Does $operatorname{Ann}(d) = operatorname{Ann}(b)$ together with $d in operatorname{Rad}(bR)$ imply $d in...
up vote
0
down vote
favorite
$DeclareMathOperator{Ann}{Ann}DeclareMathOperator{Rad}{Rad}$Let $R$ be a noetherian, reduced ring which is not an integral domain (and may also not be factorial).
Let $P$ be a minimal prime of $R$. Is it possible to find $b in R$ with $P = Ann(b)$ and $bR$ radical?
What I tried: Consider a non-zero element $d in Rad(bR)$. By assumption we have $Ann(d) subseteq Ann(b) = P$ since $P$ is prime and $d neq 0$. Moreover, since $R$ is reduced, we obtain equality, i.e. $Ann(d) = Ann(b) = P$. But I don't know how to proceed from here on.
Note that we may change $b in R$ in any way that preserves $Ann(b) = P$. We may also assume (if necessary) that $R$ is finite free over some principal ideal domain (since the original motivation is from geometry where I consider a projective curve lying finitely over the projective line).
I am grateful for any kind of help.
Edit: I edited the question since it wasn't clear what I meant.
This question is related to the following: Is $R/bR$ reduced if $R$ is reduced and $operatorname{Ann}(b)$ is a minimal prime of $R$?
abstract-algebra algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
asked 22 hours ago
windsheaf
580312
add a comment |
up vote
0
down vote
favorite
$DeclareMathOperator{Ann}{Ann}DeclareMathOperator{Rad}{Rad}$Let $R$ be a noetherian, reduced ring which is not an integral domain (and may also not be factorial).
Let $P$ be a minimal prime of $R$. Is it possible to find $b in R$ with $P = Ann(b)$ and $bR$ radical?
What I tried: Consider a non-zero element $d in Rad(bR)$. By assumption we have $Ann(d) subseteq Ann(b) = P$ since $P$ is prime and $d neq 0$. Moreover, since $R$ is reduced, we obtain equality, i.e. $Ann(d) = Ann(b) = P$. But I don't know how to proceed from here on.
Note that we may change $b in R$ in any way that preserves $Ann(b) = P$. We may also assume (if necessary) that $R$ is finite free over some principal ideal domain (since the original motivation is from geometry where I consider a projective curve lying finitely over the projective line).
I am grateful for any kind of help.
Edit: I edited the question since it wasn't clear what I meant.
This question is related to the following: Is $R/bR$ reduced if $R$ is reduced and $operatorname{Ann}(b)$ is a minimal prime of $R$?
abstract-algebra algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
asked 22 hours ago
windsheaf
580312
1
Isn't this just a rephrasing of the previous question? ($d in text{Rad}(bR)$ implies $d in bR$) if and only if ($R/bR$ is reduced).
– Christopher
22 hours ago
Your question appears to have been answered in your link. What is different between these two questions? (Aside from $R$ isn't an integral domain here, which is hardly a restriction) Take $k[x,y]/(xy)$, then $Ann(x^2)=(y)$, which is a minimal prime, but $(x^2)$ is not radical.
– jgon
22 hours ago
@jgon The question comprises whether it is always possible to choose $b in R$ such that $bR$ is radical. In your example, $(x)$ is radical.
– windsheaf
22 hours ago
@windsheaf That's not what this question says. Nowhere in this question does it even imply that that's what you're asking. I'm then a bit confused about what you're asking, are you asking if every reduced noetherian ring has a principal radical ideal? Please edit and clarify
– jgon
22 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$DeclareMathOperator{Ann}{Ann}DeclareMathOperator{Rad}{Rad}$Let $R$ be a noetherian, reduced ring which is not an integral domain (and may also not be factorial).
Let $P$ be a minimal prime of $R$. Is it possible to find $b in R$ with $P = Ann(b)$ and $bR$ radical?
What I tried: Consider a non-zero element $d in Rad(bR)$. By assumption we have $Ann(d) subseteq Ann(b) = P$ since $P$ is prime and $d neq 0$. Moreover, since $R$ is reduced, we obtain equality, i.e. $Ann(d) = Ann(b) = P$. But I don't know how to proceed from here on.
Note that we may change $b in R$ in any way that preserves $Ann(b) = P$. We may also assume (if necessary) that $R$ is finite free over some principal ideal domain (since the original motivation is from geometry where I consider a projective curve lying finitely over the projective line).
I am grateful for any kind of help.
Edit: I edited the question since it wasn't clear what I meant.
This question is related to the following: Is $R/bR$ reduced if $R$ is reduced and $operatorname{Ann}(b)$ is a minimal prime of $R$?
abstract-algebra algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
asked 22 hours ago
windsheaf
580312
$DeclareMathOperator{Ann}{Ann}DeclareMathOperator{Rad}{Rad}$Let $R$ be a noetherian, reduced ring which is not an integral domain (and may also not be factorial).
Let $P$ be a minimal prime of $R$. Is it possible to find $b in R$ with $P = Ann(b)$ and $bR$ radical?
What I tried: Consider a non-zero element $d in Rad(bR)$. By assumption we have $Ann(d) subseteq Ann(b) = P$ since $P$ is prime and $d neq 0$. Moreover, since $R$ is reduced, we obtain equality, i.e. $Ann(d) = Ann(b) = P$. But I don't know how to proceed from here on.
Note that we may change $b in R$ in any way that preserves $Ann(b) = P$. We may also assume (if necessary) that $R$ is finite free over some principal ideal domain (since the original motivation is from geometry where I consider a projective curve lying finitely over the projective line).
I am grateful for any kind of help.
Edit: I edited the question since it wasn't clear what I meant.
This question is related to the following: Is $R/bR$ reduced if $R$ is reduced and $operatorname{Ann}(b)$ is a minimal prime of $R$?
abstract-algebra algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
abstract-algebra algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
asked 22 hours ago
windsheaf
580312
asked 22 hours ago
windsheaf
580312
edited 22 hours ago
asked 22 hours ago
windsheaf
580312
asked 22 hours ago
windsheaf
580312
asked 22 hours ago
windsheaf
580312
580312
1
Isn't this just a rephrasing of the previous question? ($d in text{Rad}(bR)$ implies $d in bR$) if and only if ($R/bR$ is reduced).
– Christopher
22 hours ago
Your question appears to have been answered in your link. What is different between these two questions? (Aside from $R$ isn't an integral domain here, which is hardly a restriction) Take $k[x,y]/(xy)$, then $Ann(x^2)=(y)$, which is a minimal prime, but $(x^2)$ is not radical.
– jgon
22 hours ago
@jgon The question comprises whether it is always possible to choose $b in R$ such that $bR$ is radical. In your example, $(x)$ is radical.
– windsheaf
22 hours ago
@windsheaf That's not what this question says. Nowhere in this question does it even imply that that's what you're asking. I'm then a bit confused about what you're asking, are you asking if every reduced noetherian ring has a principal radical ideal? Please edit and clarify
– jgon
22 hours ago
add a comment |
1
Isn't this just a rephrasing of the previous question? ($d in text{Rad}(bR)$ implies $d in bR$) if and only if ($R/bR$ is reduced).
– Christopher
22 hours ago
Your question appears to have been answered in your link. What is different between these two questions? (Aside from $R$ isn't an integral domain here, which is hardly a restriction) Take $k[x,y]/(xy)$, then $Ann(x^2)=(y)$, which is a minimal prime, but $(x^2)$ is not radical.
– jgon
22 hours ago
@jgon The question comprises whether it is always possible to choose $b in R$ such that $bR$ is radical. In your example, $(x)$ is radical.
– windsheaf
22 hours ago
@windsheaf That's not what this question says. Nowhere in this question does it even imply that that's what you're asking. I'm then a bit confused about what you're asking, are you asking if every reduced noetherian ring has a principal radical ideal? Please edit and clarify
– jgon
22 hours ago
1
1
Isn't this just a rephrasing of the previous question? ($d in text{Rad}(bR)$ implies $d in bR$) if and only if ($R/bR$ is reduced).
– Christopher
22 hours ago
Isn't this just a rephrasing of the previous question? ($d in text{Rad}(bR)$ implies $d in bR$) if and only if ($R/bR$ is reduced).
– Christopher
22 hours ago
Your question appears to have been answered in your link. What is different between these two questions? (Aside from $R$ isn't an integral domain here, which is hardly a restriction) Take $k[x,y]/(xy)$, then $Ann(x^2)=(y)$, which is a minimal prime, but $(x^2)$ is not radical.
– jgon
22 hours ago
Your question appears to have been answered in your link. What is different between these two questions? (Aside from $R$ isn't an integral domain here, which is hardly a restriction) Take $k[x,y]/(xy)$, then $Ann(x^2)=(y)$, which is a minimal prime, but $(x^2)$ is not radical.
– jgon
22 hours ago
@jgon The question comprises whether it is always possible to choose $b in R$ such that $bR$ is radical. In your example, $(x)$ is radical.
– windsheaf
22 hours ago
@jgon The question comprises whether it is always possible to choose $b in R$ such that $bR$ is radical. In your example, $(x)$ is radical.
– windsheaf
22 hours ago
@windsheaf That's not what this question says. Nowhere in this question does it even imply that that's what you're asking. I'm then a bit confused about what you're asking, are you asking if every reduced noetherian ring has a principal radical ideal? Please edit and clarify
– jgon
22 hours ago
@windsheaf That's not what this question says. Nowhere in this question does it even imply that that's what you're asking. I'm then a bit confused about what you're asking, are you asking if every reduced noetherian ring has a principal radical ideal? Please edit and clarify
– jgon
22 hours ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995350%2fdoes-operatornameannd-operatornameannb-together-with-d-in-opera%23new-answer', 'question_page');
}
);
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Isn't this just a rephrasing of the previous question? ($d in text{Rad}(bR)$ implies $d in bR$) if and only if ($R/bR$ is reduced).
– Christopher
22 hours ago
Your question appears to have been answered in your link. What is different between these two questions? (Aside from $R$ isn't an integral domain here, which is hardly a restriction) Take $k[x,y]/(xy)$, then $Ann(x^2)=(y)$, which is a minimal prime, but $(x^2)$ is not radical.
– jgon
22 hours ago
@jgon The question comprises whether it is always possible to choose $b in R$ such that $bR$ is radical. In your example, $(x)$ is radical.
– windsheaf
22 hours ago
@windsheaf That's not what this question says. Nowhere in this question does it even imply that that's what you're asking. I'm then a bit confused about what you're asking, are you asking if every reduced noetherian ring has a principal radical ideal? Please edit and clarify
– jgon
22 hours ago