Mixing
Veronese embedding as ring isomorphism [duplicate]
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This question already has an answer here:
Krull dimension of $mathbb{C}[x_1, x_2, x_3, x_4]/left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3right>$
3 answers
At the level of rings, the Veronese map corresponds to isomorphisms like
$$k[w,x,y,z]/(wz - xy, wy-x^2, xz - y^2) cong k[a^3, a^2 b, a b^2, b^3].$$
This isomorphism is the statement that the relations satisfied by the monomials of degree 3 generate exactly the same ideal as $I = (wz - xy, wy-x^2, xz - y^2)$. One direction is obvious (corresponding to a well-defined map $to$): the monomials of degree 3 certainly satisfy these three relations.
The other direction seems hard. It encodes the idea that we have found all the relations, which is certainly an extremely hard general problem. How would one prove such a thing explicitly?
I believe all of this can be formulated in much higher geometric language as an isomorphism of $operatorname{Proj}$ constructions, but that language should surely encode a concrete argument.
algebraic-geometry ring-theory commutative-algebra
asked Jan 16 at 2:52
C.D.C.D.
415
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marked as duplicate by user26857 commutative-algebra
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Jan 16 at 16:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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This question already has an answer here:
Krull dimension of $mathbb{C}[x_1, x_2, x_3, x_4]/left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3right>$
3 answers
At the level of rings, the Veronese map corresponds to isomorphisms like
$$k[w,x,y,z]/(wz - xy, wy-x^2, xz - y^2) cong k[a^3, a^2 b, a b^2, b^3].$$
This isomorphism is the statement that the relations satisfied by the monomials of degree 3 generate exactly the same ideal as $I = (wz - xy, wy-x^2, xz - y^2)$. One direction is obvious (corresponding to a well-defined map $to$): the monomials of degree 3 certainly satisfy these three relations.
The other direction seems hard. It encodes the idea that we have found all the relations, which is certainly an extremely hard general problem. How would one prove such a thing explicitly?
I believe all of this can be formulated in much higher geometric language as an isomorphism of $operatorname{Proj}$ constructions, but that language should surely encode a concrete argument.
algebraic-geometry ring-theory commutative-algebra
asked Jan 16 at 2:52
C.D.C.D.
415
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marked as duplicate by user26857 commutative-algebra
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Jan 16 at 16:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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I think this post may answer your question.
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– André 3000
Jan 16 at 6:42
add a comment |
$begingroup$
This question already has an answer here:
Krull dimension of $mathbb{C}[x_1, x_2, x_3, x_4]/left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3right>$
3 answers
At the level of rings, the Veronese map corresponds to isomorphisms like
$$k[w,x,y,z]/(wz - xy, wy-x^2, xz - y^2) cong k[a^3, a^2 b, a b^2, b^3].$$
This isomorphism is the statement that the relations satisfied by the monomials of degree 3 generate exactly the same ideal as $I = (wz - xy, wy-x^2, xz - y^2)$. One direction is obvious (corresponding to a well-defined map $to$): the monomials of degree 3 certainly satisfy these three relations.
The other direction seems hard. It encodes the idea that we have found all the relations, which is certainly an extremely hard general problem. How would one prove such a thing explicitly?
I believe all of this can be formulated in much higher geometric language as an isomorphism of $operatorname{Proj}$ constructions, but that language should surely encode a concrete argument.
algebraic-geometry ring-theory commutative-algebra
asked Jan 16 at 2:52
C.D.C.D.
415
$endgroup$
This question already has an answer here:
Krull dimension of $mathbb{C}[x_1, x_2, x_3, x_4]/left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3right>$
3 answers
At the level of rings, the Veronese map corresponds to isomorphisms like
$$k[w,x,y,z]/(wz - xy, wy-x^2, xz - y^2) cong k[a^3, a^2 b, a b^2, b^3].$$
This isomorphism is the statement that the relations satisfied by the monomials of degree 3 generate exactly the same ideal as $I = (wz - xy, wy-x^2, xz - y^2)$. One direction is obvious (corresponding to a well-defined map $to$): the monomials of degree 3 certainly satisfy these three relations.
The other direction seems hard. It encodes the idea that we have found all the relations, which is certainly an extremely hard general problem. How would one prove such a thing explicitly?
I believe all of this can be formulated in much higher geometric language as an isomorphism of $operatorname{Proj}$ constructions, but that language should surely encode a concrete argument.
This question already has an answer here:
Krull dimension of $mathbb{C}[x_1, x_2, x_3, x_4]/left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3right>$
3 answers
algebraic-geometry ring-theory commutative-algebra
algebraic-geometry ring-theory commutative-algebra
asked Jan 16 at 2:52
C.D.C.D.
415
asked Jan 16 at 2:52
C.D.C.D.
415
asked Jan 16 at 2:52
C.D.C.D.
415
asked Jan 16 at 2:52
C.D.C.D.
415
asked Jan 16 at 2:52
C.D.C.D.
415
415
marked as duplicate by user26857 commutative-algebra
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Jan 16 at 16:28
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Jan 16 at 16:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
I think this post may answer your question.
$endgroup$
– André 3000
Jan 16 at 6:42
add a comment |
$begingroup$
I think this post may answer your question.
$endgroup$
– André 3000
Jan 16 at 6:42
$begingroup$
I think this post may answer your question.
$endgroup$
– André 3000
Jan 16 at 6:42
$begingroup$
I think this post may answer your question.
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– André 3000
Jan 16 at 6:42
add a comment |
1 Answer
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I suspect there are fancier more systematic ways to prove it, but here's a simple hands-on proof. By the easy direction we have a surjective homomorphism $f:Rto S$ where $R=k[w,x,y,z]$ and $S=k[a^3, a^2 b, a b^2, b^3]$, and the kernel of $f$ contains $I$. To show that the kernel of $f$ is contained in $I$, it suffices to show the following.
Claim: If monomials $m=w^i x^j y^k z^ell$ and $m'=w^{i'}x^{j'}y^{k'}z^{ell'}$ are such that $f(m)=f(m')$, then $mequiv m'pmod I$.
Why does this suffice? Well, since $f$ maps monomials to monomials, if $pin ker f$, then the monomials in $p$ must map to identical monomials in $S$ whose coefficients cancel out (since monomials in $S$ are linearly independent, that's the only way $f(p)$ can be $0$). But then the Claim would show that the monomials of $p$ also cancel out mod $I$, so $pin I$.
So, let's prove the Claim. We have $f(m)=a^{3i+2j+k}b^{j+2k+3ell}$ and similarly for $m'$, so we must have $$3i+2j+k=3i'+2j'+k'$$ and $$j+2k+3ell=j'+2k'+3ell'.$$
We will use induction on $i+i'+ell+ell'$. Let us first suppose that $i+i'+ell+ell'=0$. Then our equations are just $2j+k=2j'+k'$ and $j+2k=j'+2k'$. This easily implies $j=j'$ and $k=k'$, and so $m=m'$ and we are done.
Now suppose $i+i'+ell+ell'>0$; we may assume without loss of generality that $i>0$. If $i'>0$, we can cancel $w$ from both $m$ and $m'$ to reduce to a case where $i+i'+ell+ell'$ is smaller. If $ell>0$, we can use the relation $wzequiv xypmod{I}$ on $m$ to decrease $i$ and $ell$ by $1$ and increase $j$ and $k$ by $1$, reducing to a case where $i+i'+ell+ell'$ is smaller. Similarly if $k>0$ we can use the relation $wyequiv x^2pmod{I}$ on $m$ to reduce to a case where $i+i'+ell+ell'$ is smaller.
So, we may assume that $i'=ell=k=0$. Our two equations then reduce to $$3i+2j=2j'+k'$$ and $$j=j'+2k'+3ell'.$$ But now observe that the second equation implies $2jgeq 2j'+k'$ which makes the first equation impossible since $i>0$. Thus this case is impossible, completing the proof.
answered Jan 16 at 5:34
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
Small typo: you seem to have written "kernel of $f$ contains $I$" when you meant "kernel of $f$ is contained in $I$" the second time. Otherwise thanks! This is exactly what I was looking for.
$endgroup$
– C.D.
Jan 17 at 17:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suspect there are fancier more systematic ways to prove it, but here's a simple hands-on proof. By the easy direction we have a surjective homomorphism $f:Rto S$ where $R=k[w,x,y,z]$ and $S=k[a^3, a^2 b, a b^2, b^3]$, and the kernel of $f$ contains $I$. To show that the kernel of $f$ is contained in $I$, it suffices to show the following.
Claim: If monomials $m=w^i x^j y^k z^ell$ and $m'=w^{i'}x^{j'}y^{k'}z^{ell'}$ are such that $f(m)=f(m')$, then $mequiv m'pmod I$.
Why does this suffice? Well, since $f$ maps monomials to monomials, if $pin ker f$, then the monomials in $p$ must map to identical monomials in $S$ whose coefficients cancel out (since monomials in $S$ are linearly independent, that's the only way $f(p)$ can be $0$). But then the Claim would show that the monomials of $p$ also cancel out mod $I$, so $pin I$.
So, let's prove the Claim. We have $f(m)=a^{3i+2j+k}b^{j+2k+3ell}$ and similarly for $m'$, so we must have $$3i+2j+k=3i'+2j'+k'$$ and $$j+2k+3ell=j'+2k'+3ell'.$$
We will use induction on $i+i'+ell+ell'$. Let us first suppose that $i+i'+ell+ell'=0$. Then our equations are just $2j+k=2j'+k'$ and $j+2k=j'+2k'$. This easily implies $j=j'$ and $k=k'$, and so $m=m'$ and we are done.
Now suppose $i+i'+ell+ell'>0$; we may assume without loss of generality that $i>0$. If $i'>0$, we can cancel $w$ from both $m$ and $m'$ to reduce to a case where $i+i'+ell+ell'$ is smaller. If $ell>0$, we can use the relation $wzequiv xypmod{I}$ on $m$ to decrease $i$ and $ell$ by $1$ and increase $j$ and $k$ by $1$, reducing to a case where $i+i'+ell+ell'$ is smaller. Similarly if $k>0$ we can use the relation $wyequiv x^2pmod{I}$ on $m$ to reduce to a case where $i+i'+ell+ell'$ is smaller.
So, we may assume that $i'=ell=k=0$. Our two equations then reduce to $$3i+2j=2j'+k'$$ and $$j=j'+2k'+3ell'.$$ But now observe that the second equation implies $2jgeq 2j'+k'$ which makes the first equation impossible since $i>0$. Thus this case is impossible, completing the proof.
answered Jan 16 at 5:34
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
Small typo: you seem to have written "kernel of $f$ contains $I$" when you meant "kernel of $f$ is contained in $I$" the second time. Otherwise thanks! This is exactly what I was looking for.
$endgroup$
– C.D.
Jan 17 at 17:18
add a comment |
$begingroup$
I suspect there are fancier more systematic ways to prove it, but here's a simple hands-on proof. By the easy direction we have a surjective homomorphism $f:Rto S$ where $R=k[w,x,y,z]$ and $S=k[a^3, a^2 b, a b^2, b^3]$, and the kernel of $f$ contains $I$. To show that the kernel of $f$ is contained in $I$, it suffices to show the following.
Claim: If monomials $m=w^i x^j y^k z^ell$ and $m'=w^{i'}x^{j'}y^{k'}z^{ell'}$ are such that $f(m)=f(m')$, then $mequiv m'pmod I$.
Why does this suffice? Well, since $f$ maps monomials to monomials, if $pin ker f$, then the monomials in $p$ must map to identical monomials in $S$ whose coefficients cancel out (since monomials in $S$ are linearly independent, that's the only way $f(p)$ can be $0$). But then the Claim would show that the monomials of $p$ also cancel out mod $I$, so $pin I$.
So, let's prove the Claim. We have $f(m)=a^{3i+2j+k}b^{j+2k+3ell}$ and similarly for $m'$, so we must have $$3i+2j+k=3i'+2j'+k'$$ and $$j+2k+3ell=j'+2k'+3ell'.$$
We will use induction on $i+i'+ell+ell'$. Let us first suppose that $i+i'+ell+ell'=0$. Then our equations are just $2j+k=2j'+k'$ and $j+2k=j'+2k'$. This easily implies $j=j'$ and $k=k'$, and so $m=m'$ and we are done.
Now suppose $i+i'+ell+ell'>0$; we may assume without loss of generality that $i>0$. If $i'>0$, we can cancel $w$ from both $m$ and $m'$ to reduce to a case where $i+i'+ell+ell'$ is smaller. If $ell>0$, we can use the relation $wzequiv xypmod{I}$ on $m$ to decrease $i$ and $ell$ by $1$ and increase $j$ and $k$ by $1$, reducing to a case where $i+i'+ell+ell'$ is smaller. Similarly if $k>0$ we can use the relation $wyequiv x^2pmod{I}$ on $m$ to reduce to a case where $i+i'+ell+ell'$ is smaller.
So, we may assume that $i'=ell=k=0$. Our two equations then reduce to $$3i+2j=2j'+k'$$ and $$j=j'+2k'+3ell'.$$ But now observe that the second equation implies $2jgeq 2j'+k'$ which makes the first equation impossible since $i>0$. Thus this case is impossible, completing the proof.
answered Jan 16 at 5:34
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
Small typo: you seem to have written "kernel of $f$ contains $I$" when you meant "kernel of $f$ is contained in $I$" the second time. Otherwise thanks! This is exactly what I was looking for.
$endgroup$
– C.D.
Jan 17 at 17:18
add a comment |
$begingroup$
I suspect there are fancier more systematic ways to prove it, but here's a simple hands-on proof. By the easy direction we have a surjective homomorphism $f:Rto S$ where $R=k[w,x,y,z]$ and $S=k[a^3, a^2 b, a b^2, b^3]$, and the kernel of $f$ contains $I$. To show that the kernel of $f$ is contained in $I$, it suffices to show the following.
Claim: If monomials $m=w^i x^j y^k z^ell$ and $m'=w^{i'}x^{j'}y^{k'}z^{ell'}$ are such that $f(m)=f(m')$, then $mequiv m'pmod I$.
Why does this suffice? Well, since $f$ maps monomials to monomials, if $pin ker f$, then the monomials in $p$ must map to identical monomials in $S$ whose coefficients cancel out (since monomials in $S$ are linearly independent, that's the only way $f(p)$ can be $0$). But then the Claim would show that the monomials of $p$ also cancel out mod $I$, so $pin I$.
So, let's prove the Claim. We have $f(m)=a^{3i+2j+k}b^{j+2k+3ell}$ and similarly for $m'$, so we must have $$3i+2j+k=3i'+2j'+k'$$ and $$j+2k+3ell=j'+2k'+3ell'.$$
We will use induction on $i+i'+ell+ell'$. Let us first suppose that $i+i'+ell+ell'=0$. Then our equations are just $2j+k=2j'+k'$ and $j+2k=j'+2k'$. This easily implies $j=j'$ and $k=k'$, and so $m=m'$ and we are done.
Now suppose $i+i'+ell+ell'>0$; we may assume without loss of generality that $i>0$. If $i'>0$, we can cancel $w$ from both $m$ and $m'$ to reduce to a case where $i+i'+ell+ell'$ is smaller. If $ell>0$, we can use the relation $wzequiv xypmod{I}$ on $m$ to decrease $i$ and $ell$ by $1$ and increase $j$ and $k$ by $1$, reducing to a case where $i+i'+ell+ell'$ is smaller. Similarly if $k>0$ we can use the relation $wyequiv x^2pmod{I}$ on $m$ to reduce to a case where $i+i'+ell+ell'$ is smaller.
So, we may assume that $i'=ell=k=0$. Our two equations then reduce to $$3i+2j=2j'+k'$$ and $$j=j'+2k'+3ell'.$$ But now observe that the second equation implies $2jgeq 2j'+k'$ which makes the first equation impossible since $i>0$. Thus this case is impossible, completing the proof.
answered Jan 16 at 5:34
Eric WofseyEric Wofsey
187k14215344
$endgroup$
I suspect there are fancier more systematic ways to prove it, but here's a simple hands-on proof. By the easy direction we have a surjective homomorphism $f:Rto S$ where $R=k[w,x,y,z]$ and $S=k[a^3, a^2 b, a b^2, b^3]$, and the kernel of $f$ contains $I$. To show that the kernel of $f$ is contained in $I$, it suffices to show the following.
Claim: If monomials $m=w^i x^j y^k z^ell$ and $m'=w^{i'}x^{j'}y^{k'}z^{ell'}$ are such that $f(m)=f(m')$, then $mequiv m'pmod I$.
Why does this suffice? Well, since $f$ maps monomials to monomials, if $pin ker f$, then the monomials in $p$ must map to identical monomials in $S$ whose coefficients cancel out (since monomials in $S$ are linearly independent, that's the only way $f(p)$ can be $0$). But then the Claim would show that the monomials of $p$ also cancel out mod $I$, so $pin I$.
So, let's prove the Claim. We have $f(m)=a^{3i+2j+k}b^{j+2k+3ell}$ and similarly for $m'$, so we must have $$3i+2j+k=3i'+2j'+k'$$ and $$j+2k+3ell=j'+2k'+3ell'.$$
We will use induction on $i+i'+ell+ell'$. Let us first suppose that $i+i'+ell+ell'=0$. Then our equations are just $2j+k=2j'+k'$ and $j+2k=j'+2k'$. This easily implies $j=j'$ and $k=k'$, and so $m=m'$ and we are done.
Now suppose $i+i'+ell+ell'>0$; we may assume without loss of generality that $i>0$. If $i'>0$, we can cancel $w$ from both $m$ and $m'$ to reduce to a case where $i+i'+ell+ell'$ is smaller. If $ell>0$, we can use the relation $wzequiv xypmod{I}$ on $m$ to decrease $i$ and $ell$ by $1$ and increase $j$ and $k$ by $1$, reducing to a case where $i+i'+ell+ell'$ is smaller. Similarly if $k>0$ we can use the relation $wyequiv x^2pmod{I}$ on $m$ to reduce to a case where $i+i'+ell+ell'$ is smaller.
So, we may assume that $i'=ell=k=0$. Our two equations then reduce to $$3i+2j=2j'+k'$$ and $$j=j'+2k'+3ell'.$$ But now observe that the second equation implies $2jgeq 2j'+k'$ which makes the first equation impossible since $i>0$. Thus this case is impossible, completing the proof.
answered Jan 16 at 5:34
Eric WofseyEric Wofsey
187k14215344
edited Jan 17 at 17:19
answered Jan 16 at 5:34
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 5:34
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 5:34
Eric WofseyEric Wofsey
187k14215344
187k14215344
$begingroup$
Small typo: you seem to have written "kernel of $f$ contains $I$" when you meant "kernel of $f$ is contained in $I$" the second time. Otherwise thanks! This is exactly what I was looking for.
$endgroup$
– C.D.
Jan 17 at 17:18
add a comment |
$begingroup$
Small typo: you seem to have written "kernel of $f$ contains $I$" when you meant "kernel of $f$ is contained in $I$" the second time. Otherwise thanks! This is exactly what I was looking for.
$endgroup$
– C.D.
Jan 17 at 17:18
$begingroup$
Small typo: you seem to have written "kernel of $f$ contains $I$" when you meant "kernel of $f$ is contained in $I$" the second time. Otherwise thanks! This is exactly what I was looking for.
$endgroup$
– C.D.
Jan 17 at 17:18
$begingroup$
Small typo: you seem to have written "kernel of $f$ contains $I$" when you meant "kernel of $f$ is contained in $I$" the second time. Otherwise thanks! This is exactly what I was looking for.
$endgroup$
– C.D.
Jan 17 at 17:18
add a comment |
$begingroup$
I think this post may answer your question.
$endgroup$
– André 3000
Jan 16 at 6:42