 18 '13 at 14:45
$begingroup$
Thank you very much.I could prove that if R is a domain,any principal ideal is isomorhphic to R.But I couldn't prove that if I is a free R module ,then I is a principai ideal .Please give me an advice.....
$endgroup$
– tha
Jun 18 '13 at 14:45
1
1
$begingroup$
@smith OK, I added some more to the solution for your comment's question.
$endgroup$
– rschwieb
Jun 18 '13 at 16:48
$begingroup$
@smith OK, I added some more to the solution for your comment's question.
$endgroup$
– rschwieb
Jun 18 '13 at 16:48
add a comment |
$begingroup$
It is true that, if $R$ is a commutative ring and $I$ is an ideal of $R$, then $I$ is free iff $I$ is principal and generated by a non-zerodivisor.
Proof: Say $I$ is free. By way of contradiction, suppose $I$ has an $R$-basis containing more than one element. Let $e_1$ and $e_2$ be distinct elements of this basis. Then we have $e_2e_1-e_1e_2=0$, which is impossible, since the $e_i$ are linearly independent (this is where we use commutativity). Thus, $I$ has finite rank, and its rank is $1$. Say $I$ is generated over $R$ by $e in R$; notice that $e$ must be a non-zerodivisor, or else ${e}$ would not be linearly independent.
Conversely, suppose $I$ is a principal ideal generated by a non-zerodivisor $e$. Then the map $R to I$ given by $r mapsto re$ is an isomorphism of $R$-modules.
$endgroup$
add a comment |
$begingroup$
It is true that, if $R$ is a commutative ring and $I$ is an ideal of $R$, then $I$ is free iff $I$ is principal and generated by a non-zerodivisor.
Proof: Say $I$ is free. By way of contradiction, suppose $I$ has an $R$-basis containing more than one element. Let $e_1$ and $e_2$ be distinct elements of this basis. Then we have $e_2e_1-e_1e_2=0$, which is impossible, since the $e_i$ are linearly independent (this is where we use commutativity). Thus, $I$ has finite rank, and its rank is $1$. Say $I$ is generated over $R$ by $e in R$; notice that $e$ must be a non-zerodivisor, or else ${e}$ would not be linearly independent.
Conversely, suppose $I$ is a principal ideal generated by a non-zerodivisor $e$. Then the map $R to I$ given by $r mapsto re$ is an isomorphism of $R$-modules.
$endgroup$
add a comment |
$begingroup$
It is true that, if $R$ is a commutative ring and $I$ is an ideal of $R$, then $I$ is free iff $I$ is principal and generated by a non-zerodivisor.
Proof: Say $I$ is free. By way of contradiction, suppose $I$ has an $R$-basis containing more than one element. Let $e_1$ and $e_2$ be distinct elements of this basis. Then we have $e_2e_1-e_1e_2=0$, which is impossible, since the $e_i$ are linearly independent (this is where we use commutativity). Thus, $I$ has finite rank, and its rank is $1$. Say $I$ is generated over $R$ by $e in R$; notice that $e$ must be a non-zerodivisor, or else ${e}$ would not be linearly independent.
Conversely, suppose $I$ is a principal ideal generated by a non-zerodivisor $e$. Then the map $R to I$ given by $r mapsto re$ is an isomorphism of $R$-modules.
$endgroup$
It is true that, if $R$ is a commutative ring and $I$ is an ideal of $R$, then $I$ is free iff $I$ is principal and generated by a non-zerodivisor.
Proof: Say $I$ is free. By way of contradiction, suppose $I$ has an $R$-basis containing more than one element. Let $e_1$ and $e_2$ be distinct elements of this basis. Then we have $e_2e_1-e_1e_2=0$, which is impossible, since the $e_i$ are linearly independent (this is where we use commutativity). Thus, $I$ has finite rank, and its rank is $1$. Say $I$ is generated over $R$ by $e in R$; notice that $e$ must be a non-zerodivisor, or else ${e}$ would not be linearly independent.
Conversely, suppose $I$ is a principal ideal generated by a non-zerodivisor $e$. Then the map $R to I$ given by $r mapsto re$ is an isomorphism of $R$-modules.
edited Jun 18 '13 at 20:33
answered Jun 18 '13 at 14:53
user55407
add a comment |
add a comment |
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$begingroup$
I'm not sure about the context the question came from. I get the idea that it might have been intended to be a fairly simple question, but even then, it really ought to explicitly exclude the zero ideal.
$endgroup$
– rschwieb
Jun 18 '13 at 14:12