$M$ is free $R$-module $iff$ $M$ has $R$-basis
We will define the free $R$-modules.
Definition. Let $R$ be a ring with $1_R$ and $F$ an left $R$-module. We call $F$ free $R$-module, if $$F=bigoplus_{iin I}
R_i$$ where $R_i:=langle b_i rangle cong _RR, forall i in I $
and $I$ is a set of indexes (finite or infinite).
We will try to prove the following theorem.
Theorem. Let $R$ be a ring with $1_R$ and $M$ an left $R$-module. The following are equivalent.
$M$ is a left, free $R$-module.
$M$ has basis.
Proof. $1.implies 2.$ According to our definition, $M=bigoplus_{iin I} R_i$, where $R_i:=langle b_i rangle cong _RR, forall i in I $.
and $I$ is a set of indexes (finite or infinite).
We define the set
$$S:={e_i:=(delta_{i,lambda} )_{lambdain I}: iin I}subseteq bigoplus_{iin I}
R,$$
where $delta_{i,lambda} =1_R$, if $lambda=i$ and $delta_{i,lambda} =0_R$ otherwise. Then, it's easy to observe that the set $S$ is $R$-basis for $bigoplus_{iin I} R$ has an $R$-basis and $bigoplus_{iin I} R_i cong bigoplus_{iin I} R = M $, thus $M$ has an $R$-basis.
$2.implies 1.$ We suppose that $S:={e_i in M :i in I } subseteq M$ is an $R$-basis for $M$. Obviously, the key is to define an $R$ module homomorphism
$$phi:Mlongrightarrow bigoplus_{iin I} R.$$
And now my questions.
Questions.
(1) Is the first part okey?
(2) The index set $I$ may be infinite, so does $S$. Then, does every element $min M=langle S rangle$ have a unique expression, as finite linear $R$-combination, in the form $m=sum_{k=1}^{n} r_k e_k$, where $nin Bbb N , r_k in R, e_k in S$?
(3) Which the $R$-module isomorphism?
Thank you.
abstract-algebra modules free-modules
asked Nov 29 '18 at 21:30
ChrisChris
836411
|
show 4 more comments
We will define the free $R$-modules.
Definition. Let $R$ be a ring with $1_R$ and $F$ an left $R$-module. We call $F$ free $R$-module, if $$F=bigoplus_{iin I}
R_i$$ where $R_i:=langle b_i rangle cong _RR, forall i in I $
and $I$ is a set of indexes (finite or infinite).
We will try to prove the following theorem.
Theorem. Let $R$ be a ring with $1_R$ and $M$ an left $R$-module. The following are equivalent.
$M$ is a left, free $R$-module.
$M$ has basis.
Proof. $1.implies 2.$ According to our definition, $M=bigoplus_{iin I} R_i$, where $R_i:=langle b_i rangle cong _RR, forall i in I $.
and $I$ is a set of indexes (finite or infinite).
We define the set
$$S:={e_i:=(delta_{i,lambda} )_{lambdain I}: iin I}subseteq bigoplus_{iin I}
R,$$
where $delta_{i,lambda} =1_R$, if $lambda=i$ and $delta_{i,lambda} =0_R$ otherwise. Then, it's easy to observe that the set $S$ is $R$-basis for $bigoplus_{iin I} R$ has an $R$-basis and $bigoplus_{iin I} R_i cong bigoplus_{iin I} R = M $, thus $M$ has an $R$-basis.
$2.implies 1.$ We suppose that $S:={e_i in M :i in I } subseteq M$ is an $R$-basis for $M$. Obviously, the key is to define an $R$ module homomorphism
$$phi:Mlongrightarrow bigoplus_{iin I} R.$$
And now my questions.
Questions.
(1) Is the first part okey?
(2) The index set $I$ may be infinite, so does $S$. Then, does every element $min M=langle S rangle$ have a unique expression, as finite linear $R$-combination, in the form $m=sum_{k=1}^{n} r_k e_k$, where $nin Bbb N , r_k in R, e_k in S$?
(3) Which the $R$-module isomorphism?
Thank you.
abstract-algebra modules free-modules
asked Nov 29 '18 at 21:30
ChrisChris
836411
1
I'm guessing that the definition is using the internal direct sum (which is a relation between submodules, despite the functional and equality notation) and not the external direct sum which has formal elements $sum_{i in I} lambda_i e_i$.
– Daniel Schepler
Nov 29 '18 at 21:52
@DanielSchepler Yes, indeed. But the two direct sums are in some way equivalent. So, it'doesn't really matter. Do you agree? Note that this definitions is taken form Rotman's book, Advanced Modern Algebra, AMS, 2012.
– Chris
Nov 29 '18 at 21:59
Well, even when you have a canonical isomorphism $alpha$, it's usual to make applications of that isomorphism explicit - so instead of $S := { e_i ldots }$ you would say something like $S := { alpha(e_i) ldots }$. So, under the canonical isomorphism of the external direct sum $bigoplus_{iin I} R$ with $R$ under the hypothesis of $R$ being the internal direct sum of submodules $langle b_i rangle simeq R$, what would $e_i$ be sent to?
– Daniel Schepler
Nov 29 '18 at 22:50
Okay, let's take the external. It's easy then to do the transfromation. But my question is, shall we sent $m:=sum_{i=1}^{n}r_i e_i longmapsto phi(m)= phi (sum_{i=1}^{n}r_i e_i) :=(r_1,...,r_n) $? And should these sums be finite?
– Chris
Nov 29 '18 at 22:58
1
I'd say more specifically, my suggestion is to prove $M = bigoplus_{iin I} langle b_i rangle$ and $langle b_i rangle simeq_R R$ for all $i in I$ $Leftrightarrow$ ${ b_i mid i in i }$ is a basis for $M$. (The part requiring $langle b_i rangle simeq_R R$ is essential to rule out modules like $R / I$ where $I$ is a proper nontrivial ideal.)
– Daniel Schepler
Nov 30 '18 at 0:37
|
show 4 more comments
We will define the free $R$-modules.
Definition. Let $R$ be a ring with $1_R$ and $F$ an left $R$-module. We call $F$ free $R$-module, if $$F=bigoplus_{iin I}
R_i$$ where $R_i:=langle b_i rangle cong _RR, forall i in I $
and $I$ is a set of indexes (finite or infinite).
We will try to prove the following theorem.
Theorem. Let $R$ be a ring with $1_R$ and $M$ an left $R$-module. The following are equivalent.
$M$ is a left, free $R$-module.
$M$ has basis.
Proof. $1.implies 2.$ According to our definition, $M=bigoplus_{iin I} R_i$, where $R_i:=langle b_i rangle cong _RR, forall i in I $.
and $I$ is a set of indexes (finite or infinite).
We define the set
$$S:={e_i:=(delta_{i,lambda} )_{lambdain I}: iin I}subseteq bigoplus_{iin I}
R,$$
where $delta_{i,lambda} =1_R$, if $lambda=i$ and $delta_{i,lambda} =0_R$ otherwise. Then, it's easy to observe that the set $S$ is $R$-basis for $bigoplus_{iin I} R$ has an $R$-basis and $bigoplus_{iin I} R_i cong bigoplus_{iin I} R = M $, thus $M$ has an $R$-basis.
$2.implies 1.$ We suppose that $S:={e_i in M :i in I } subseteq M$ is an $R$-basis for $M$. Obviously, the key is to define an $R$ module homomorphism
$$phi:Mlongrightarrow bigoplus_{iin I} R.$$
And now my questions.
Questions.
(1) Is the first part okey?
(2) The index set $I$ may be infinite, so does $S$. Then, does every element $min M=langle S rangle$ have a unique expression, as finite linear $R$-combination, in the form $m=sum_{k=1}^{n} r_k e_k$, where $nin Bbb N , r_k in R, e_k in S$?
(3) Which the $R$-module isomorphism?
Thank you.
abstract-algebra modules free-modules
asked Nov 29 '18 at 21:30
ChrisChris
836411
We will define the free $R$-modules.
Definition. Let $R$ be a ring with $1_R$ and $F$ an left $R$-module. We call $F$ free $R$-module, if $$F=bigoplus_{iin I}
R_i$$ where $R_i:=langle b_i rangle cong _RR, forall i in I $
and $I$ is a set of indexes (finite or infinite).
We will try to prove the following theorem.
Theorem. Let $R$ be a ring with $1_R$ and $M$ an left $R$-module. The following are equivalent.
$M$ is a left, free $R$-module.
$M$ has basis.
Proof. $1.implies 2.$ According to our definition, $M=bigoplus_{iin I} R_i$, where $R_i:=langle b_i rangle cong _RR, forall i in I $.
and $I$ is a set of indexes (finite or infinite).
We define the set
$$S:={e_i:=(delta_{i,lambda} )_{lambdain I}: iin I}subseteq bigoplus_{iin I}
R,$$
where $delta_{i,lambda} =1_R$, if $lambda=i$ and $delta_{i,lambda} =0_R$ otherwise. Then, it's easy to observe that the set $S$ is $R$-basis for $bigoplus_{iin I} R$ has an $R$-basis and $bigoplus_{iin I} R_i cong bigoplus_{iin I} R = M $, thus $M$ has an $R$-basis.
$2.implies 1.$ We suppose that $S:={e_i in M :i in I } subseteq M$ is an $R$-basis for $M$. Obviously, the key is to define an $R$ module homomorphism
$$phi:Mlongrightarrow bigoplus_{iin I} R.$$
And now my questions.
Questions.
(1) Is the first part okey?
(2) The index set $I$ may be infinite, so does $S$. Then, does every element $min M=langle S rangle$ have a unique expression, as finite linear $R$-combination, in the form $m=sum_{k=1}^{n} r_k e_k$, where $nin Bbb N , r_k in R, e_k in S$?
(3) Which the $R$-module isomorphism?
Thank you.
abstract-algebra modules free-modules
abstract-algebra modules free-modules
asked Nov 29 '18 at 21:30
ChrisChris
836411
asked Nov 29 '18 at 21:30
ChrisChris
836411
edited Nov 29 '18 at 23:04
Chris
asked Nov 29 '18 at 21:30
ChrisChris
836411
asked Nov 29 '18 at 21:30
ChrisChris
836411
asked Nov 29 '18 at 21:30
ChrisChris
836411
836411
1
I'm guessing that the definition is using the internal direct sum (which is a relation between submodules, despite the functional and equality notation) and not the external direct sum which has formal elements $sum_{i in I} lambda_i e_i$.
– Daniel Schepler
Nov 29 '18 at 21:52
@DanielSchepler Yes, indeed. But the two direct sums are in some way equivalent. So, it'doesn't really matter. Do you agree? Note that this definitions is taken form Rotman's book, Advanced Modern Algebra, AMS, 2012.
– Chris
Nov 29 '18 at 21:59
Well, even when you have a canonical isomorphism $alpha$, it's usual to make applications of that isomorphism explicit - so instead of $S := { e_i ldots }$ you would say something like $S := { alpha(e_i) ldots }$. So, under the canonical isomorphism of the external direct sum $bigoplus_{iin I} R$ with $R$ under the hypothesis of $R$ being the internal direct sum of submodules $langle b_i rangle simeq R$, what would $e_i$ be sent to?
– Daniel Schepler
Nov 29 '18 at 22:50
Okay, let's take the external. It's easy then to do the transfromation. But my question is, shall we sent $m:=sum_{i=1}^{n}r_i e_i longmapsto phi(m)= phi (sum_{i=1}^{n}r_i e_i) :=(r_1,...,r_n) $? And should these sums be finite?
– Chris
Nov 29 '18 at 22:58
1
I'd say more specifically, my suggestion is to prove $M = bigoplus_{iin I} langle b_i rangle$ and $langle b_i rangle simeq_R R$ for all $i in I$ $Leftrightarrow$ ${ b_i mid i in i }$ is a basis for $M$. (The part requiring $langle b_i rangle simeq_R R$ is essential to rule out modules like $R / I$ where $I$ is a proper nontrivial ideal.)
– Daniel Schepler
Nov 30 '18 at 0:37
|
show 4 more comments
1
I'm guessing that the definition is using the internal direct sum (which is a relation between submodules, despite the functional and equality notation) and not the external direct sum which has formal elements $sum_{i in I} lambda_i e_i$.
– Daniel Schepler
Nov 29 '18 at 21:52
@DanielSchepler Yes, indeed. But the two direct sums are in some way equivalent. So, it'doesn't really matter. Do you agree? Note that this definitions is taken form Rotman's book, Advanced Modern Algebra, AMS, 2012.
– Chris
Nov 29 '18 at 21:59
Well, even when you have a canonical isomorphism $alpha$, it's usual to make applications of that isomorphism explicit - so instead of $S := { e_i ldots }$ you would say something like $S := { alpha(e_i) ldots }$. So, under the canonical isomorphism of the external direct sum $bigoplus_{iin I} R$ with $R$ under the hypothesis of $R$ being the internal direct sum of submodules $langle b_i rangle simeq R$, what would $e_i$ be sent to?
– Daniel Schepler
Nov 29 '18 at 22:50
Okay, let's take the external. It's easy then to do the transfromation. But my question is, shall we sent $m:=sum_{i=1}^{n}r_i e_i longmapsto phi(m)= phi (sum_{i=1}^{n}r_i e_i) :=(r_1,...,r_n) $? And should these sums be finite?
– Chris
Nov 29 '18 at 22:58
1
I'd say more specifically, my suggestion is to prove $M = bigoplus_{iin I} langle b_i rangle$ and $langle b_i rangle simeq_R R$ for all $i in I$ $Leftrightarrow$ ${ b_i mid i in i }$ is a basis for $M$. (The part requiring $langle b_i rangle simeq_R R$ is essential to rule out modules like $R / I$ where $I$ is a proper nontrivial ideal.)
– Daniel Schepler
Nov 30 '18 at 0:37
1
1
I'm guessing that the definition is using the internal direct sum (which is a relation between submodules, despite the functional and equality notation) and not the external direct sum which has formal elements $sum_{i in I} lambda_i e_i$.
– Daniel Schepler
Nov 29 '18 at 21:52
I'm guessing that the definition is using the internal direct sum (which is a relation between submodules, despite the functional and equality notation) and not the external direct sum which has formal elements $sum_{i in I} lambda_i e_i$.
– Daniel Schepler
Nov 29 '18 at 21:52
@DanielSchepler Yes, indeed. But the two direct sums are in some way equivalent. So, it'doesn't really matter. Do you agree? Note that this definitions is taken form Rotman's book, Advanced Modern Algebra, AMS, 2012.
– Chris
Nov 29 '18 at 21:59
@DanielSchepler Yes, indeed. But the two direct sums are in some way equivalent. So, it'doesn't really matter. Do you agree? Note that this definitions is taken form Rotman's book, Advanced Modern Algebra, AMS, 2012.
– Chris
Nov 29 '18 at 21:59
Well, even when you have a canonical isomorphism $alpha$, it's usual to make applications of that isomorphism explicit - so instead of $S := { e_i ldots }$ you would say something like $S := { alpha(e_i) ldots }$. So, under the canonical isomorphism of the external direct sum $bigoplus_{iin I} R$ with $R$ under the hypothesis of $R$ being the internal direct sum of submodules $langle b_i rangle simeq R$, what would $e_i$ be sent to?
– Daniel Schepler
Nov 29 '18 at 22:50
Well, even when you have a canonical isomorphism $alpha$, it's usual to make applications of that isomorphism explicit - so instead of $S := { e_i ldots }$ you would say something like $S := { alpha(e_i) ldots }$. So, under the canonical isomorphism of the external direct sum $bigoplus_{iin I} R$ with $R$ under the hypothesis of $R$ being the internal direct sum of submodules $langle b_i rangle simeq R$, what would $e_i$ be sent to?
– Daniel Schepler
Nov 29 '18 at 22:50
Okay, let's take the external. It's easy then to do the transfromation. But my question is, shall we sent $m:=sum_{i=1}^{n}r_i e_i longmapsto phi(m)= phi (sum_{i=1}^{n}r_i e_i) :=(r_1,...,r_n) $? And should these sums be finite?
– Chris
Nov 29 '18 at 22:58
Okay, let's take the external. It's easy then to do the transfromation. But my question is, shall we sent $m:=sum_{i=1}^{n}r_i e_i longmapsto phi(m)= phi (sum_{i=1}^{n}r_i e_i) :=(r_1,...,r_n) $? And should these sums be finite?
– Chris
Nov 29 '18 at 22:58
1
1
I'd say more specifically, my suggestion is to prove $M = bigoplus_{iin I} langle b_i rangle$ and $langle b_i rangle simeq_R R$ for all $i in I$ $Leftrightarrow$ ${ b_i mid i in i }$ is a basis for $M$. (The part requiring $langle b_i rangle simeq_R R$ is essential to rule out modules like $R / I$ where $I$ is a proper nontrivial ideal.)
– Daniel Schepler
Nov 30 '18 at 0:37
I'd say more specifically, my suggestion is to prove $M = bigoplus_{iin I} langle b_i rangle$ and $langle b_i rangle simeq_R R$ for all $i in I$ $Leftrightarrow$ ${ b_i mid i in i }$ is a basis for $M$. (The part requiring $langle b_i rangle simeq_R R$ is essential to rule out modules like $R / I$ where $I$ is a proper nontrivial ideal.)
– Daniel Schepler
Nov 30 '18 at 0:37
|
show 4 more comments
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1
I'm guessing that the definition is using the internal direct sum (which is a relation between submodules, despite the functional and equality notation) and not the external direct sum which has formal elements $sum_{i in I} lambda_i e_i$.
– Daniel Schepler
Nov 29 '18 at 21:52
@DanielSchepler Yes, indeed. But the two direct sums are in some way equivalent. So, it'doesn't really matter. Do you agree? Note that this definitions is taken form Rotman's book, Advanced Modern Algebra, AMS, 2012.
– Chris
Nov 29 '18 at 21:59
Well, even when you have a canonical isomorphism $alpha$, it's usual to make applications of that isomorphism explicit - so instead of $S := { e_i ldots }$ you would say something like $S := { alpha(e_i) ldots }$. So, under the canonical isomorphism of the external direct sum $bigoplus_{iin I} R$ with $R$ under the hypothesis of $R$ being the internal direct sum of submodules $langle b_i rangle simeq R$, what would $e_i$ be sent to?
– Daniel Schepler
Nov 29 '18 at 22:50
Okay, let's take the external. It's easy then to do the transfromation. But my question is, shall we sent $m:=sum_{i=1}^{n}r_i e_i longmapsto phi(m)= phi (sum_{i=1}^{n}r_i e_i) :=(r_1,...,r_n) $? And should these sums be finite?
– Chris
Nov 29 '18 at 22:58
1
I'd say more specifically, my suggestion is to prove $M = bigoplus_{iin I} langle b_i rangle$ and $langle b_i rangle simeq_R R$ for all $i in I$ $Leftrightarrow$ ${ b_i mid i in i }$ is a basis for $M$. (The part requiring $langle b_i rangle simeq_R R$ is essential to rule out modules like $R / I$ where $I$ is a proper nontrivial ideal.)
– Daniel Schepler
Nov 30 '18 at 0:37