Mixing
Characterizing commutative infinitely generated algebras having zero divisors
Let $R$ be a commutative $k$-algebra, where $k$ is a field of characteristic zero. Assume that $R$ satisfies the following two conditions:
(i) $R$ is not affine (= $R$ is infinitely generated as a $k$-algebra).
(ii) $R$ is not an integral domain (= $R$ has zero divisors).
After receiving an easy answer to this question, I would like to generalize it to the following question:
Is it possible to characterize all such $R$'s?
If I am not wrong, all such $R$'s are of the following form:
$frac{k[x_i]_{i in mathbb{N}}}{I}$, for an appropriate ideal $I$, for example $I=(x_{17}^2)$.
How to find all such $I$'s?
Is it true that a general such $I$ is an ideal generated by at least one reducible polynomial? (clearly, if $h=h_1h_2 in I$, then $bar{h_1}bar{h_2}=bar{0}$).
Thank you very much!
Edit: Perhaps the power series ring in infinitely many variables (which is an integral domain) quotiented by an appropriate ideal shows that I was wrong, and not all such $R$'s are quotients of polynomial rings in infinitely many variables.
algebraic-geometry commutative-algebra
asked Nov 22 '18 at 1:12
user237522user237522
2,1411617
add a comment |
Let $R$ be a commutative $k$-algebra, where $k$ is a field of characteristic zero. Assume that $R$ satisfies the following two conditions:
(i) $R$ is not affine (= $R$ is infinitely generated as a $k$-algebra).
(ii) $R$ is not an integral domain (= $R$ has zero divisors).
After receiving an easy answer to this question, I would like to generalize it to the following question:
Is it possible to characterize all such $R$'s?
If I am not wrong, all such $R$'s are of the following form:
$frac{k[x_i]_{i in mathbb{N}}}{I}$, for an appropriate ideal $I$, for example $I=(x_{17}^2)$.
How to find all such $I$'s?
Is it true that a general such $I$ is an ideal generated by at least one reducible polynomial? (clearly, if $h=h_1h_2 in I$, then $bar{h_1}bar{h_2}=bar{0}$).
Thank you very much!
Edit: Perhaps the power series ring in infinitely many variables (which is an integral domain) quotiented by an appropriate ideal shows that I was wrong, and not all such $R$'s are quotients of polynomial rings in infinitely many variables.
algebraic-geometry commutative-algebra
asked Nov 22 '18 at 1:12
user237522user237522
2,1411617
Well in your question you only ask for infinitely generated, but all your $R$'s are countably generated, so any uncountably generated ring lies outside the class you suggest.
– Alex J Best
Nov 22 '18 at 1:19
Thanks, you are right. I actually meant to ask about $frac{k[x_i]_{i in A}}{I}$, for an arbitrary set $A$, countably generated or not.
– user237522
Nov 22 '18 at 1:22
@AlexJBest, if my question still interests you, please see my edit (in which I claim that there exist such $R$'s of another form as well).
– user237522
Nov 22 '18 at 2:18
3
Every commutative ring whatsoever is a quotient of a polynomial ring in enough variables (for example, it suffices to take one variable for each element of the ring). But asking for a classification is totally hopeless.
– Qiaochu Yuan
Nov 22 '18 at 2:46
@QiaochuYuan, thank you.
– user237522
Nov 22 '18 at 3:16
add a comment |
Let $R$ be a commutative $k$-algebra, where $k$ is a field of characteristic zero. Assume that $R$ satisfies the following two conditions:
(i) $R$ is not affine (= $R$ is infinitely generated as a $k$-algebra).
(ii) $R$ is not an integral domain (= $R$ has zero divisors).
After receiving an easy answer to this question, I would like to generalize it to the following question:
Is it possible to characterize all such $R$'s?
If I am not wrong, all such $R$'s are of the following form:
$frac{k[x_i]_{i in mathbb{N}}}{I}$, for an appropriate ideal $I$, for example $I=(x_{17}^2)$.
How to find all such $I$'s?
Is it true that a general such $I$ is an ideal generated by at least one reducible polynomial? (clearly, if $h=h_1h_2 in I$, then $bar{h_1}bar{h_2}=bar{0}$).
Thank you very much!
Edit: Perhaps the power series ring in infinitely many variables (which is an integral domain) quotiented by an appropriate ideal shows that I was wrong, and not all such $R$'s are quotients of polynomial rings in infinitely many variables.
algebraic-geometry commutative-algebra
asked Nov 22 '18 at 1:12
user237522user237522
2,1411617
Let $R$ be a commutative $k$-algebra, where $k$ is a field of characteristic zero. Assume that $R$ satisfies the following two conditions:
(i) $R$ is not affine (= $R$ is infinitely generated as a $k$-algebra).
(ii) $R$ is not an integral domain (= $R$ has zero divisors).
After receiving an easy answer to this question, I would like to generalize it to the following question:
Is it possible to characterize all such $R$'s?
If I am not wrong, all such $R$'s are of the following form:
$frac{k[x_i]_{i in mathbb{N}}}{I}$, for an appropriate ideal $I$, for example $I=(x_{17}^2)$.
How to find all such $I$'s?
Is it true that a general such $I$ is an ideal generated by at least one reducible polynomial? (clearly, if $h=h_1h_2 in I$, then $bar{h_1}bar{h_2}=bar{0}$).
Thank you very much!
Edit: Perhaps the power series ring in infinitely many variables (which is an integral domain) quotiented by an appropriate ideal shows that I was wrong, and not all such $R$'s are quotients of polynomial rings in infinitely many variables.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 22 '18 at 1:12
user237522user237522
2,1411617
asked Nov 22 '18 at 1:12
user237522user237522
2,1411617
edited Nov 22 '18 at 2:17
user237522
asked Nov 22 '18 at 1:12
user237522user237522
2,1411617
asked Nov 22 '18 at 1:12
user237522user237522
2,1411617
asked Nov 22 '18 at 1:12
user237522user237522
2,1411617
2,1411617
Well in your question you only ask for infinitely generated, but all your $R$'s are countably generated, so any uncountably generated ring lies outside the class you suggest.
– Alex J Best
Nov 22 '18 at 1:19
Thanks, you are right. I actually meant to ask about $frac{k[x_i]_{i in A}}{I}$, for an arbitrary set $A$, countably generated or not.
– user237522
Nov 22 '18 at 1:22
@AlexJBest, if my question still interests you, please see my edit (in which I claim that there exist such $R$'s of another form as well).
– user237522
Nov 22 '18 at 2:18
3
Every commutative ring whatsoever is a quotient of a polynomial ring in enough variables (for example, it suffices to take one variable for each element of the ring). But asking for a classification is totally hopeless.
– Qiaochu Yuan
Nov 22 '18 at 2:46
@QiaochuYuan, thank you.
– user237522
Nov 22 '18 at 3:16
add a comment |
Well in your question you only ask for infinitely generated, but all your $R$'s are countably generated, so any uncountably generated ring lies outside the class you suggest.
– Alex J Best
Nov 22 '18 at 1:19
Thanks, you are right. I actually meant to ask about $frac{k[x_i]_{i in A}}{I}$, for an arbitrary set $A$, countably generated or not.
– user237522
Nov 22 '18 at 1:22
@AlexJBest, if my question still interests you, please see my edit (in which I claim that there exist such $R$'s of another form as well).
– user237522
Nov 22 '18 at 2:18
3
Every commutative ring whatsoever is a quotient of a polynomial ring in enough variables (for example, it suffices to take one variable for each element of the ring). But asking for a classification is totally hopeless.
– Qiaochu Yuan
Nov 22 '18 at 2:46
@QiaochuYuan, thank you.
– user237522
Nov 22 '18 at 3:16
Well in your question you only ask for infinitely generated, but all your $R$'s are countably generated, so any uncountably generated ring lies outside the class you suggest.
– Alex J Best
Nov 22 '18 at 1:19
Well in your question you only ask for infinitely generated, but all your $R$'s are countably generated, so any uncountably generated ring lies outside the class you suggest.
– Alex J Best
Nov 22 '18 at 1:19
Thanks, you are right. I actually meant to ask about $frac{k[x_i]_{i in A}}{I}$, for an arbitrary set $A$, countably generated or not.
– user237522
Nov 22 '18 at 1:22
Thanks, you are right. I actually meant to ask about $frac{k[x_i]_{i in A}}{I}$, for an arbitrary set $A$, countably generated or not.
– user237522
Nov 22 '18 at 1:22
@AlexJBest, if my question still interests you, please see my edit (in which I claim that there exist such $R$'s of another form as well).
– user237522
Nov 22 '18 at 2:18
@AlexJBest, if my question still interests you, please see my edit (in which I claim that there exist such $R$'s of another form as well).
– user237522
Nov 22 '18 at 2:18
3
3
Every commutative ring whatsoever is a quotient of a polynomial ring in enough variables (for example, it suffices to take one variable for each element of the ring). But asking for a classification is totally hopeless.
– Qiaochu Yuan
Nov 22 '18 at 2:46
Every commutative ring whatsoever is a quotient of a polynomial ring in enough variables (for example, it suffices to take one variable for each element of the ring). But asking for a classification is totally hopeless.
– Qiaochu Yuan
Nov 22 '18 at 2:46
@QiaochuYuan, thank you.
– user237522
Nov 22 '18 at 3:16
@QiaochuYuan, thank you.
– user237522
Nov 22 '18 at 3:16
add a comment |
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Well in your question you only ask for infinitely generated, but all your $R$'s are countably generated, so any uncountably generated ring lies outside the class you suggest.
– Alex J Best
Nov 22 '18 at 1:19
Thanks, you are right. I actually meant to ask about $frac{k[x_i]_{i in A}}{I}$, for an arbitrary set $A$, countably generated or not.
– user237522
Nov 22 '18 at 1:22
@AlexJBest, if my question still interests you, please see my edit (in which I claim that there exist such $R$'s of another form as well).
– user237522
Nov 22 '18 at 2:18
3
Every commutative ring whatsoever is a quotient of a polynomial ring in enough variables (for example, it suffices to take one variable for each element of the ring). But asking for a classification is totally hopeless.
– Qiaochu Yuan
Nov 22 '18 at 2:46
@QiaochuYuan, thank you.
– user237522
Nov 22 '18 at 3:16