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Hartshorne Problem 1.2.14 on Segre Embedding
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This is a problem in Hartshorne concerning showing that the image of $Bbb{P}^n times Bbb{P}^m$ under the Segre embedding $psi$ is actually irreducible. Now I have shown with some effort that $psi(Bbb{P}^n times Bbb{P}^m)$ is actually equal to $V(mathfrak{a})$ where $mathfrak{a}$ is the ideal generated by the set of all monomials
$$Big{z_{ij}z_{kl} - z_{il}z_{kj} hspace{1mm} Big| hspace{1mm} i,k = 0,ldots, n; hspace{2mm} j,l = 0,ldots,mBig}.$$
My main problem now is in showing that $mathfrak{a}$ is actually equal to the kernel of the ring homomorphism $$varphi : k[z_{ij}] to k[x_0,ldots,x_n,y_0,ldots,y_m]$$
that sends $z_{ij}$ to $x_iy_j$.
I have spent quite a few hours playing around with monomial orderings and trying to show that $mathfrak{a} supseteq ker varphi$ but to no avail. Of course the other inclusion is immediate.
Is there anything I can do apart from playing around with monomial orderings to try and show that the kernel of $varphi$ is equal to $mathfrak{a}$? Perhaps maybe something along the lines of inducting on $n$, those this does not look promising.
Note: Please do not close this question; my question is different from the other questions on this site concerning the Segre embedding. Also, I can show my work concerning how I arrived at the conclusion that $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^n)$.
algebraic-geometry commutative-algebra invariant-theory algebraic-combinatorics
edited yesterday
darij grinberg
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This is a problem in Hartshorne concerning showing that the image of $Bbb{P}^n times Bbb{P}^m$ under the Segre embedding $psi$ is actually irreducible. Now I have shown with some effort that $psi(Bbb{P}^n times Bbb{P}^m)$ is actually equal to $V(mathfrak{a})$ where $mathfrak{a}$ is the ideal generated by the set of all monomials
$$Big{z_{ij}z_{kl} - z_{il}z_{kj} hspace{1mm} Big| hspace{1mm} i,k = 0,ldots, n; hspace{2mm} j,l = 0,ldots,mBig}.$$
My main problem now is in showing that $mathfrak{a}$ is actually equal to the kernel of the ring homomorphism $$varphi : k[z_{ij}] to k[x_0,ldots,x_n,y_0,ldots,y_m]$$
that sends $z_{ij}$ to $x_iy_j$.
I have spent quite a few hours playing around with monomial orderings and trying to show that $mathfrak{a} supseteq ker varphi$ but to no avail. Of course the other inclusion is immediate.
Is there anything I can do apart from playing around with monomial orderings to try and show that the kernel of $varphi$ is equal to $mathfrak{a}$? Perhaps maybe something along the lines of inducting on $n$, those this does not look promising.
Note: Please do not close this question; my question is different from the other questions on this site concerning the Segre embedding. Also, I can show my work concerning how I arrived at the conclusion that $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^n)$.
algebraic-geometry commutative-algebra invariant-theory algebraic-combinatorics
edited yesterday
darij grinberg
9,81032961
If I remember it correctly, you don't even need to show that $mathfrak{a}=ker(varphi)$, you can prove $V(ker(varphi))=operatorname{im}(psi)$ directly. Could you show us roughly how you proved $V(mathfrak{a})=operatorname{im}(psi)$?
– Nils Matthes
Apr 7 '13 at 14:37
@NilsMatthes I have posted an answer below.
– user38268
Apr 8 '13 at 5:22
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up vote
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up vote
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down vote
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This is a problem in Hartshorne concerning showing that the image of $Bbb{P}^n times Bbb{P}^m$ under the Segre embedding $psi$ is actually irreducible. Now I have shown with some effort that $psi(Bbb{P}^n times Bbb{P}^m)$ is actually equal to $V(mathfrak{a})$ where $mathfrak{a}$ is the ideal generated by the set of all monomials
$$Big{z_{ij}z_{kl} - z_{il}z_{kj} hspace{1mm} Big| hspace{1mm} i,k = 0,ldots, n; hspace{2mm} j,l = 0,ldots,mBig}.$$
My main problem now is in showing that $mathfrak{a}$ is actually equal to the kernel of the ring homomorphism $$varphi : k[z_{ij}] to k[x_0,ldots,x_n,y_0,ldots,y_m]$$
that sends $z_{ij}$ to $x_iy_j$.
I have spent quite a few hours playing around with monomial orderings and trying to show that $mathfrak{a} supseteq ker varphi$ but to no avail. Of course the other inclusion is immediate.
Is there anything I can do apart from playing around with monomial orderings to try and show that the kernel of $varphi$ is equal to $mathfrak{a}$? Perhaps maybe something along the lines of inducting on $n$, those this does not look promising.
Note: Please do not close this question; my question is different from the other questions on this site concerning the Segre embedding. Also, I can show my work concerning how I arrived at the conclusion that $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^n)$.
algebraic-geometry commutative-algebra invariant-theory algebraic-combinatorics
edited yesterday
darij grinberg
9,81032961
This is a problem in Hartshorne concerning showing that the image of $Bbb{P}^n times Bbb{P}^m$ under the Segre embedding $psi$ is actually irreducible. Now I have shown with some effort that $psi(Bbb{P}^n times Bbb{P}^m)$ is actually equal to $V(mathfrak{a})$ where $mathfrak{a}$ is the ideal generated by the set of all monomials
$$Big{z_{ij}z_{kl} - z_{il}z_{kj} hspace{1mm} Big| hspace{1mm} i,k = 0,ldots, n; hspace{2mm} j,l = 0,ldots,mBig}.$$
My main problem now is in showing that $mathfrak{a}$ is actually equal to the kernel of the ring homomorphism $$varphi : k[z_{ij}] to k[x_0,ldots,x_n,y_0,ldots,y_m]$$
that sends $z_{ij}$ to $x_iy_j$.
I have spent quite a few hours playing around with monomial orderings and trying to show that $mathfrak{a} supseteq ker varphi$ but to no avail. Of course the other inclusion is immediate.
Is there anything I can do apart from playing around with monomial orderings to try and show that the kernel of $varphi$ is equal to $mathfrak{a}$? Perhaps maybe something along the lines of inducting on $n$, those this does not look promising.
Note: Please do not close this question; my question is different from the other questions on this site concerning the Segre embedding. Also, I can show my work concerning how I arrived at the conclusion that $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^n)$.
algebraic-geometry commutative-algebra invariant-theory algebraic-combinatorics
algebraic-geometry commutative-algebra invariant-theory algebraic-combinatorics
edited yesterday
darij grinberg
9,81032961
edited yesterday
darij grinberg
9,81032961
edited yesterday
darij grinberg
9,81032961
edited yesterday
darij grinberg
9,81032961
edited yesterday
darij grinberg
9,81032961
9,81032961
asked Apr 7 '13 at 13:38
user38268
If I remember it correctly, you don't even need to show that $mathfrak{a}=ker(varphi)$, you can prove $V(ker(varphi))=operatorname{im}(psi)$ directly. Could you show us roughly how you proved $V(mathfrak{a})=operatorname{im}(psi)$?
– Nils Matthes
Apr 7 '13 at 14:37
@NilsMatthes I have posted an answer below.
– user38268
Apr 8 '13 at 5:22
add a comment |
If I remember it correctly, you don't even need to show that $mathfrak{a}=ker(varphi)$, you can prove $V(ker(varphi))=operatorname{im}(psi)$ directly. Could you show us roughly how you proved $V(mathfrak{a})=operatorname{im}(psi)$?
– Nils Matthes
Apr 7 '13 at 14:37
@NilsMatthes I have posted an answer below.
– user38268
Apr 8 '13 at 5:22
If I remember it correctly, you don't even need to show that $mathfrak{a}=ker(varphi)$, you can prove $V(ker(varphi))=operatorname{im}(psi)$ directly. Could you show us roughly how you proved $V(mathfrak{a})=operatorname{im}(psi)$?
– Nils Matthes
Apr 7 '13 at 14:37
If I remember it correctly, you don't even need to show that $mathfrak{a}=ker(varphi)$, you can prove $V(ker(varphi))=operatorname{im}(psi)$ directly. Could you show us roughly how you proved $V(mathfrak{a})=operatorname{im}(psi)$?
– Nils Matthes
Apr 7 '13 at 14:37
@NilsMatthes I have posted an answer below.
– user38268
Apr 8 '13 at 5:22
@NilsMatthes I have posted an answer below.
– user38268
Apr 8 '13 at 5:22
add a comment |
3 Answers
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I outlined a proof by Bjorn Poonen at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2450856#p2450856 . As Nils Matthes has noticed, computing the kernel is not necessary to solve Hartshorne's problem, though (in my opinion) it is more interesting than the problem itself.
$newcommand{kk}{mathbb{k}}$
$newcommand{Ker}{operatorname{Ker}}$
For the sake of self-containedness, let me repost the proof linked above here (with improved notations). I begin by restating the problem:
Theorem 1. Let $kk$ be a commutative ring with $1$. Let $n$ and $m$ be two nonnegative integers. Let $M=left{0,1,ldots ,mright}$ and $N=left{0,1,ldots ,nright}$.
Let $R$ be the polynomial ring $kkleft[Z_{i,j} mid i in M text{ and } j in N right]$.
Let $S$ be the polynomial ring $kkleft[x_0, x_1, ldots, x_m, y_0, y_1, ldots, y_n right]$.
Let $phi : R to S$ be the unique $kk$-algebra homomorphism that sends each $Z_{i,j}$ to $x_i y_j$.
Let $W$ be the ideal of $R$ generated by all elements of the type $Z_{a,b} Z_{c,d} - Z_{a,d} Z_{c,b}$ with $a in M$, $b in N$, $c in M$ and $d in N$.
Then, $Ker phi = W$.
To prove this, we shall use the following easy algebraic lemma:
Lemma 2. Let $C$ be a $kk$-module. Let $A$ and $B$ be two submodules of $C$ such that $C=A+B$. Let $psi$ be a $kk$-module map from $C$ to another $kk$-module $D$ such that $psimid_A$ is injective and $Ker psi supseteq B$. Then, $Ker psi = B$.
Proof of Lemma 2. Let $c in Ker psi$. Thus, $c in Ker psi subseteq C = A + B$; hence, we can write $c$ in the form $c = a + b$ for some $a in A$ and $b in B$. Consider these $a$ and $b$. We have $b in B subseteq Ker psi$, so that $psileft(bright) = 0$. Applying the map $psi$ to the equality $c = a + b$, we obtain $psileft(cright) = psileft(a + bright) = psileft(aright) + psileft(bright)$ (since $psi$ is a $kk$-module map). Comparing this with $psileft(cright) = 0$ (which follows from $c in Ker psi$), we obtain $0 = psileft(aright) + underbrace{psileft(bright)}_{= 0} = psileft(aright)$, so that $psileft(aright) = 0 = psileft(0right)$. Since $psimid_A$ is injective, this entails $a = 0$ (because both $a$ and $0$ belong to $A$). Thus, $c = underbrace{a}_{=0} + b = b in B$.
Now, forget that we fixed $c$. We thus have shown that $c in B$ for each $c in Ker psi$. Thus, $Ker psi subseteq B$. Combining this with $Ker psi supseteq B$, we obtain $Ker psi = B$. This proves Lemma 2. $blacksquare$
Proof of Theorem 1 (Bjorn Poonen) (sketched). We notice that $Ker phisupseteq W$ is very easy to prove (in fact, a trivial computation shows that $Ker phi$ contains $Z_{a,b}Z_{c,d}-Z_{a,d}Z_{c,b}$ for all $a$, $b$, $c$, $d$).
We order the set $Mtimes N$ lexicographically.
If $k$ is a nonnegative integer, then $S_k$ shall denote the symmetric group consisting of all permutations of $left{1,2,ldots,kright}$.
Every $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots ,left(a_k,b_kright)right) in left(Mtimes Nright)^k$ and every permutation $sigma in S_k$ satisfy
begin{equation}
Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}} mod W .
label{darij.pf.thm1.1}
tag{1}
end{equation}
(In fact, this is obvious from the definition of $W$ when $sigma$ is a transposition, and hence, by induction, it also holds for every permutation $sigma$, because every permutation is a composition of transpositions.)
Now, let $T$ be the $kk$-submodule of $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generated by all products of the form $Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}$ with $k$ being a nonnegative integer and $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in left(Mtimes Nright)^k$ being a $k$-tuple satisfying $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$. It is easy to see that the map $left.phimid_Tright. : T to kkleft[x_0,x_1,ldots,x_m,y_0,y_1,ldots,y_nright]$ is injective. (In fact, if $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in T$, then
begin{align}
left(phimid_Tright)left(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right)
&= phileft(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right) \
&= x_{a_1}y_{c_1}x_{a_2}y_{c_2}cdots x_{a_k}y_{c_k} \
&=x_{a_1}x_{a_2}cdots x_{a_k}y_{c_1}y_{c_2}cdots y_{c_k}
end{align}
is a monomial from which we can recover the $k$-tuple $left(a_1,a_2,ldots ,a_kright)$ up to order and the $k$-tuple $left(c_1,c_2,ldots ,c_kright)$ up to order; but since the order of each of these two $k$-tuples is predetermined by the condition that $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we can therefore recover these two $k$-tuples completely; hence, the map $phimid_T$ sends distinct monomials to distinct monomials, and thus is injective.)
Next we are going to show that:
begin{equation}
text{every monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ lies in $T+W$.}
label{darij.pf.thm1.2}
tag{2}
end{equation}
[Proof of eqref{darij.pf.thm1.2}: Let $mu$ be any monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$. Then, $mu=Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k}$ for some nonnegative integer $k$ and some $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)in left(Mtimes Nright)^k$ such that $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$. Consider such a $k$ and such a $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)$. Since $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$, we have $a_1leq a_2leq cdotsleq a_k$ (since our order is lexicographic). Clearly there exists a permutation $sigmain S_k$ such that $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$. Consider such a $sigma$. Let $c_i=b_{sigma i}$ for every $iinleft{1,2,ldots,kright}$. Hence, the chain of inequalities $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$ rewrites as $c_1leq c_2leq cdotsleq c_k$. Also,
begin{align}
mu
&= Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} \
&equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}}
qquad left( text{by eqref{darij.pf.thm1.1}} right) \
&= Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k} mod W
end{align}
(since $b_{sigma i}=c_i$ for every $iinleft{1,2,ldots,kright}$).
But since $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we have $Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k}in T$ (by the definition of $T$), so this rewrites as follows:
begin{align}
mu
&equiv left(text{an element of }Tright)mod W .
end{align}
In other words, $muin T+W$. Since this holds for every monomial $mu$ in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, this proves eqref{darij.pf.thm1.2}.]
Since the monomials in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generate the $kk$-module $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, and since $T+W$ is a submodule of this $kk$-module, we obtain $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]=T+W$ from eqref{darij.pf.thm1.2}.
Applying Lemma 2 to $C=kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, $A=T$, $B=W$ and $psi=phi$, we thus conclude that $Ker phi = W$. This proves Theorem 1. $blacksquare$
There is yet another way to prove Theorem 1 -- namely, by revealing it to be a particular case of the Second Fundamental Theorem of Invariant Theory for GL. See https://mathoverflow.net/questions/202005/a-vector-version-of-the-segre-embedding-what-is-the-kernel-of-the-ring-map for this generalization. (Another place where this generalization appears with proof is Theorem 5.1 of J. Désarménien, Joseph P. S. Kung, Gian-Carlo Rota, Invariant Theory, Young Bitableaux, and Combinatorics, unofficial re-edition 2017; you just need to set $d = 1$, and realize that every standard $left(mathcal{X},mathcal{U}right)$-bideterminant of shape strictly longer than $left(dright)$ contains at least one row of length $geq 2$, which is easily seen to place it inside the ideal $W$.)
answered Apr 8 '13 at 5:42
darij grinberg
9,81032961
3
Dear Darij, thanks for your answer. I posted an answer below showing that we don't need to prove explicitly that the kernel is a prime ideal to show that the image of the Segre embedding is a projective variety. It should be correct. Regards,
– user38268
Apr 8 '13 at 5:55
Regarding OP's original solution - why is the observation that $operatorname{Im}psi$ equals the vanishing locus of the ideal $I$ generated by the mentioned monomials, along with the observation $I$ is prime not enough?
– Arrow
2 days ago
@Arrow: how do you prove $I$ is prime?
– darij grinberg
2 days ago
thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities?
– user347489
yesterday
@user347489: No, I don't. Is anything wrong with the argument?
– darij grinberg
yesterday
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Indeed as Nils Matthes suggested we don't know need to go through such a mess and just use the hint of Hartshorne.
Define a map $varphi : k[T_{00} , ldots, T_{nm}] to k[X_0,ldots,X_n,Y_0,ldots,Y_m]$ that sends $T_{ij}$ to $X_iY_j$. Let $mathfrak{a} := ker varphi$. We claim that $psi (Bbb{P}^n times Bbb{P}^m) = V(mathfrak{a})$. For one inclusion if a point
$$a = [a_{00} : ldots : a_{nm}] in V(mathfrak{a})$$
then in particular $a$ is a zero of all the polynomials $T_{ij}T_{kl} - T_{il}T_{kj}$. But this means that $a in psi(Bbb{P}^n times Bbb{P}^m)$. The reverse inclusion follows immediately from the definition of $varphi$. Thus $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^m)$ and the projective Nullstellensatz implies that $psi(Bbb{P}^n times Bbb{P}^m)$ is a projective variety.
edited Oct 30 '14 at 10:16
user26857
39.1k123882
1
Shouldn't you prove that $mathfrak a$ is actually a homogeneous ideal?
– Andrea Gagna
May 15 '14 at 15:46
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It's easy to show the preimage under the Segre map of an algebraic set is an algebraic subset of $mathbb{P}^n times mathbb{P}^m.$ If $V$ were reducible, we could write $V = V_1 cup V_2,$ where $V_ineq V$ are closed. Then $$ mathbb{P}^n times mathbb{P}^m = S^{-1}(V) = S^{-1} (V_1) bigcup S^{-1}(V_2)$$ where $S^{-1}(V_i)$ are closed. Picking $x_i in Vsetminus V_i$ we have $S^{-1}(x_i)cap S^{-1}(V_i)=emptyset$ so $S^{-1}(V_i) subset S^{-1}(V)$ is a strict inclusion, contradicting that $mathbb{P}^n times mathbb{P}^m$ is irreducible.
answered Jun 3 '13 at 9:20
Ragib Zaman
28.3k34889
3
Of course we do, it's the same condition as always in any topological space: A space $X$ is reducible if $X= X_1 cup X_2$ for some closed proper subsets $X_1, X_2$ and irreducible if not reducible. A subset $V$ of $mathbb{P}^n times mathbb{P}^m$ is closed if $V=V(S)$ where $S$ is a set of biforms in $k[X_0, cdots, X_n, Y_0, cdots, Y_m].$ This already establishes what it means for $mathbb{P}^n times mathbb{P}^m$ to be irreducible.
– Ragib Zaman
Jun 3 '13 at 10:11
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I outlined a proof by Bjorn Poonen at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2450856#p2450856 . As Nils Matthes has noticed, computing the kernel is not necessary to solve Hartshorne's problem, though (in my opinion) it is more interesting than the problem itself.
$newcommand{kk}{mathbb{k}}$
$newcommand{Ker}{operatorname{Ker}}$
For the sake of self-containedness, let me repost the proof linked above here (with improved notations). I begin by restating the problem:
Theorem 1. Let $kk$ be a commutative ring with $1$. Let $n$ and $m$ be two nonnegative integers. Let $M=left{0,1,ldots ,mright}$ and $N=left{0,1,ldots ,nright}$.
Let $R$ be the polynomial ring $kkleft[Z_{i,j} mid i in M text{ and } j in N right]$.
Let $S$ be the polynomial ring $kkleft[x_0, x_1, ldots, x_m, y_0, y_1, ldots, y_n right]$.
Let $phi : R to S$ be the unique $kk$-algebra homomorphism that sends each $Z_{i,j}$ to $x_i y_j$.
Let $W$ be the ideal of $R$ generated by all elements of the type $Z_{a,b} Z_{c,d} - Z_{a,d} Z_{c,b}$ with $a in M$, $b in N$, $c in M$ and $d in N$.
Then, $Ker phi = W$.
To prove this, we shall use the following easy algebraic lemma:
Lemma 2. Let $C$ be a $kk$-module. Let $A$ and $B$ be two submodules of $C$ such that $C=A+B$. Let $psi$ be a $kk$-module map from $C$ to another $kk$-module $D$ such that $psimid_A$ is injective and $Ker psi supseteq B$. Then, $Ker psi = B$.
Proof of Lemma 2. Let $c in Ker psi$. Thus, $c in Ker psi subseteq C = A + B$; hence, we can write $c$ in the form $c = a + b$ for some $a in A$ and $b in B$. Consider these $a$ and $b$. We have $b in B subseteq Ker psi$, so that $psileft(bright) = 0$. Applying the map $psi$ to the equality $c = a + b$, we obtain $psileft(cright) = psileft(a + bright) = psileft(aright) + psileft(bright)$ (since $psi$ is a $kk$-module map). Comparing this with $psileft(cright) = 0$ (which follows from $c in Ker psi$), we obtain $0 = psileft(aright) + underbrace{psileft(bright)}_{= 0} = psileft(aright)$, so that $psileft(aright) = 0 = psileft(0right)$. Since $psimid_A$ is injective, this entails $a = 0$ (because both $a$ and $0$ belong to $A$). Thus, $c = underbrace{a}_{=0} + b = b in B$.
Now, forget that we fixed $c$. We thus have shown that $c in B$ for each $c in Ker psi$. Thus, $Ker psi subseteq B$. Combining this with $Ker psi supseteq B$, we obtain $Ker psi = B$. This proves Lemma 2. $blacksquare$
Proof of Theorem 1 (Bjorn Poonen) (sketched). We notice that $Ker phisupseteq W$ is very easy to prove (in fact, a trivial computation shows that $Ker phi$ contains $Z_{a,b}Z_{c,d}-Z_{a,d}Z_{c,b}$ for all $a$, $b$, $c$, $d$).
We order the set $Mtimes N$ lexicographically.
If $k$ is a nonnegative integer, then $S_k$ shall denote the symmetric group consisting of all permutations of $left{1,2,ldots,kright}$.
Every $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots ,left(a_k,b_kright)right) in left(Mtimes Nright)^k$ and every permutation $sigma in S_k$ satisfy
begin{equation}
Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}} mod W .
label{darij.pf.thm1.1}
tag{1}
end{equation}
(In fact, this is obvious from the definition of $W$ when $sigma$ is a transposition, and hence, by induction, it also holds for every permutation $sigma$, because every permutation is a composition of transpositions.)
Now, let $T$ be the $kk$-submodule of $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generated by all products of the form $Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}$ with $k$ being a nonnegative integer and $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in left(Mtimes Nright)^k$ being a $k$-tuple satisfying $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$. It is easy to see that the map $left.phimid_Tright. : T to kkleft[x_0,x_1,ldots,x_m,y_0,y_1,ldots,y_nright]$ is injective. (In fact, if $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in T$, then
begin{align}
left(phimid_Tright)left(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right)
&= phileft(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right) \
&= x_{a_1}y_{c_1}x_{a_2}y_{c_2}cdots x_{a_k}y_{c_k} \
&=x_{a_1}x_{a_2}cdots x_{a_k}y_{c_1}y_{c_2}cdots y_{c_k}
end{align}
is a monomial from which we can recover the $k$-tuple $left(a_1,a_2,ldots ,a_kright)$ up to order and the $k$-tuple $left(c_1,c_2,ldots ,c_kright)$ up to order; but since the order of each of these two $k$-tuples is predetermined by the condition that $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we can therefore recover these two $k$-tuples completely; hence, the map $phimid_T$ sends distinct monomials to distinct monomials, and thus is injective.)
Next we are going to show that:
begin{equation}
text{every monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ lies in $T+W$.}
label{darij.pf.thm1.2}
tag{2}
end{equation}
[Proof of eqref{darij.pf.thm1.2}: Let $mu$ be any monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$. Then, $mu=Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k}$ for some nonnegative integer $k$ and some $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)in left(Mtimes Nright)^k$ such that $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$. Consider such a $k$ and such a $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)$. Since $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$, we have $a_1leq a_2leq cdotsleq a_k$ (since our order is lexicographic). Clearly there exists a permutation $sigmain S_k$ such that $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$. Consider such a $sigma$. Let $c_i=b_{sigma i}$ for every $iinleft{1,2,ldots,kright}$. Hence, the chain of inequalities $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$ rewrites as $c_1leq c_2leq cdotsleq c_k$. Also,
begin{align}
mu
&= Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} \
&equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}}
qquad left( text{by eqref{darij.pf.thm1.1}} right) \
&= Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k} mod W
end{align}
(since $b_{sigma i}=c_i$ for every $iinleft{1,2,ldots,kright}$).
But since $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we have $Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k}in T$ (by the definition of $T$), so this rewrites as follows:
begin{align}
mu
&equiv left(text{an element of }Tright)mod W .
end{align}
In other words, $muin T+W$. Since this holds for every monomial $mu$ in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, this proves eqref{darij.pf.thm1.2}.]
Since the monomials in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generate the $kk$-module $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, and since $T+W$ is a submodule of this $kk$-module, we obtain $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]=T+W$ from eqref{darij.pf.thm1.2}.
Applying Lemma 2 to $C=kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, $A=T$, $B=W$ and $psi=phi$, we thus conclude that $Ker phi = W$. This proves Theorem 1. $blacksquare$
There is yet another way to prove Theorem 1 -- namely, by revealing it to be a particular case of the Second Fundamental Theorem of Invariant Theory for GL. See https://mathoverflow.net/questions/202005/a-vector-version-of-the-segre-embedding-what-is-the-kernel-of-the-ring-map for this generalization. (Another place where this generalization appears with proof is Theorem 5.1 of J. Désarménien, Joseph P. S. Kung, Gian-Carlo Rota, Invariant Theory, Young Bitableaux, and Combinatorics, unofficial re-edition 2017; you just need to set $d = 1$, and realize that every standard $left(mathcal{X},mathcal{U}right)$-bideterminant of shape strictly longer than $left(dright)$ contains at least one row of length $geq 2$, which is easily seen to place it inside the ideal $W$.)
answered Apr 8 '13 at 5:42
darij grinberg
9,81032961
3
Dear Darij, thanks for your answer. I posted an answer below showing that we don't need to prove explicitly that the kernel is a prime ideal to show that the image of the Segre embedding is a projective variety. It should be correct. Regards,
– user38268
Apr 8 '13 at 5:55
Regarding OP's original solution - why is the observation that $operatorname{Im}psi$ equals the vanishing locus of the ideal $I$ generated by the mentioned monomials, along with the observation $I$ is prime not enough?
– Arrow
2 days ago
@Arrow: how do you prove $I$ is prime?
– darij grinberg
2 days ago
thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities?
– user347489
yesterday
@user347489: No, I don't. Is anything wrong with the argument?
– darij grinberg
yesterday
add a comment |
up vote
9
down vote
I outlined a proof by Bjorn Poonen at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2450856#p2450856 . As Nils Matthes has noticed, computing the kernel is not necessary to solve Hartshorne's problem, though (in my opinion) it is more interesting than the problem itself.
$newcommand{kk}{mathbb{k}}$
$newcommand{Ker}{operatorname{Ker}}$
For the sake of self-containedness, let me repost the proof linked above here (with improved notations). I begin by restating the problem:
Theorem 1. Let $kk$ be a commutative ring with $1$. Let $n$ and $m$ be two nonnegative integers. Let $M=left{0,1,ldots ,mright}$ and $N=left{0,1,ldots ,nright}$.
Let $R$ be the polynomial ring $kkleft[Z_{i,j} mid i in M text{ and } j in N right]$.
Let $S$ be the polynomial ring $kkleft[x_0, x_1, ldots, x_m, y_0, y_1, ldots, y_n right]$.
Let $phi : R to S$ be the unique $kk$-algebra homomorphism that sends each $Z_{i,j}$ to $x_i y_j$.
Let $W$ be the ideal of $R$ generated by all elements of the type $Z_{a,b} Z_{c,d} - Z_{a,d} Z_{c,b}$ with $a in M$, $b in N$, $c in M$ and $d in N$.
Then, $Ker phi = W$.
To prove this, we shall use the following easy algebraic lemma:
Lemma 2. Let $C$ be a $kk$-module. Let $A$ and $B$ be two submodules of $C$ such that $C=A+B$. Let $psi$ be a $kk$-module map from $C$ to another $kk$-module $D$ such that $psimid_A$ is injective and $Ker psi supseteq B$. Then, $Ker psi = B$.
Proof of Lemma 2. Let $c in Ker psi$. Thus, $c in Ker psi subseteq C = A + B$; hence, we can write $c$ in the form $c = a + b$ for some $a in A$ and $b in B$. Consider these $a$ and $b$. We have $b in B subseteq Ker psi$, so that $psileft(bright) = 0$. Applying the map $psi$ to the equality $c = a + b$, we obtain $psileft(cright) = psileft(a + bright) = psileft(aright) + psileft(bright)$ (since $psi$ is a $kk$-module map). Comparing this with $psileft(cright) = 0$ (which follows from $c in Ker psi$), we obtain $0 = psileft(aright) + underbrace{psileft(bright)}_{= 0} = psileft(aright)$, so that $psileft(aright) = 0 = psileft(0right)$. Since $psimid_A$ is injective, this entails $a = 0$ (because both $a$ and $0$ belong to $A$). Thus, $c = underbrace{a}_{=0} + b = b in B$.
Now, forget that we fixed $c$. We thus have shown that $c in B$ for each $c in Ker psi$. Thus, $Ker psi subseteq B$. Combining this with $Ker psi supseteq B$, we obtain $Ker psi = B$. This proves Lemma 2. $blacksquare$
Proof of Theorem 1 (Bjorn Poonen) (sketched). We notice that $Ker phisupseteq W$ is very easy to prove (in fact, a trivial computation shows that $Ker phi$ contains $Z_{a,b}Z_{c,d}-Z_{a,d}Z_{c,b}$ for all $a$, $b$, $c$, $d$).
We order the set $Mtimes N$ lexicographically.
If $k$ is a nonnegative integer, then $S_k$ shall denote the symmetric group consisting of all permutations of $left{1,2,ldots,kright}$.
Every $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots ,left(a_k,b_kright)right) in left(Mtimes Nright)^k$ and every permutation $sigma in S_k$ satisfy
begin{equation}
Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}} mod W .
label{darij.pf.thm1.1}
tag{1}
end{equation}
(In fact, this is obvious from the definition of $W$ when $sigma$ is a transposition, and hence, by induction, it also holds for every permutation $sigma$, because every permutation is a composition of transpositions.)
Now, let $T$ be the $kk$-submodule of $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generated by all products of the form $Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}$ with $k$ being a nonnegative integer and $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in left(Mtimes Nright)^k$ being a $k$-tuple satisfying $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$. It is easy to see that the map $left.phimid_Tright. : T to kkleft[x_0,x_1,ldots,x_m,y_0,y_1,ldots,y_nright]$ is injective. (In fact, if $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in T$, then
begin{align}
left(phimid_Tright)left(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right)
&= phileft(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right) \
&= x_{a_1}y_{c_1}x_{a_2}y_{c_2}cdots x_{a_k}y_{c_k} \
&=x_{a_1}x_{a_2}cdots x_{a_k}y_{c_1}y_{c_2}cdots y_{c_k}
end{align}
is a monomial from which we can recover the $k$-tuple $left(a_1,a_2,ldots ,a_kright)$ up to order and the $k$-tuple $left(c_1,c_2,ldots ,c_kright)$ up to order; but since the order of each of these two $k$-tuples is predetermined by the condition that $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we can therefore recover these two $k$-tuples completely; hence, the map $phimid_T$ sends distinct monomials to distinct monomials, and thus is injective.)
Next we are going to show that:
begin{equation}
text{every monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ lies in $T+W$.}
label{darij.pf.thm1.2}
tag{2}
end{equation}
[Proof of eqref{darij.pf.thm1.2}: Let $mu$ be any monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$. Then, $mu=Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k}$ for some nonnegative integer $k$ and some $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)in left(Mtimes Nright)^k$ such that $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$. Consider such a $k$ and such a $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)$. Since $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$, we have $a_1leq a_2leq cdotsleq a_k$ (since our order is lexicographic). Clearly there exists a permutation $sigmain S_k$ such that $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$. Consider such a $sigma$. Let $c_i=b_{sigma i}$ for every $iinleft{1,2,ldots,kright}$. Hence, the chain of inequalities $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$ rewrites as $c_1leq c_2leq cdotsleq c_k$. Also,
begin{align}
mu
&= Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} \
&equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}}
qquad left( text{by eqref{darij.pf.thm1.1}} right) \
&= Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k} mod W
end{align}
(since $b_{sigma i}=c_i$ for every $iinleft{1,2,ldots,kright}$).
But since $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we have $Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k}in T$ (by the definition of $T$), so this rewrites as follows:
begin{align}
mu
&equiv left(text{an element of }Tright)mod W .
end{align}
In other words, $muin T+W$. Since this holds for every monomial $mu$ in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, this proves eqref{darij.pf.thm1.2}.]
Since the monomials in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generate the $kk$-module $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, and since $T+W$ is a submodule of this $kk$-module, we obtain $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]=T+W$ from eqref{darij.pf.thm1.2}.
Applying Lemma 2 to $C=kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, $A=T$, $B=W$ and $psi=phi$, we thus conclude that $Ker phi = W$. This proves Theorem 1. $blacksquare$
There is yet another way to prove Theorem 1 -- namely, by revealing it to be a particular case of the Second Fundamental Theorem of Invariant Theory for GL. See https://mathoverflow.net/questions/202005/a-vector-version-of-the-segre-embedding-what-is-the-kernel-of-the-ring-map for this generalization. (Another place where this generalization appears with proof is Theorem 5.1 of J. Désarménien, Joseph P. S. Kung, Gian-Carlo Rota, Invariant Theory, Young Bitableaux, and Combinatorics, unofficial re-edition 2017; you just need to set $d = 1$, and realize that every standard $left(mathcal{X},mathcal{U}right)$-bideterminant of shape strictly longer than $left(dright)$ contains at least one row of length $geq 2$, which is easily seen to place it inside the ideal $W$.)
answered Apr 8 '13 at 5:42
darij grinberg
9,81032961
3
Dear Darij, thanks for your answer. I posted an answer below showing that we don't need to prove explicitly that the kernel is a prime ideal to show that the image of the Segre embedding is a projective variety. It should be correct. Regards,
– user38268
Apr 8 '13 at 5:55
Regarding OP's original solution - why is the observation that $operatorname{Im}psi$ equals the vanishing locus of the ideal $I$ generated by the mentioned monomials, along with the observation $I$ is prime not enough?
– Arrow
2 days ago
@Arrow: how do you prove $I$ is prime?
– darij grinberg
2 days ago
thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities?
– user347489
yesterday
@user347489: No, I don't. Is anything wrong with the argument?
– darij grinberg
yesterday
add a comment |
up vote
9
down vote
up vote
9
down vote
I outlined a proof by Bjorn Poonen at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2450856#p2450856 . As Nils Matthes has noticed, computing the kernel is not necessary to solve Hartshorne's problem, though (in my opinion) it is more interesting than the problem itself.
$newcommand{kk}{mathbb{k}}$
$newcommand{Ker}{operatorname{Ker}}$
For the sake of self-containedness, let me repost the proof linked above here (with improved notations). I begin by restating the problem:
Theorem 1. Let $kk$ be a commutative ring with $1$. Let $n$ and $m$ be two nonnegative integers. Let $M=left{0,1,ldots ,mright}$ and $N=left{0,1,ldots ,nright}$.
Let $R$ be the polynomial ring $kkleft[Z_{i,j} mid i in M text{ and } j in N right]$.
Let $S$ be the polynomial ring $kkleft[x_0, x_1, ldots, x_m, y_0, y_1, ldots, y_n right]$.
Let $phi : R to S$ be the unique $kk$-algebra homomorphism that sends each $Z_{i,j}$ to $x_i y_j$.
Let $W$ be the ideal of $R$ generated by all elements of the type $Z_{a,b} Z_{c,d} - Z_{a,d} Z_{c,b}$ with $a in M$, $b in N$, $c in M$ and $d in N$.
Then, $Ker phi = W$.
To prove this, we shall use the following easy algebraic lemma:
Lemma 2. Let $C$ be a $kk$-module. Let $A$ and $B$ be two submodules of $C$ such that $C=A+B$. Let $psi$ be a $kk$-module map from $C$ to another $kk$-module $D$ such that $psimid_A$ is injective and $Ker psi supseteq B$. Then, $Ker psi = B$.
Proof of Lemma 2. Let $c in Ker psi$. Thus, $c in Ker psi subseteq C = A + B$; hence, we can write $c$ in the form $c = a + b$ for some $a in A$ and $b in B$. Consider these $a$ and $b$. We have $b in B subseteq Ker psi$, so that $psileft(bright) = 0$. Applying the map $psi$ to the equality $c = a + b$, we obtain $psileft(cright) = psileft(a + bright) = psileft(aright) + psileft(bright)$ (since $psi$ is a $kk$-module map). Comparing this with $psileft(cright) = 0$ (which follows from $c in Ker psi$), we obtain $0 = psileft(aright) + underbrace{psileft(bright)}_{= 0} = psileft(aright)$, so that $psileft(aright) = 0 = psileft(0right)$. Since $psimid_A$ is injective, this entails $a = 0$ (because both $a$ and $0$ belong to $A$). Thus, $c = underbrace{a}_{=0} + b = b in B$.
Now, forget that we fixed $c$. We thus have shown that $c in B$ for each $c in Ker psi$. Thus, $Ker psi subseteq B$. Combining this with $Ker psi supseteq B$, we obtain $Ker psi = B$. This proves Lemma 2. $blacksquare$
Proof of Theorem 1 (Bjorn Poonen) (sketched). We notice that $Ker phisupseteq W$ is very easy to prove (in fact, a trivial computation shows that $Ker phi$ contains $Z_{a,b}Z_{c,d}-Z_{a,d}Z_{c,b}$ for all $a$, $b$, $c$, $d$).
We order the set $Mtimes N$ lexicographically.
If $k$ is a nonnegative integer, then $S_k$ shall denote the symmetric group consisting of all permutations of $left{1,2,ldots,kright}$.
Every $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots ,left(a_k,b_kright)right) in left(Mtimes Nright)^k$ and every permutation $sigma in S_k$ satisfy
begin{equation}
Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}} mod W .
label{darij.pf.thm1.1}
tag{1}
end{equation}
(In fact, this is obvious from the definition of $W$ when $sigma$ is a transposition, and hence, by induction, it also holds for every permutation $sigma$, because every permutation is a composition of transpositions.)
Now, let $T$ be the $kk$-submodule of $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generated by all products of the form $Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}$ with $k$ being a nonnegative integer and $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in left(Mtimes Nright)^k$ being a $k$-tuple satisfying $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$. It is easy to see that the map $left.phimid_Tright. : T to kkleft[x_0,x_1,ldots,x_m,y_0,y_1,ldots,y_nright]$ is injective. (In fact, if $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in T$, then
begin{align}
left(phimid_Tright)left(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right)
&= phileft(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right) \
&= x_{a_1}y_{c_1}x_{a_2}y_{c_2}cdots x_{a_k}y_{c_k} \
&=x_{a_1}x_{a_2}cdots x_{a_k}y_{c_1}y_{c_2}cdots y_{c_k}
end{align}
is a monomial from which we can recover the $k$-tuple $left(a_1,a_2,ldots ,a_kright)$ up to order and the $k$-tuple $left(c_1,c_2,ldots ,c_kright)$ up to order; but since the order of each of these two $k$-tuples is predetermined by the condition that $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we can therefore recover these two $k$-tuples completely; hence, the map $phimid_T$ sends distinct monomials to distinct monomials, and thus is injective.)
Next we are going to show that:
begin{equation}
text{every monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ lies in $T+W$.}
label{darij.pf.thm1.2}
tag{2}
end{equation}
[Proof of eqref{darij.pf.thm1.2}: Let $mu$ be any monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$. Then, $mu=Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k}$ for some nonnegative integer $k$ and some $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)in left(Mtimes Nright)^k$ such that $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$. Consider such a $k$ and such a $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)$. Since $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$, we have $a_1leq a_2leq cdotsleq a_k$ (since our order is lexicographic). Clearly there exists a permutation $sigmain S_k$ such that $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$. Consider such a $sigma$. Let $c_i=b_{sigma i}$ for every $iinleft{1,2,ldots,kright}$. Hence, the chain of inequalities $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$ rewrites as $c_1leq c_2leq cdotsleq c_k$. Also,
begin{align}
mu
&= Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} \
&equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}}
qquad left( text{by eqref{darij.pf.thm1.1}} right) \
&= Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k} mod W
end{align}
(since $b_{sigma i}=c_i$ for every $iinleft{1,2,ldots,kright}$).
But since $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we have $Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k}in T$ (by the definition of $T$), so this rewrites as follows:
begin{align}
mu
&equiv left(text{an element of }Tright)mod W .
end{align}
In other words, $muin T+W$. Since this holds for every monomial $mu$ in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, this proves eqref{darij.pf.thm1.2}.]
Since the monomials in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generate the $kk$-module $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, and since $T+W$ is a submodule of this $kk$-module, we obtain $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]=T+W$ from eqref{darij.pf.thm1.2}.
Applying Lemma 2 to $C=kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, $A=T$, $B=W$ and $psi=phi$, we thus conclude that $Ker phi = W$. This proves Theorem 1. $blacksquare$
There is yet another way to prove Theorem 1 -- namely, by revealing it to be a particular case of the Second Fundamental Theorem of Invariant Theory for GL. See https://mathoverflow.net/questions/202005/a-vector-version-of-the-segre-embedding-what-is-the-kernel-of-the-ring-map for this generalization. (Another place where this generalization appears with proof is Theorem 5.1 of J. Désarménien, Joseph P. S. Kung, Gian-Carlo Rota, Invariant Theory, Young Bitableaux, and Combinatorics, unofficial re-edition 2017; you just need to set $d = 1$, and realize that every standard $left(mathcal{X},mathcal{U}right)$-bideterminant of shape strictly longer than $left(dright)$ contains at least one row of length $geq 2$, which is easily seen to place it inside the ideal $W$.)
answered Apr 8 '13 at 5:42
darij grinberg
9,81032961
I outlined a proof by Bjorn Poonen at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2450856#p2450856 . As Nils Matthes has noticed, computing the kernel is not necessary to solve Hartshorne's problem, though (in my opinion) it is more interesting than the problem itself.
$newcommand{kk}{mathbb{k}}$
$newcommand{Ker}{operatorname{Ker}}$
For the sake of self-containedness, let me repost the proof linked above here (with improved notations). I begin by restating the problem:
Theorem 1. Let $kk$ be a commutative ring with $1$. Let $n$ and $m$ be two nonnegative integers. Let $M=left{0,1,ldots ,mright}$ and $N=left{0,1,ldots ,nright}$.
Let $R$ be the polynomial ring $kkleft[Z_{i,j} mid i in M text{ and } j in N right]$.
Let $S$ be the polynomial ring $kkleft[x_0, x_1, ldots, x_m, y_0, y_1, ldots, y_n right]$.
Let $phi : R to S$ be the unique $kk$-algebra homomorphism that sends each $Z_{i,j}$ to $x_i y_j$.
Let $W$ be the ideal of $R$ generated by all elements of the type $Z_{a,b} Z_{c,d} - Z_{a,d} Z_{c,b}$ with $a in M$, $b in N$, $c in M$ and $d in N$.
Then, $Ker phi = W$.
To prove this, we shall use the following easy algebraic lemma:
Lemma 2. Let $C$ be a $kk$-module. Let $A$ and $B$ be two submodules of $C$ such that $C=A+B$. Let $psi$ be a $kk$-module map from $C$ to another $kk$-module $D$ such that $psimid_A$ is injective and $Ker psi supseteq B$. Then, $Ker psi = B$.
Proof of Lemma 2. Let $c in Ker psi$. Thus, $c in Ker psi subseteq C = A + B$; hence, we can write $c$ in the form $c = a + b$ for some $a in A$ and $b in B$. Consider these $a$ and $b$. We have $b in B subseteq Ker psi$, so that $psileft(bright) = 0$. Applying the map $psi$ to the equality $c = a + b$, we obtain $psileft(cright) = psileft(a + bright) = psileft(aright) + psileft(bright)$ (since $psi$ is a $kk$-module map). Comparing this with $psileft(cright) = 0$ (which follows from $c in Ker psi$), we obtain $0 = psileft(aright) + underbrace{psileft(bright)}_{= 0} = psileft(aright)$, so that $psileft(aright) = 0 = psileft(0right)$. Since $psimid_A$ is injective, this entails $a = 0$ (because both $a$ and $0$ belong to $A$). Thus, $c = underbrace{a}_{=0} + b = b in B$.
Now, forget that we fixed $c$. We thus have shown that $c in B$ for each $c in Ker psi$. Thus, $Ker psi subseteq B$. Combining this with $Ker psi supseteq B$, we obtain $Ker psi = B$. This proves Lemma 2. $blacksquare$
Proof of Theorem 1 (Bjorn Poonen) (sketched). We notice that $Ker phisupseteq W$ is very easy to prove (in fact, a trivial computation shows that $Ker phi$ contains $Z_{a,b}Z_{c,d}-Z_{a,d}Z_{c,b}$ for all $a$, $b$, $c$, $d$).
We order the set $Mtimes N$ lexicographically.
If $k$ is a nonnegative integer, then $S_k$ shall denote the symmetric group consisting of all permutations of $left{1,2,ldots,kright}$.
Every $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots ,left(a_k,b_kright)right) in left(Mtimes Nright)^k$ and every permutation $sigma in S_k$ satisfy
begin{equation}
Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}} mod W .
label{darij.pf.thm1.1}
tag{1}
end{equation}
(In fact, this is obvious from the definition of $W$ when $sigma$ is a transposition, and hence, by induction, it also holds for every permutation $sigma$, because every permutation is a composition of transpositions.)
Now, let $T$ be the $kk$-submodule of $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generated by all products of the form $Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}$ with $k$ being a nonnegative integer and $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in left(Mtimes Nright)^k$ being a $k$-tuple satisfying $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$. It is easy to see that the map $left.phimid_Tright. : T to kkleft[x_0,x_1,ldots,x_m,y_0,y_1,ldots,y_nright]$ is injective. (In fact, if $left(left(a_1,c_1right),left(a_2,c_2right),ldots,left(a_k,c_kright)right)in T$, then
begin{align}
left(phimid_Tright)left(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right)
&= phileft(Z_{a_1,c_1}Z_{a_2,c_2}cdots Z_{a_k,c_k}right) \
&= x_{a_1}y_{c_1}x_{a_2}y_{c_2}cdots x_{a_k}y_{c_k} \
&=x_{a_1}x_{a_2}cdots x_{a_k}y_{c_1}y_{c_2}cdots y_{c_k}
end{align}
is a monomial from which we can recover the $k$-tuple $left(a_1,a_2,ldots ,a_kright)$ up to order and the $k$-tuple $left(c_1,c_2,ldots ,c_kright)$ up to order; but since the order of each of these two $k$-tuples is predetermined by the condition that $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we can therefore recover these two $k$-tuples completely; hence, the map $phimid_T$ sends distinct monomials to distinct monomials, and thus is injective.)
Next we are going to show that:
begin{equation}
text{every monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ lies in $T+W$.}
label{darij.pf.thm1.2}
tag{2}
end{equation}
[Proof of eqref{darij.pf.thm1.2}: Let $mu$ be any monomial in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$. Then, $mu=Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k}$ for some nonnegative integer $k$ and some $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)in left(Mtimes Nright)^k$ such that $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$. Consider such a $k$ and such a $k$-tuple $left(left(a_1,b_1right),left(a_2,b_2right),ldots,left(a_k,b_kright)right)$. Since $left(a_1,b_1right)leqleft(a_2,b_2right)leq cdotsleq left(a_k,b_kright)$, we have $a_1leq a_2leq cdotsleq a_k$ (since our order is lexicographic). Clearly there exists a permutation $sigmain S_k$ such that $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$. Consider such a $sigma$. Let $c_i=b_{sigma i}$ for every $iinleft{1,2,ldots,kright}$. Hence, the chain of inequalities $b_{sigma 1}leq b_{sigma 2}leq cdots leq b_{sigma k}$ rewrites as $c_1leq c_2leq cdotsleq c_k$. Also,
begin{align}
mu
&= Z_{a_1,b_1}Z_{a_2,b_2}cdots Z_{a_k,b_k} \
&equiv Z_{a_1,b_{sigma 1}}Z_{a_2,b_{sigma 2}}cdots Z_{a_k,b_{sigma k}}
qquad left( text{by eqref{darij.pf.thm1.1}} right) \
&= Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k} mod W
end{align}
(since $b_{sigma i}=c_i$ for every $iinleft{1,2,ldots,kright}$).
But since $a_1leq a_2leq cdotsleq a_k$ and $c_1leq c_2leq cdotsleq c_k$, we have $Z_{a_1,c_1} Z_{a_2,c_2} cdots Z_{a_k,c_k}in T$ (by the definition of $T$), so this rewrites as follows:
begin{align}
mu
&equiv left(text{an element of }Tright)mod W .
end{align}
In other words, $muin T+W$. Since this holds for every monomial $mu$ in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, this proves eqref{darij.pf.thm1.2}.]
Since the monomials in $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$ generate the $kk$-module $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, and since $T+W$ is a submodule of this $kk$-module, we obtain $kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]=T+W$ from eqref{darij.pf.thm1.2}.
Applying Lemma 2 to $C=kkleft[Z_{i,j}mid left(i,jright)in Mtimes Nright]$, $A=T$, $B=W$ and $psi=phi$, we thus conclude that $Ker phi = W$. This proves Theorem 1. $blacksquare$
There is yet another way to prove Theorem 1 -- namely, by revealing it to be a particular case of the Second Fundamental Theorem of Invariant Theory for GL. See https://mathoverflow.net/questions/202005/a-vector-version-of-the-segre-embedding-what-is-the-kernel-of-the-ring-map for this generalization. (Another place where this generalization appears with proof is Theorem 5.1 of J. Désarménien, Joseph P. S. Kung, Gian-Carlo Rota, Invariant Theory, Young Bitableaux, and Combinatorics, unofficial re-edition 2017; you just need to set $d = 1$, and realize that every standard $left(mathcal{X},mathcal{U}right)$-bideterminant of shape strictly longer than $left(dright)$ contains at least one row of length $geq 2$, which is easily seen to place it inside the ideal $W$.)
answered Apr 8 '13 at 5:42
darij grinberg
9,81032961
edited yesterday
answered Apr 8 '13 at 5:42
darij grinberg
9,81032961
answered Apr 8 '13 at 5:42
darij grinberg
9,81032961
answered Apr 8 '13 at 5:42
darij grinberg
9,81032961
9,81032961
3
Dear Darij, thanks for your answer. I posted an answer below showing that we don't need to prove explicitly that the kernel is a prime ideal to show that the image of the Segre embedding is a projective variety. It should be correct. Regards,
– user38268
Apr 8 '13 at 5:55
Regarding OP's original solution - why is the observation that $operatorname{Im}psi$ equals the vanishing locus of the ideal $I$ generated by the mentioned monomials, along with the observation $I$ is prime not enough?
– Arrow
2 days ago
@Arrow: how do you prove $I$ is prime?
– darij grinberg
2 days ago
thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities?
– user347489
yesterday
@user347489: No, I don't. Is anything wrong with the argument?
– darij grinberg
yesterday
add a comment |
3
Dear Darij, thanks for your answer. I posted an answer below showing that we don't need to prove explicitly that the kernel is a prime ideal to show that the image of the Segre embedding is a projective variety. It should be correct. Regards,
– user38268
Apr 8 '13 at 5:55
Regarding OP's original solution - why is the observation that $operatorname{Im}psi$ equals the vanishing locus of the ideal $I$ generated by the mentioned monomials, along with the observation $I$ is prime not enough?
– Arrow
2 days ago
@Arrow: how do you prove $I$ is prime?
– darij grinberg
2 days ago
thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities?
– user347489
yesterday
@user347489: No, I don't. Is anything wrong with the argument?
– darij grinberg
yesterday
3
3
Dear Darij, thanks for your answer. I posted an answer below showing that we don't need to prove explicitly that the kernel is a prime ideal to show that the image of the Segre embedding is a projective variety. It should be correct. Regards,
– user38268
Apr 8 '13 at 5:55
Dear Darij, thanks for your answer. I posted an answer below showing that we don't need to prove explicitly that the kernel is a prime ideal to show that the image of the Segre embedding is a projective variety. It should be correct. Regards,
– user38268
Apr 8 '13 at 5:55
Regarding OP's original solution - why is the observation that $operatorname{Im}psi$ equals the vanishing locus of the ideal $I$ generated by the mentioned monomials, along with the observation $I$ is prime not enough?
– Arrow
2 days ago
Regarding OP's original solution - why is the observation that $operatorname{Im}psi$ equals the vanishing locus of the ideal $I$ generated by the mentioned monomials, along with the observation $I$ is prime not enough?
– Arrow
2 days ago
@Arrow: how do you prove $I$ is prime?
– darij grinberg
2 days ago
@Arrow: how do you prove $I$ is prime?
– darij grinberg
2 days ago
thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities?
– user347489
yesterday
thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities?
– user347489
yesterday
@user347489: No, I don't. Is anything wrong with the argument?
– darij grinberg
yesterday
@user347489: No, I don't. Is anything wrong with the argument?
– darij grinberg
yesterday
add a comment |
up vote
6
down vote
accepted
Indeed as Nils Matthes suggested we don't know need to go through such a mess and just use the hint of Hartshorne.
Define a map $varphi : k[T_{00} , ldots, T_{nm}] to k[X_0,ldots,X_n,Y_0,ldots,Y_m]$ that sends $T_{ij}$ to $X_iY_j$. Let $mathfrak{a} := ker varphi$. We claim that $psi (Bbb{P}^n times Bbb{P}^m) = V(mathfrak{a})$. For one inclusion if a point
$$a = [a_{00} : ldots : a_{nm}] in V(mathfrak{a})$$
then in particular $a$ is a zero of all the polynomials $T_{ij}T_{kl} - T_{il}T_{kj}$. But this means that $a in psi(Bbb{P}^n times Bbb{P}^m)$. The reverse inclusion follows immediately from the definition of $varphi$. Thus $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^m)$ and the projective Nullstellensatz implies that $psi(Bbb{P}^n times Bbb{P}^m)$ is a projective variety.
edited Oct 30 '14 at 10:16
user26857
39.1k123882
1
Shouldn't you prove that $mathfrak a$ is actually a homogeneous ideal?
– Andrea Gagna
May 15 '14 at 15:46
add a comment |
up vote
6
down vote
accepted
Indeed as Nils Matthes suggested we don't know need to go through such a mess and just use the hint of Hartshorne.
Define a map $varphi : k[T_{00} , ldots, T_{nm}] to k[X_0,ldots,X_n,Y_0,ldots,Y_m]$ that sends $T_{ij}$ to $X_iY_j$. Let $mathfrak{a} := ker varphi$. We claim that $psi (Bbb{P}^n times Bbb{P}^m) = V(mathfrak{a})$. For one inclusion if a point
$$a = [a_{00} : ldots : a_{nm}] in V(mathfrak{a})$$
then in particular $a$ is a zero of all the polynomials $T_{ij}T_{kl} - T_{il}T_{kj}$. But this means that $a in psi(Bbb{P}^n times Bbb{P}^m)$. The reverse inclusion follows immediately from the definition of $varphi$. Thus $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^m)$ and the projective Nullstellensatz implies that $psi(Bbb{P}^n times Bbb{P}^m)$ is a projective variety.
edited Oct 30 '14 at 10:16
user26857
39.1k123882
1
Shouldn't you prove that $mathfrak a$ is actually a homogeneous ideal?
– Andrea Gagna
May 15 '14 at 15:46
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Indeed as Nils Matthes suggested we don't know need to go through such a mess and just use the hint of Hartshorne.
Define a map $varphi : k[T_{00} , ldots, T_{nm}] to k[X_0,ldots,X_n,Y_0,ldots,Y_m]$ that sends $T_{ij}$ to $X_iY_j$. Let $mathfrak{a} := ker varphi$. We claim that $psi (Bbb{P}^n times Bbb{P}^m) = V(mathfrak{a})$. For one inclusion if a point
$$a = [a_{00} : ldots : a_{nm}] in V(mathfrak{a})$$
then in particular $a$ is a zero of all the polynomials $T_{ij}T_{kl} - T_{il}T_{kj}$. But this means that $a in psi(Bbb{P}^n times Bbb{P}^m)$. The reverse inclusion follows immediately from the definition of $varphi$. Thus $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^m)$ and the projective Nullstellensatz implies that $psi(Bbb{P}^n times Bbb{P}^m)$ is a projective variety.
edited Oct 30 '14 at 10:16
user26857
39.1k123882
Indeed as Nils Matthes suggested we don't know need to go through such a mess and just use the hint of Hartshorne.
Define a map $varphi : k[T_{00} , ldots, T_{nm}] to k[X_0,ldots,X_n,Y_0,ldots,Y_m]$ that sends $T_{ij}$ to $X_iY_j$. Let $mathfrak{a} := ker varphi$. We claim that $psi (Bbb{P}^n times Bbb{P}^m) = V(mathfrak{a})$. For one inclusion if a point
$$a = [a_{00} : ldots : a_{nm}] in V(mathfrak{a})$$
then in particular $a$ is a zero of all the polynomials $T_{ij}T_{kl} - T_{il}T_{kj}$. But this means that $a in psi(Bbb{P}^n times Bbb{P}^m)$. The reverse inclusion follows immediately from the definition of $varphi$. Thus $V(mathfrak{a}) = psi(Bbb{P}^n times Bbb{P}^m)$ and the projective Nullstellensatz implies that $psi(Bbb{P}^n times Bbb{P}^m)$ is a projective variety.
edited Oct 30 '14 at 10:16
user26857
39.1k123882
edited Oct 30 '14 at 10:16
user26857
39.1k123882
edited Oct 30 '14 at 10:16
user26857
39.1k123882
edited Oct 30 '14 at 10:16
user26857
39.1k123882
39.1k123882
answered Apr 8 '13 at 5:22
user38268
1
Shouldn't you prove that $mathfrak a$ is actually a homogeneous ideal?
– Andrea Gagna
May 15 '14 at 15:46
add a comment |
1
Shouldn't you prove that $mathfrak a$ is actually a homogeneous ideal?
– Andrea Gagna
May 15 '14 at 15:46
1
1
Shouldn't you prove that $mathfrak a$ is actually a homogeneous ideal?
– Andrea Gagna
May 15 '14 at 15:46
Shouldn't you prove that $mathfrak a$ is actually a homogeneous ideal?
– Andrea Gagna
May 15 '14 at 15:46
add a comment |
up vote
4
down vote
It's easy to show the preimage under the Segre map of an algebraic set is an algebraic subset of $mathbb{P}^n times mathbb{P}^m.$ If $V$ were reducible, we could write $V = V_1 cup V_2,$ where $V_ineq V$ are closed. Then $$ mathbb{P}^n times mathbb{P}^m = S^{-1}(V) = S^{-1} (V_1) bigcup S^{-1}(V_2)$$ where $S^{-1}(V_i)$ are closed. Picking $x_i in Vsetminus V_i$ we have $S^{-1}(x_i)cap S^{-1}(V_i)=emptyset$ so $S^{-1}(V_i) subset S^{-1}(V)$ is a strict inclusion, contradicting that $mathbb{P}^n times mathbb{P}^m$ is irreducible.
answered Jun 3 '13 at 9:20
Ragib Zaman
28.3k34889
3
Of course we do, it's the same condition as always in any topological space: A space $X$ is reducible if $X= X_1 cup X_2$ for some closed proper subsets $X_1, X_2$ and irreducible if not reducible. A subset $V$ of $mathbb{P}^n times mathbb{P}^m$ is closed if $V=V(S)$ where $S$ is a set of biforms in $k[X_0, cdots, X_n, Y_0, cdots, Y_m].$ This already establishes what it means for $mathbb{P}^n times mathbb{P}^m$ to be irreducible.
– Ragib Zaman
Jun 3 '13 at 10:11
add a comment |
up vote
4
down vote
It's easy to show the preimage under the Segre map of an algebraic set is an algebraic subset of $mathbb{P}^n times mathbb{P}^m.$ If $V$ were reducible, we could write $V = V_1 cup V_2,$ where $V_ineq V$ are closed. Then $$ mathbb{P}^n times mathbb{P}^m = S^{-1}(V) = S^{-1} (V_1) bigcup S^{-1}(V_2)$$ where $S^{-1}(V_i)$ are closed. Picking $x_i in Vsetminus V_i$ we have $S^{-1}(x_i)cap S^{-1}(V_i)=emptyset$ so $S^{-1}(V_i) subset S^{-1}(V)$ is a strict inclusion, contradicting that $mathbb{P}^n times mathbb{P}^m$ is irreducible.
answered Jun 3 '13 at 9:20
Ragib Zaman
28.3k34889
3
Of course we do, it's the same condition as always in any topological space: A space $X$ is reducible if $X= X_1 cup X_2$ for some closed proper subsets $X_1, X_2$ and irreducible if not reducible. A subset $V$ of $mathbb{P}^n times mathbb{P}^m$ is closed if $V=V(S)$ where $S$ is a set of biforms in $k[X_0, cdots, X_n, Y_0, cdots, Y_m].$ This already establishes what it means for $mathbb{P}^n times mathbb{P}^m$ to be irreducible.
– Ragib Zaman
Jun 3 '13 at 10:11
add a comment |
up vote
4
down vote
up vote
4
down vote
It's easy to show the preimage under the Segre map of an algebraic set is an algebraic subset of $mathbb{P}^n times mathbb{P}^m.$ If $V$ were reducible, we could write $V = V_1 cup V_2,$ where $V_ineq V$ are closed. Then $$ mathbb{P}^n times mathbb{P}^m = S^{-1}(V) = S^{-1} (V_1) bigcup S^{-1}(V_2)$$ where $S^{-1}(V_i)$ are closed. Picking $x_i in Vsetminus V_i$ we have $S^{-1}(x_i)cap S^{-1}(V_i)=emptyset$ so $S^{-1}(V_i) subset S^{-1}(V)$ is a strict inclusion, contradicting that $mathbb{P}^n times mathbb{P}^m$ is irreducible.
answered Jun 3 '13 at 9:20
Ragib Zaman
28.3k34889
It's easy to show the preimage under the Segre map of an algebraic set is an algebraic subset of $mathbb{P}^n times mathbb{P}^m.$ If $V$ were reducible, we could write $V = V_1 cup V_2,$ where $V_ineq V$ are closed. Then $$ mathbb{P}^n times mathbb{P}^m = S^{-1}(V) = S^{-1} (V_1) bigcup S^{-1}(V_2)$$ where $S^{-1}(V_i)$ are closed. Picking $x_i in Vsetminus V_i$ we have $S^{-1}(x_i)cap S^{-1}(V_i)=emptyset$ so $S^{-1}(V_i) subset S^{-1}(V)$ is a strict inclusion, contradicting that $mathbb{P}^n times mathbb{P}^m$ is irreducible.
answered Jun 3 '13 at 9:20
Ragib Zaman
28.3k34889
answered Jun 3 '13 at 9:20
Ragib Zaman
28.3k34889
answered Jun 3 '13 at 9:20
Ragib Zaman
28.3k34889
answered Jun 3 '13 at 9:20
Ragib Zaman
28.3k34889
28.3k34889
3
Of course we do, it's the same condition as always in any topological space: A space $X$ is reducible if $X= X_1 cup X_2$ for some closed proper subsets $X_1, X_2$ and irreducible if not reducible. A subset $V$ of $mathbb{P}^n times mathbb{P}^m$ is closed if $V=V(S)$ where $S$ is a set of biforms in $k[X_0, cdots, X_n, Y_0, cdots, Y_m].$ This already establishes what it means for $mathbb{P}^n times mathbb{P}^m$ to be irreducible.
– Ragib Zaman
Jun 3 '13 at 10:11
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3
Of course we do, it's the same condition as always in any topological space: A space $X$ is reducible if $X= X_1 cup X_2$ for some closed proper subsets $X_1, X_2$ and irreducible if not reducible. A subset $V$ of $mathbb{P}^n times mathbb{P}^m$ is closed if $V=V(S)$ where $S$ is a set of biforms in $k[X_0, cdots, X_n, Y_0, cdots, Y_m].$ This already establishes what it means for $mathbb{P}^n times mathbb{P}^m$ to be irreducible.
– Ragib Zaman
Jun 3 '13 at 10:11
3
3
Of course we do, it's the same condition as always in any topological space: A space $X$ is reducible if $X= X_1 cup X_2$ for some closed proper subsets $X_1, X_2$ and irreducible if not reducible. A subset $V$ of $mathbb{P}^n times mathbb{P}^m$ is closed if $V=V(S)$ where $S$ is a set of biforms in $k[X_0, cdots, X_n, Y_0, cdots, Y_m].$ This already establishes what it means for $mathbb{P}^n times mathbb{P}^m$ to be irreducible.
– Ragib Zaman
Jun 3 '13 at 10:11
Of course we do, it's the same condition as always in any topological space: A space $X$ is reducible if $X= X_1 cup X_2$ for some closed proper subsets $X_1, X_2$ and irreducible if not reducible. A subset $V$ of $mathbb{P}^n times mathbb{P}^m$ is closed if $V=V(S)$ where $S$ is a set of biforms in $k[X_0, cdots, X_n, Y_0, cdots, Y_m].$ This already establishes what it means for $mathbb{P}^n times mathbb{P}^m$ to be irreducible.
– Ragib Zaman
Jun 3 '13 at 10:11
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If I remember it correctly, you don't even need to show that $mathfrak{a}=ker(varphi)$, you can prove $V(ker(varphi))=operatorname{im}(psi)$ directly. Could you show us roughly how you proved $V(mathfrak{a})=operatorname{im}(psi)$?
– Nils Matthes
Apr 7 '13 at 14:37
@NilsMatthes I have posted an answer below.
– user38268
Apr 8 '13 at 5:22