If two ideals agree on $Klhd R$ and yield the same quotients $pi_K(I)=pi_K(J)$, are they equal?
$begingroup$
Inspired by my lectures on the basics of commutative algebra, I tried to investigate whether the noetherian property still holds when taking extensions (here meaning: if $Ilhd R$ and $R/I$ both satisfy the a.c.c., does $R$ as well?). I started to wonder the following:
Let $I,J,Klhd R$, and $pi_K: R to R/K$ the canonical projection. If $Icap K = Jcap K$ and $pi_K(I) = pi_K (J)$, does this already imply $I = J$?
We can represent this graphically via the short exact sequence of $R$-Modules $$
0 to Kstackrel{iota}{to} R stackrel{pi_K}{to} R/K to 0
$$ where $iota$ is just the embedding of $K$ into $R$, and notice that the intersections $Icap K, Jcap K$ is basically just the preimages under $iota$.
I tried to consider all these objects as $R$-moduls and play around with isomorphism theorems, but I only got statements for quotiens or intersections of $I, J, I+J$ or similar, but not these objects directly.
Is it even true?
ring-theory commutative-algebra ideals
edited Jan 14 at 15:06
user26857
39.4k124183
asked Jan 14 at 0:38
LukeLuke
1,2721023
$endgroup$
add a comment |
$begingroup$
Inspired by my lectures on the basics of commutative algebra, I tried to investigate whether the noetherian property still holds when taking extensions (here meaning: if $Ilhd R$ and $R/I$ both satisfy the a.c.c., does $R$ as well?). I started to wonder the following:
Let $I,J,Klhd R$, and $pi_K: R to R/K$ the canonical projection. If $Icap K = Jcap K$ and $pi_K(I) = pi_K (J)$, does this already imply $I = J$?
We can represent this graphically via the short exact sequence of $R$-Modules $$
0 to Kstackrel{iota}{to} R stackrel{pi_K}{to} R/K to 0
$$ where $iota$ is just the embedding of $K$ into $R$, and notice that the intersections $Icap K, Jcap K$ is basically just the preimages under $iota$.
I tried to consider all these objects as $R$-moduls and play around with isomorphism theorems, but I only got statements for quotiens or intersections of $I, J, I+J$ or similar, but not these objects directly.
Is it even true?
ring-theory commutative-algebra ideals
edited Jan 14 at 15:06
user26857
39.4k124183
asked Jan 14 at 0:38
LukeLuke
1,2721023
$endgroup$
add a comment |
$begingroup$
Inspired by my lectures on the basics of commutative algebra, I tried to investigate whether the noetherian property still holds when taking extensions (here meaning: if $Ilhd R$ and $R/I$ both satisfy the a.c.c., does $R$ as well?). I started to wonder the following:
Let $I,J,Klhd R$, and $pi_K: R to R/K$ the canonical projection. If $Icap K = Jcap K$ and $pi_K(I) = pi_K (J)$, does this already imply $I = J$?
We can represent this graphically via the short exact sequence of $R$-Modules $$
0 to Kstackrel{iota}{to} R stackrel{pi_K}{to} R/K to 0
$$ where $iota$ is just the embedding of $K$ into $R$, and notice that the intersections $Icap K, Jcap K$ is basically just the preimages under $iota$.
I tried to consider all these objects as $R$-moduls and play around with isomorphism theorems, but I only got statements for quotiens or intersections of $I, J, I+J$ or similar, but not these objects directly.
Is it even true?
ring-theory commutative-algebra ideals
edited Jan 14 at 15:06
user26857
39.4k124183
asked Jan 14 at 0:38
LukeLuke
1,2721023
$endgroup$
Inspired by my lectures on the basics of commutative algebra, I tried to investigate whether the noetherian property still holds when taking extensions (here meaning: if $Ilhd R$ and $R/I$ both satisfy the a.c.c., does $R$ as well?). I started to wonder the following:
Let $I,J,Klhd R$, and $pi_K: R to R/K$ the canonical projection. If $Icap K = Jcap K$ and $pi_K(I) = pi_K (J)$, does this already imply $I = J$?
We can represent this graphically via the short exact sequence of $R$-Modules $$
0 to Kstackrel{iota}{to} R stackrel{pi_K}{to} R/K to 0
$$ where $iota$ is just the embedding of $K$ into $R$, and notice that the intersections $Icap K, Jcap K$ is basically just the preimages under $iota$.
I tried to consider all these objects as $R$-moduls and play around with isomorphism theorems, but I only got statements for quotiens or intersections of $I, J, I+J$ or similar, but not these objects directly.
Is it even true?
ring-theory commutative-algebra ideals
ring-theory commutative-algebra ideals
edited Jan 14 at 15:06
user26857
39.4k124183
asked Jan 14 at 0:38
LukeLuke
1,2721023
edited Jan 14 at 15:06
user26857
39.4k124183
asked Jan 14 at 0:38
LukeLuke
1,2721023
edited Jan 14 at 15:06
user26857
39.4k124183
edited Jan 14 at 15:06
user26857
39.4k124183
edited Jan 14 at 15:06
user26857
39.4k124183
39.4k124183
asked Jan 14 at 0:38
LukeLuke
1,2721023
asked Jan 14 at 0:38
LukeLuke
1,2721023
asked Jan 14 at 0:38
LukeLuke
1,2721023
1,2721023
add a comment |
add a comment |
1 Answer
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$begingroup$
It's not true. For instance, let $R=k[x,y]/(x^2,xy,y^2)$ for some field $k$, and let $K=(x)$, $I=(y)$, and $J=(x+y)$. Then $R/Kcong k[y]/(y^2)$ and $pi_K(I)=pi_K(J)=(y)$, and also $Icap K=Jcap K=0$, but $Ineq J$.
It is true, though, if you know that $Isubseteq J$, which is what you need for proving that Noetherianness is preserved by extensions. Indeed, given $jin J$, there exists $iin I$ such that $pi_K(i)=pi_K(j)$. Then $i-jin K$, and also $i-jin J$ since $i,jin J$ (here we use the assumption that $Isubseteq J$). Thus $i-jin Jcap K=Icap K$ and so $j=i-(i-j)in I$.
answered Jan 14 at 1:24
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
Thanks! Do you have an example where $R$ is a domain? If I see correctly, $k[x,y]/(x^2,xy,y^2)$ is essentially a three-dimensional $k$-algebra with trivial multiplication, so everything that's not in the $k$ part is a zero divisor.
$endgroup$
– Luke
Jan 14 at 1:39
$begingroup$
You can get an example where $R$ is a domain by just pulling all of these ideals back to $k[x,y]$.
$endgroup$
– Eric Wofsey
Jan 14 at 2:29
$begingroup$
That's a good point, thanks again.
$endgroup$
– Luke
Jan 14 at 9:31
add a comment |
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$begingroup$
It's not true. For instance, let $R=k[x,y]/(x^2,xy,y^2)$ for some field $k$, and let $K=(x)$, $I=(y)$, and $J=(x+y)$. Then $R/Kcong k[y]/(y^2)$ and $pi_K(I)=pi_K(J)=(y)$, and also $Icap K=Jcap K=0$, but $Ineq J$.
It is true, though, if you know that $Isubseteq J$, which is what you need for proving that Noetherianness is preserved by extensions. Indeed, given $jin J$, there exists $iin I$ such that $pi_K(i)=pi_K(j)$. Then $i-jin K$, and also $i-jin J$ since $i,jin J$ (here we use the assumption that $Isubseteq J$). Thus $i-jin Jcap K=Icap K$ and so $j=i-(i-j)in I$.
answered Jan 14 at 1:24
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
Thanks! Do you have an example where $R$ is a domain? If I see correctly, $k[x,y]/(x^2,xy,y^2)$ is essentially a three-dimensional $k$-algebra with trivial multiplication, so everything that's not in the $k$ part is a zero divisor.
$endgroup$
– Luke
Jan 14 at 1:39
$begingroup$
You can get an example where $R$ is a domain by just pulling all of these ideals back to $k[x,y]$.
$endgroup$
– Eric Wofsey
Jan 14 at 2:29
$begingroup$
That's a good point, thanks again.
$endgroup$
– Luke
Jan 14 at 9:31
add a comment |
$begingroup$
It's not true. For instance, let $R=k[x,y]/(x^2,xy,y^2)$ for some field $k$, and let $K=(x)$, $I=(y)$, and $J=(x+y)$. Then $R/Kcong k[y]/(y^2)$ and $pi_K(I)=pi_K(J)=(y)$, and also $Icap K=Jcap K=0$, but $Ineq J$.
It is true, though, if you know that $Isubseteq J$, which is what you need for proving that Noetherianness is preserved by extensions. Indeed, given $jin J$, there exists $iin I$ such that $pi_K(i)=pi_K(j)$. Then $i-jin K$, and also $i-jin J$ since $i,jin J$ (here we use the assumption that $Isubseteq J$). Thus $i-jin Jcap K=Icap K$ and so $j=i-(i-j)in I$.
answered Jan 14 at 1:24
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
Thanks! Do you have an example where $R$ is a domain? If I see correctly, $k[x,y]/(x^2,xy,y^2)$ is essentially a three-dimensional $k$-algebra with trivial multiplication, so everything that's not in the $k$ part is a zero divisor.
$endgroup$
– Luke
Jan 14 at 1:39
$begingroup$
You can get an example where $R$ is a domain by just pulling all of these ideals back to $k[x,y]$.
$endgroup$
– Eric Wofsey
Jan 14 at 2:29
$begingroup$
That's a good point, thanks again.
$endgroup$
– Luke
Jan 14 at 9:31
add a comment |
$begingroup$
It's not true. For instance, let $R=k[x,y]/(x^2,xy,y^2)$ for some field $k$, and let $K=(x)$, $I=(y)$, and $J=(x+y)$. Then $R/Kcong k[y]/(y^2)$ and $pi_K(I)=pi_K(J)=(y)$, and also $Icap K=Jcap K=0$, but $Ineq J$.
It is true, though, if you know that $Isubseteq J$, which is what you need for proving that Noetherianness is preserved by extensions. Indeed, given $jin J$, there exists $iin I$ such that $pi_K(i)=pi_K(j)$. Then $i-jin K$, and also $i-jin J$ since $i,jin J$ (here we use the assumption that $Isubseteq J$). Thus $i-jin Jcap K=Icap K$ and so $j=i-(i-j)in I$.
answered Jan 14 at 1:24
Eric WofseyEric Wofsey
190k14216348
$endgroup$
It's not true. For instance, let $R=k[x,y]/(x^2,xy,y^2)$ for some field $k$, and let $K=(x)$, $I=(y)$, and $J=(x+y)$. Then $R/Kcong k[y]/(y^2)$ and $pi_K(I)=pi_K(J)=(y)$, and also $Icap K=Jcap K=0$, but $Ineq J$.
It is true, though, if you know that $Isubseteq J$, which is what you need for proving that Noetherianness is preserved by extensions. Indeed, given $jin J$, there exists $iin I$ such that $pi_K(i)=pi_K(j)$. Then $i-jin K$, and also $i-jin J$ since $i,jin J$ (here we use the assumption that $Isubseteq J$). Thus $i-jin Jcap K=Icap K$ and so $j=i-(i-j)in I$.
answered Jan 14 at 1:24
Eric WofseyEric Wofsey
190k14216348
answered Jan 14 at 1:24
Eric WofseyEric Wofsey
190k14216348
answered Jan 14 at 1:24
Eric WofseyEric Wofsey
190k14216348
answered Jan 14 at 1:24
Eric WofseyEric Wofsey
190k14216348
190k14216348
$begingroup$
Thanks! Do you have an example where $R$ is a domain? If I see correctly, $k[x,y]/(x^2,xy,y^2)$ is essentially a three-dimensional $k$-algebra with trivial multiplication, so everything that's not in the $k$ part is a zero divisor.
$endgroup$
– Luke
Jan 14 at 1:39
$begingroup$
You can get an example where $R$ is a domain by just pulling all of these ideals back to $k[x,y]$.
$endgroup$
– Eric Wofsey
Jan 14 at 2:29
$begingroup$
That's a good point, thanks again.
$endgroup$
– Luke
Jan 14 at 9:31
add a comment |
$begingroup$
Thanks! Do you have an example where $R$ is a domain? If I see correctly, $k[x,y]/(x^2,xy,y^2)$ is essentially a three-dimensional $k$-algebra with trivial multiplication, so everything that's not in the $k$ part is a zero divisor.
$endgroup$
– Luke
Jan 14 at 1:39
$begingroup$
You can get an example where $R$ is a domain by just pulling all of these ideals back to $k[x,y]$.
$endgroup$
– Eric Wofsey
Jan 14 at 2:29
$begingroup$
That's a good point, thanks again.
$endgroup$
– Luke
Jan 14 at 9:31
$begingroup$
Thanks! Do you have an example where $R$ is a domain? If I see correctly, $k[x,y]/(x^2,xy,y^2)$ is essentially a three-dimensional $k$-algebra with trivial multiplication, so everything that's not in the $k$ part is a zero divisor.
$endgroup$
– Luke
Jan 14 at 1:39
$begingroup$
Thanks! Do you have an example where $R$ is a domain? If I see correctly, $k[x,y]/(x^2,xy,y^2)$ is essentially a three-dimensional $k$-algebra with trivial multiplication, so everything that's not in the $k$ part is a zero divisor.
$endgroup$
– Luke
Jan 14 at 1:39
$begingroup$
You can get an example where $R$ is a domain by just pulling all of these ideals back to $k[x,y]$.
$endgroup$
– Eric Wofsey
Jan 14 at 2:29
$begingroup$
You can get an example where $R$ is a domain by just pulling all of these ideals back to $k[x,y]$.
$endgroup$
– Eric Wofsey
Jan 14 at 2:29
$begingroup$
That's a good point, thanks again.
$endgroup$
– Luke
Jan 14 at 9:31
$begingroup$
That's a good point, thanks again.
$endgroup$
– Luke
Jan 14 at 9:31
add a comment |
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