Mixing
When the element-wise product of two ideals produces an ideal
$begingroup$
Consider the ring $R=mathbb C[X,Y]$. For every two ideals $I,J$ of $R$, define $I*J:={ij : iin I, jin J}$.
Now definitely, $I*J=J*I$ always holds. If $I$ is principal, then actually $I*J$ is an ideal of $R$.
My question is:
If $I$ is a proper, non-zero ideal of $R=mathbb C[X,Y]$ such that for every ideal $J$ of $R$, $I*J$ is also an ideal of $R$ (i.e. $I*J=IJ$), then does it imply that $I$ is principal ? Or at least $I$ is contained in a principal prime ideal ? If neither of these are true, then can we characterize all ideals $I$ with the said property in some other way ?
Some thoughts towards possibly showing $I$ is principal : To show $I$ is principal, enough to show $I$ is free, then by Quillen-Suslin, enough to show $I$ is projective, and since we are in Noetherian, finitely generated case, enough to show $I$ is flat over $R$. So it is enough to show $I otimes_R J cong IJ =I*J$ for every ideal $J$. No idea how to show that though ...
algebraic-geometry polynomials ring-theory commutative-algebra ideals
asked Jan 29 at 23:39
user102248user102248
30619
$endgroup$
add a comment |
$begingroup$
Consider the ring $R=mathbb C[X,Y]$. For every two ideals $I,J$ of $R$, define $I*J:={ij : iin I, jin J}$.
Now definitely, $I*J=J*I$ always holds. If $I$ is principal, then actually $I*J$ is an ideal of $R$.
My question is:
If $I$ is a proper, non-zero ideal of $R=mathbb C[X,Y]$ such that for every ideal $J$ of $R$, $I*J$ is also an ideal of $R$ (i.e. $I*J=IJ$), then does it imply that $I$ is principal ? Or at least $I$ is contained in a principal prime ideal ? If neither of these are true, then can we characterize all ideals $I$ with the said property in some other way ?
Some thoughts towards possibly showing $I$ is principal : To show $I$ is principal, enough to show $I$ is free, then by Quillen-Suslin, enough to show $I$ is projective, and since we are in Noetherian, finitely generated case, enough to show $I$ is flat over $R$. So it is enough to show $I otimes_R J cong IJ =I*J$ for every ideal $J$. No idea how to show that though ...
algebraic-geometry polynomials ring-theory commutative-algebra ideals
asked Jan 29 at 23:39
user102248user102248
30619
$endgroup$
$begingroup$
If $I$ is contained in a principal prime ideal, then it would be same as saying it has height $1$, which in the context of your ring, would be same as saying $I$ is contained in a proper principal ideal ...
$endgroup$
– user521337
Jan 30 at 18:14
$begingroup$
(contd. from last comment ) ... probably that doesn't help in any way though ...
$endgroup$
– user521337
Jan 30 at 22:19
$begingroup$
since $I*J=IJ$. so $IJ$ contains no irreducible element for every proper ideal $J$, so taking $J=IK$, we get $I^2K$ does not contain any irreducible, hence does not contain any prime ideal for every ideal $K$ ...
$endgroup$
– user521337
Jan 31 at 22:11
$begingroup$
so in particular $I^2$ does not contain any non-zero prime ideal ...
$endgroup$
– user521337
Jan 31 at 22:32
add a comment |
$begingroup$
Consider the ring $R=mathbb C[X,Y]$. For every two ideals $I,J$ of $R$, define $I*J:={ij : iin I, jin J}$.
Now definitely, $I*J=J*I$ always holds. If $I$ is principal, then actually $I*J$ is an ideal of $R$.
My question is:
If $I$ is a proper, non-zero ideal of $R=mathbb C[X,Y]$ such that for every ideal $J$ of $R$, $I*J$ is also an ideal of $R$ (i.e. $I*J=IJ$), then does it imply that $I$ is principal ? Or at least $I$ is contained in a principal prime ideal ? If neither of these are true, then can we characterize all ideals $I$ with the said property in some other way ?
Some thoughts towards possibly showing $I$ is principal : To show $I$ is principal, enough to show $I$ is free, then by Quillen-Suslin, enough to show $I$ is projective, and since we are in Noetherian, finitely generated case, enough to show $I$ is flat over $R$. So it is enough to show $I otimes_R J cong IJ =I*J$ for every ideal $J$. No idea how to show that though ...
algebraic-geometry polynomials ring-theory commutative-algebra ideals
asked Jan 29 at 23:39
user102248user102248
30619
$endgroup$
Consider the ring $R=mathbb C[X,Y]$. For every two ideals $I,J$ of $R$, define $I*J:={ij : iin I, jin J}$.
Now definitely, $I*J=J*I$ always holds. If $I$ is principal, then actually $I*J$ is an ideal of $R$.
My question is:
If $I$ is a proper, non-zero ideal of $R=mathbb C[X,Y]$ such that for every ideal $J$ of $R$, $I*J$ is also an ideal of $R$ (i.e. $I*J=IJ$), then does it imply that $I$ is principal ? Or at least $I$ is contained in a principal prime ideal ? If neither of these are true, then can we characterize all ideals $I$ with the said property in some other way ?
Some thoughts towards possibly showing $I$ is principal : To show $I$ is principal, enough to show $I$ is free, then by Quillen-Suslin, enough to show $I$ is projective, and since we are in Noetherian, finitely generated case, enough to show $I$ is flat over $R$. So it is enough to show $I otimes_R J cong IJ =I*J$ for every ideal $J$. No idea how to show that though ...
algebraic-geometry polynomials ring-theory commutative-algebra ideals
algebraic-geometry polynomials ring-theory commutative-algebra ideals
asked Jan 29 at 23:39
user102248user102248
30619
asked Jan 29 at 23:39
user102248user102248
30619
edited Feb 1 at 13:34
user102248
asked Jan 29 at 23:39
user102248user102248
30619
asked Jan 29 at 23:39
user102248user102248
30619
asked Jan 29 at 23:39
user102248user102248
30619
30619
$begingroup$
If $I$ is contained in a principal prime ideal, then it would be same as saying it has height $1$, which in the context of your ring, would be same as saying $I$ is contained in a proper principal ideal ...
$endgroup$
– user521337
Jan 30 at 18:14
$begingroup$
(contd. from last comment ) ... probably that doesn't help in any way though ...
$endgroup$
– user521337
Jan 30 at 22:19
$begingroup$
since $I*J=IJ$. so $IJ$ contains no irreducible element for every proper ideal $J$, so taking $J=IK$, we get $I^2K$ does not contain any irreducible, hence does not contain any prime ideal for every ideal $K$ ...
$endgroup$
– user521337
Jan 31 at 22:11
$begingroup$
so in particular $I^2$ does not contain any non-zero prime ideal ...
$endgroup$
– user521337
Jan 31 at 22:32
add a comment |
$begingroup$
If $I$ is contained in a principal prime ideal, then it would be same as saying it has height $1$, which in the context of your ring, would be same as saying $I$ is contained in a proper principal ideal ...
$endgroup$
– user521337
Jan 30 at 18:14
$begingroup$
(contd. from last comment ) ... probably that doesn't help in any way though ...
$endgroup$
– user521337
Jan 30 at 22:19
$begingroup$
since $I*J=IJ$. so $IJ$ contains no irreducible element for every proper ideal $J$, so taking $J=IK$, we get $I^2K$ does not contain any irreducible, hence does not contain any prime ideal for every ideal $K$ ...
$endgroup$
– user521337
Jan 31 at 22:11
$begingroup$
so in particular $I^2$ does not contain any non-zero prime ideal ...
$endgroup$
– user521337
Jan 31 at 22:32
$begingroup$
If $I$ is contained in a principal prime ideal, then it would be same as saying it has height $1$, which in the context of your ring, would be same as saying $I$ is contained in a proper principal ideal ...
$endgroup$
– user521337
Jan 30 at 18:14
$begingroup$
If $I$ is contained in a principal prime ideal, then it would be same as saying it has height $1$, which in the context of your ring, would be same as saying $I$ is contained in a proper principal ideal ...
$endgroup$
– user521337
Jan 30 at 18:14
$begingroup$
(contd. from last comment ) ... probably that doesn't help in any way though ...
$endgroup$
– user521337
Jan 30 at 22:19
$begingroup$
(contd. from last comment ) ... probably that doesn't help in any way though ...
$endgroup$
– user521337
Jan 30 at 22:19
$begingroup$
since $I*J=IJ$. so $IJ$ contains no irreducible element for every proper ideal $J$, so taking $J=IK$, we get $I^2K$ does not contain any irreducible, hence does not contain any prime ideal for every ideal $K$ ...
$endgroup$
– user521337
Jan 31 at 22:11
$begingroup$
since $I*J=IJ$. so $IJ$ contains no irreducible element for every proper ideal $J$, so taking $J=IK$, we get $I^2K$ does not contain any irreducible, hence does not contain any prime ideal for every ideal $K$ ...
$endgroup$
– user521337
Jan 31 at 22:11
$begingroup$
so in particular $I^2$ does not contain any non-zero prime ideal ...
$endgroup$
– user521337
Jan 31 at 22:32
$begingroup$
so in particular $I^2$ does not contain any non-zero prime ideal ...
$endgroup$
– user521337
Jan 31 at 22:32
add a comment |
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$begingroup$
If $I$ is contained in a principal prime ideal, then it would be same as saying it has height $1$, which in the context of your ring, would be same as saying $I$ is contained in a proper principal ideal ...
$endgroup$
– user521337
Jan 30 at 18:14
$begingroup$
(contd. from last comment ) ... probably that doesn't help in any way though ...
$endgroup$
– user521337
Jan 30 at 22:19
$begingroup$
since $I*J=IJ$. so $IJ$ contains no irreducible element for every proper ideal $J$, so taking $J=IK$, we get $I^2K$ does not contain any irreducible, hence does not contain any prime ideal for every ideal $K$ ...
$endgroup$
– user521337
Jan 31 at 22:11
$begingroup$
so in particular $I^2$ does not contain any non-zero prime ideal ...
$endgroup$
– user521337
Jan 31 at 22:32