$mathfrak{m}= (x_1,x_2,x_3,…)$ as an $R$-module ($R= k[x_1,x_2,x_3,…]$, $k$ field)
Let $R= k[x_1,x_2,x_3,...]$, where $k$ is a field.
Consider the $R$-module $mathfrak{m}= (x_1,x_2,x_3,...)$.
I would like to check:
If $mathfrak{m}$ is free $R$-module.
I have seen that it is not a free module because it is not generated by a single element (obviously) and any two non-zero elements in $mathfrak{m}$ are linearly dependent. If $f,g in mathfrak{m}$, then $fg-gf=0$.If $mathfrak{m}$ is projective $R$-module. Any help?
- If $mathfrak{m}$ is injective $R$-module. Any help?
Thank you.
abstract-algebra commutative-algebra modules homological-algebra
edited Dec 28 '18 at 23:17
user26857
39.3k124083
asked Dec 28 '18 at 18:24
idriskameniidriskameni
13312
|
show 2 more comments
Let $R= k[x_1,x_2,x_3,...]$, where $k$ is a field.
Consider the $R$-module $mathfrak{m}= (x_1,x_2,x_3,...)$.
I would like to check:
If $mathfrak{m}$ is free $R$-module.
I have seen that it is not a free module because it is not generated by a single element (obviously) and any two non-zero elements in $mathfrak{m}$ are linearly dependent. If $f,g in mathfrak{m}$, then $fg-gf=0$.If $mathfrak{m}$ is projective $R$-module. Any help?
- If $mathfrak{m}$ is injective $R$-module. Any help?
Thank you.
abstract-algebra commutative-algebra modules homological-algebra
edited Dec 28 '18 at 23:17
user26857
39.3k124083
asked Dec 28 '18 at 18:24
idriskameniidriskameni
13312
1
Do you know the answers for finitely many variables? In that case, a 'compactness' argument should allow you to work out the answer here, I think.
– Pedro Tamaroff♦
Dec 28 '18 at 18:29
I do not know it.
– idriskameni
Dec 28 '18 at 18:47
1
Great, then begin with those. Keyword is "Koszul resolution".
– Pedro Tamaroff♦
Dec 28 '18 at 18:51
1
$R$ is a UFD, and in a UFD an ideal is projective iff it is principal. Moreover, if $mathfrak m$ is injective then it is divisible, which is not the case.
– user26857
Dec 28 '18 at 23:02
@user26857 Perhaps you could turn your comment into an answer? It seems to be the quickest way to go around here, and it would be unpleasant for it to go unnoticed.
– Pedro Tamaroff♦
Dec 28 '18 at 23:49
|
show 2 more comments
Let $R= k[x_1,x_2,x_3,...]$, where $k$ is a field.
Consider the $R$-module $mathfrak{m}= (x_1,x_2,x_3,...)$.
I would like to check:
If $mathfrak{m}$ is free $R$-module.
I have seen that it is not a free module because it is not generated by a single element (obviously) and any two non-zero elements in $mathfrak{m}$ are linearly dependent. If $f,g in mathfrak{m}$, then $fg-gf=0$.If $mathfrak{m}$ is projective $R$-module. Any help?
- If $mathfrak{m}$ is injective $R$-module. Any help?
Thank you.
abstract-algebra commutative-algebra modules homological-algebra
edited Dec 28 '18 at 23:17
user26857
39.3k124083
asked Dec 28 '18 at 18:24
idriskameniidriskameni
13312
Let $R= k[x_1,x_2,x_3,...]$, where $k$ is a field.
Consider the $R$-module $mathfrak{m}= (x_1,x_2,x_3,...)$.
I would like to check:
If $mathfrak{m}$ is free $R$-module.
I have seen that it is not a free module because it is not generated by a single element (obviously) and any two non-zero elements in $mathfrak{m}$ are linearly dependent. If $f,g in mathfrak{m}$, then $fg-gf=0$.If $mathfrak{m}$ is projective $R$-module. Any help?
- If $mathfrak{m}$ is injective $R$-module. Any help?
Thank you.
abstract-algebra commutative-algebra modules homological-algebra
abstract-algebra commutative-algebra modules homological-algebra
edited Dec 28 '18 at 23:17
user26857
39.3k124083
asked Dec 28 '18 at 18:24
idriskameniidriskameni
13312
edited Dec 28 '18 at 23:17
user26857
39.3k124083
asked Dec 28 '18 at 18:24
idriskameniidriskameni
13312
edited Dec 28 '18 at 23:17
user26857
39.3k124083
edited Dec 28 '18 at 23:17
user26857
39.3k124083
edited Dec 28 '18 at 23:17
user26857
39.3k124083
39.3k124083
asked Dec 28 '18 at 18:24
idriskameniidriskameni
13312
asked Dec 28 '18 at 18:24
idriskameniidriskameni
13312
asked Dec 28 '18 at 18:24
idriskameniidriskameni
13312
13312
1
Do you know the answers for finitely many variables? In that case, a 'compactness' argument should allow you to work out the answer here, I think.
– Pedro Tamaroff♦
Dec 28 '18 at 18:29
I do not know it.
– idriskameni
Dec 28 '18 at 18:47
1
Great, then begin with those. Keyword is "Koszul resolution".
– Pedro Tamaroff♦
Dec 28 '18 at 18:51
1
$R$ is a UFD, and in a UFD an ideal is projective iff it is principal. Moreover, if $mathfrak m$ is injective then it is divisible, which is not the case.
– user26857
Dec 28 '18 at 23:02
@user26857 Perhaps you could turn your comment into an answer? It seems to be the quickest way to go around here, and it would be unpleasant for it to go unnoticed.
– Pedro Tamaroff♦
Dec 28 '18 at 23:49
|
show 2 more comments
1
Do you know the answers for finitely many variables? In that case, a 'compactness' argument should allow you to work out the answer here, I think.
– Pedro Tamaroff♦
Dec 28 '18 at 18:29
I do not know it.
– idriskameni
Dec 28 '18 at 18:47
1
Great, then begin with those. Keyword is "Koszul resolution".
– Pedro Tamaroff♦
Dec 28 '18 at 18:51
1
$R$ is a UFD, and in a UFD an ideal is projective iff it is principal. Moreover, if $mathfrak m$ is injective then it is divisible, which is not the case.
– user26857
Dec 28 '18 at 23:02
@user26857 Perhaps you could turn your comment into an answer? It seems to be the quickest way to go around here, and it would be unpleasant for it to go unnoticed.
– Pedro Tamaroff♦
Dec 28 '18 at 23:49
1
1
Do you know the answers for finitely many variables? In that case, a 'compactness' argument should allow you to work out the answer here, I think.
– Pedro Tamaroff♦
Dec 28 '18 at 18:29
Do you know the answers for finitely many variables? In that case, a 'compactness' argument should allow you to work out the answer here, I think.
– Pedro Tamaroff♦
Dec 28 '18 at 18:29
I do not know it.
– idriskameni
Dec 28 '18 at 18:47
I do not know it.
– idriskameni
Dec 28 '18 at 18:47
1
1
Great, then begin with those. Keyword is "Koszul resolution".
– Pedro Tamaroff♦
Dec 28 '18 at 18:51
Great, then begin with those. Keyword is "Koszul resolution".
– Pedro Tamaroff♦
Dec 28 '18 at 18:51
1
1
$R$ is a UFD, and in a UFD an ideal is projective iff it is principal. Moreover, if $mathfrak m$ is injective then it is divisible, which is not the case.
– user26857
Dec 28 '18 at 23:02
$R$ is a UFD, and in a UFD an ideal is projective iff it is principal. Moreover, if $mathfrak m$ is injective then it is divisible, which is not the case.
– user26857
Dec 28 '18 at 23:02
@user26857 Perhaps you could turn your comment into an answer? It seems to be the quickest way to go around here, and it would be unpleasant for it to go unnoticed.
– Pedro Tamaroff♦
Dec 28 '18 at 23:49
@user26857 Perhaps you could turn your comment into an answer? It seems to be the quickest way to go around here, and it would be unpleasant for it to go unnoticed.
– Pedro Tamaroff♦
Dec 28 '18 at 23:49
|
show 2 more comments
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1
Do you know the answers for finitely many variables? In that case, a 'compactness' argument should allow you to work out the answer here, I think.
– Pedro Tamaroff♦
Dec 28 '18 at 18:29
I do not know it.
– idriskameni
Dec 28 '18 at 18:47
1
Great, then begin with those. Keyword is "Koszul resolution".
– Pedro Tamaroff♦
Dec 28 '18 at 18:51
1
$R$ is a UFD, and in a UFD an ideal is projective iff it is principal. Moreover, if $mathfrak m$ is injective then it is divisible, which is not the case.
– user26857
Dec 28 '18 at 23:02
@user26857 Perhaps you could turn your comment into an answer? It seems to be the quickest way to go around here, and it would be unpleasant for it to go unnoticed.
– Pedro Tamaroff♦
Dec 28 '18 at 23:49