Every $mathbb{Z}/6mathbb{Z}$-module is projective
$begingroup$
I have to prove that every $mathbb{Z}/6mathbb{Z}$-module is projective.
I've found already this question Prove that every $mathbb{Z}/6mathbb{Z}$-module is projective and injective. Find a $mathbb{Z}/4mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).
My idea is to use the fact that $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} bigoplus mathbb{Z}/3mathbb{Z}$, these two submodules are projective and they are fields, hence all $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}/3mathbb{Z}$-modules are free. But I don't know if taken any $mathbb{Z}/6mathbb{Z}$-module I can "decompose" it in a sum of a $mathbb{Z}/2mathbb{Z}$-module and a $mathbb{Z}/3mathbb{Z}$-module.
abstract-algebra modules projective-module
edited Jan 1 at 22:46
user26857
39.5k124283
asked Jan 1 at 21:10
user289143user289143
985313
$endgroup$
add a comment |
$begingroup$
I have to prove that every $mathbb{Z}/6mathbb{Z}$-module is projective.
I've found already this question Prove that every $mathbb{Z}/6mathbb{Z}$-module is projective and injective. Find a $mathbb{Z}/4mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).
My idea is to use the fact that $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} bigoplus mathbb{Z}/3mathbb{Z}$, these two submodules are projective and they are fields, hence all $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}/3mathbb{Z}$-modules are free. But I don't know if taken any $mathbb{Z}/6mathbb{Z}$-module I can "decompose" it in a sum of a $mathbb{Z}/2mathbb{Z}$-module and a $mathbb{Z}/3mathbb{Z}$-module.
abstract-algebra modules projective-module
edited Jan 1 at 22:46
user26857
39.5k124283
asked Jan 1 at 21:10
user289143user289143
985313
$endgroup$
add a comment |
$begingroup$
I have to prove that every $mathbb{Z}/6mathbb{Z}$-module is projective.
I've found already this question Prove that every $mathbb{Z}/6mathbb{Z}$-module is projective and injective. Find a $mathbb{Z}/4mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).
My idea is to use the fact that $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} bigoplus mathbb{Z}/3mathbb{Z}$, these two submodules are projective and they are fields, hence all $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}/3mathbb{Z}$-modules are free. But I don't know if taken any $mathbb{Z}/6mathbb{Z}$-module I can "decompose" it in a sum of a $mathbb{Z}/2mathbb{Z}$-module and a $mathbb{Z}/3mathbb{Z}$-module.
abstract-algebra modules projective-module
edited Jan 1 at 22:46
user26857
39.5k124283
asked Jan 1 at 21:10
user289143user289143
985313
$endgroup$
I have to prove that every $mathbb{Z}/6mathbb{Z}$-module is projective.
I've found already this question Prove that every $mathbb{Z}/6mathbb{Z}$-module is projective and injective. Find a $mathbb{Z}/4mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).
My idea is to use the fact that $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} bigoplus mathbb{Z}/3mathbb{Z}$, these two submodules are projective and they are fields, hence all $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}/3mathbb{Z}$-modules are free. But I don't know if taken any $mathbb{Z}/6mathbb{Z}$-module I can "decompose" it in a sum of a $mathbb{Z}/2mathbb{Z}$-module and a $mathbb{Z}/3mathbb{Z}$-module.
abstract-algebra modules projective-module
abstract-algebra modules projective-module
edited Jan 1 at 22:46
user26857
39.5k124283
asked Jan 1 at 21:10
user289143user289143
985313
edited Jan 1 at 22:46
user26857
39.5k124283
asked Jan 1 at 21:10
user289143user289143
985313
edited Jan 1 at 22:46
user26857
39.5k124283
edited Jan 1 at 22:46
user26857
39.5k124283
edited Jan 1 at 22:46
user26857
39.5k124283
39.5k124283
asked Jan 1 at 21:10
user289143user289143
985313
asked Jan 1 at 21:10
user289143user289143
985313
asked Jan 1 at 21:10
user289143user289143
985313
985313
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.
If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
so $2Mcap 3M=0$.
On the other hand, $m=3m-2m$, so $2M+3M=M$.
Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.
Thus the claim you wanted to prove is true.
Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.
answered Jan 1 at 21:40
jgonjgon
16.1k32143
$endgroup$
$begingroup$
@user289143 Yes of course, thanks for catching that :)
$endgroup$
– jgon
Jan 1 at 21:45
add a comment |
$begingroup$
Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.
answered Jan 1 at 23:19
BenBen
4,313617
$endgroup$
1
$begingroup$
“the proof should be based just on the equivalent definitions of projective modules”
$endgroup$
– egreg
Jan 1 at 23:26
$begingroup$
@egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
$endgroup$
– Pedro Tamaroff♦
Jan 1 at 23:39
$begingroup$
Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
$endgroup$
– jgon
Jan 2 at 3:07
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.
If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
so $2Mcap 3M=0$.
On the other hand, $m=3m-2m$, so $2M+3M=M$.
Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.
Thus the claim you wanted to prove is true.
Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.
answered Jan 1 at 21:40
jgonjgon
16.1k32143
$endgroup$
$begingroup$
@user289143 Yes of course, thanks for catching that :)
$endgroup$
– jgon
Jan 1 at 21:45
add a comment |
$begingroup$
Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.
If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
so $2Mcap 3M=0$.
On the other hand, $m=3m-2m$, so $2M+3M=M$.
Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.
Thus the claim you wanted to prove is true.
Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.
answered Jan 1 at 21:40
jgonjgon
16.1k32143
$endgroup$
$begingroup$
@user289143 Yes of course, thanks for catching that :)
$endgroup$
– jgon
Jan 1 at 21:45
add a comment |
$begingroup$
Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.
If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
so $2Mcap 3M=0$.
On the other hand, $m=3m-2m$, so $2M+3M=M$.
Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.
Thus the claim you wanted to prove is true.
Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.
answered Jan 1 at 21:40
jgonjgon
16.1k32143
$endgroup$
Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.
If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
so $2Mcap 3M=0$.
On the other hand, $m=3m-2m$, so $2M+3M=M$.
Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.
Thus the claim you wanted to prove is true.
Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.
answered Jan 1 at 21:40
jgonjgon
16.1k32143
edited Jan 1 at 21:45
answered Jan 1 at 21:40
jgonjgon
16.1k32143
answered Jan 1 at 21:40
jgonjgon
16.1k32143
answered Jan 1 at 21:40
jgonjgon
16.1k32143
16.1k32143
$begingroup$
@user289143 Yes of course, thanks for catching that :)
$endgroup$
– jgon
Jan 1 at 21:45
add a comment |
$begingroup$
@user289143 Yes of course, thanks for catching that :)
$endgroup$
– jgon
Jan 1 at 21:45
$begingroup$
@user289143 Yes of course, thanks for catching that :)
$endgroup$
– jgon
Jan 1 at 21:45
$begingroup$
@user289143 Yes of course, thanks for catching that :)
$endgroup$
– jgon
Jan 1 at 21:45
add a comment |
$begingroup$
Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.
answered Jan 1 at 23:19
BenBen
4,313617
$endgroup$
1
$begingroup$
“the proof should be based just on the equivalent definitions of projective modules”
$endgroup$
– egreg
Jan 1 at 23:26
$begingroup$
@egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
$endgroup$
– Pedro Tamaroff♦
Jan 1 at 23:39
$begingroup$
Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
$endgroup$
– jgon
Jan 2 at 3:07
add a comment |
$begingroup$
Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.
answered Jan 1 at 23:19
BenBen
4,313617
$endgroup$
1
$begingroup$
“the proof should be based just on the equivalent definitions of projective modules”
$endgroup$
– egreg
Jan 1 at 23:26
$begingroup$
@egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
$endgroup$
– Pedro Tamaroff♦
Jan 1 at 23:39
$begingroup$
Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
$endgroup$
– jgon
Jan 2 at 3:07
add a comment |
$begingroup$
Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.
answered Jan 1 at 23:19
BenBen
4,313617
$endgroup$
Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.
answered Jan 1 at 23:19
BenBen
4,313617
answered Jan 1 at 23:19
BenBen
4,313617
answered Jan 1 at 23:19
BenBen
4,313617
answered Jan 1 at 23:19
BenBen
4,313617
4,313617
1
$begingroup$
“the proof should be based just on the equivalent definitions of projective modules”
$endgroup$
– egreg
Jan 1 at 23:26
$begingroup$
@egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
$endgroup$
– Pedro Tamaroff♦
Jan 1 at 23:39
$begingroup$
Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
$endgroup$
– jgon
Jan 2 at 3:07
add a comment |
1
$begingroup$
“the proof should be based just on the equivalent definitions of projective modules”
$endgroup$
– egreg
Jan 1 at 23:26
$begingroup$
@egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
$endgroup$
– Pedro Tamaroff♦
Jan 1 at 23:39
$begingroup$
Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
$endgroup$
– jgon
Jan 2 at 3:07
1
1
$begingroup$
“the proof should be based just on the equivalent definitions of projective modules”
$endgroup$
– egreg
Jan 1 at 23:26
$begingroup$
“the proof should be based just on the equivalent definitions of projective modules”
$endgroup$
– egreg
Jan 1 at 23:26
$begingroup$
@egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
$endgroup$
– Pedro Tamaroff♦
Jan 1 at 23:39
$begingroup$
@egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
$endgroup$
– Pedro Tamaroff♦
Jan 1 at 23:39
$begingroup$
Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
$endgroup$
– jgon
Jan 2 at 3:07
$begingroup$
Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
$endgroup$
– jgon
Jan 2 at 3:07
add a comment |
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