# Hartshorne Problem 1.2.14 on Segre Embedding

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

$$\Big\{z_{ij}z_{kl} - z_{il}z_{kj} \hspace{1mm} \Big| \hspace{1mm} i,k = 0,\ldots, n; \hspace{2mm} j,l = 0,\ldots,m\Big\}.$$

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 $$V(\mathfrak{a}) = \psi(\Bbb{P}^n \times \Bbb{P}^n)$$.

• 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

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.


Theorem 1. Let $$\kk$$ be a commutative ring with $$1$$. Let $$n$$ and $$m$$ be two nonnegative integers. Let $$M=\left\{0,1,\ldots ,m\right\}$$ and $$N=\left\{0,1,\ldots ,n\right\}$$.

Let $$R$$ be the polynomial ring $$\kk\left[Z_{i,j} \mid i \in M \text{ and } j \in N \right]$$.

Let $$S$$ be the polynomial ring $$\kk\left[x_0, x_1, \ldots, x_m, y_0, y_1, \ldots, y_n \right]$$.

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 $$\psi\mid_A$$ 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 $$\psi\left(b\right) = 0$$. Applying the map $$\psi$$ to the equality $$c = a + b$$, we obtain $$\psi\left(c\right) = \psi\left(a + b\right) = \psi\left(a\right) + \psi\left(b\right)$$ (since $$\psi$$ is a $$\kk$$-module map). Comparing this with $$\psi\left(c\right) = 0$$ (which follows from $$c \in \Ker \psi$$), we obtain $$0 = \psi\left(a\right) + \underbrace{\psi\left(b\right)}_{= 0} = \psi\left(a\right)$$, so that $$\psi\left(a\right) = 0 = \psi\left(0\right)$$. Since $$\psi\mid_A$$ 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 $$\Ker \phi\supseteq W$$ 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 $$M\times N$$ lexicographically.

If $$k$$ is a nonnegative integer, then $$S_k$$ shall denote the symmetric group consisting of all permutations of $$\left\{1,2,\ldots,k\right\}$$. Every $$k$$-tuple $$\left(\left(a_1,b_1\right),\left(a_2,b_2\right),\ldots ,\left(a_k,b_k\right)\right) \in \left(M\times N\right)^k$$ and every permutation $$\sigma \in S_k$$ satisfy

$$$$Z_{a_1,b_1}Z_{a_2,b_2}\cdots Z_{a_k,b_k} \equiv Z_{a_1,b_{\sigma 1}}Z_{a_2,b_{\sigma 2}}\cdots Z_{a_k,b_{\sigma k}} \mod W . \label{darij.pf.thm1.1} \tag{1}$$$$

(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 $$\kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$$ 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 $$\left(\left(a_1,c_1\right),\left(a_2,c_2\right),\ldots,\left(a_k,c_k\right)\right)\in \left(M\times N\right)^k$$ being a $$k$$-tuple satisfying $$a_1\leq a_2\leq \cdots\leq a_k$$ and $$c_1\leq c_2\leq \cdots\leq c_k$$. It is easy to see that the map $$\left.\phi\mid_T\right. : T \to \kk\left[x_0,x_1,\ldots,x_m,y_0,y_1,\ldots,y_n\right]$$ is injective. (In fact, if $$\left(\left(a_1,c_1\right),\left(a_2,c_2\right),\ldots,\left(a_k,c_k\right)\right)\in T$$, then \begin{align} \left(\phi\mid_T\right)\left(Z_{a_1,c_1}Z_{a_2,c_2}\cdots Z_{a_k,c_k}\right) &= \phi\left(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 $$\left(a_1,a_2,\ldots ,a_k\right)$$ up to order and the $$k$$-tuple $$\left(c_1,c_2,\ldots ,c_k\right)$$ up to order; but since the order of each of these two $$k$$-tuples is predetermined by the condition that $$a_1\leq a_2\leq \cdots\leq a_k$$ and $$c_1\leq c_2\leq \cdots\leq c_k$$, we can therefore recover these two $$k$$-tuples completely; hence, the map $$\phi\mid_T$$ sends distinct monomials to distinct monomials, and thus is injective.)

Next we are going to show that:

$$$$\text{every monomial in \kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right] lies in T+W.} \label{darij.pf.thm1.2} \tag{2}$$$$

[Proof of \eqref{darij.pf.thm1.2}: Let $$\mu$$ be any monomial in $$\kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$$. 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 $$\left(\left(a_1,b_1\right),\left(a_2,b_2\right),\ldots,\left(a_k,b_k\right)\right)\in \left(M\times N\right)^k$$ such that $$\left(a_1,b_1\right)\leq\left(a_2,b_2\right)\leq \cdots\leq \left(a_k,b_k\right)$$. Consider such a $$k$$ and such a $$k$$-tuple $$\left(\left(a_1,b_1\right),\left(a_2,b_2\right),\ldots,\left(a_k,b_k\right)\right)$$. Since $$\left(a_1,b_1\right)\leq\left(a_2,b_2\right)\leq \cdots\leq \left(a_k,b_k\right)$$, we have $$a_1\leq a_2\leq \cdots\leq a_k$$ (since our order is lexicographic). Clearly there exists a permutation $$\sigma\in S_k$$ 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 $$i\in\left\{1,2,\ldots,k\right\}$$. Hence, the chain of inequalities $$b_{\sigma 1}\leq b_{\sigma 2}\leq \cdots \leq b_{\sigma k}$$ rewrites as $$c_1\leq c_2\leq \cdots\leq c_k$$. 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 $$i\in\left\{1,2,\ldots,k\right\}$$).

But since $$a_1\leq a_2\leq \cdots\leq a_k$$ and $$c_1\leq c_2\leq \cdots\leq c_k$$, 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 }T\right)\mod W . \end{align} In other words, $$\mu\in T+W$$. Since this holds for every monomial $$\mu$$ in $$\kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$$, this proves \eqref{darij.pf.thm1.2}.]

Since the monomials in $$\kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$$ generate the $$\kk$$-module $$\kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$$, and since $$T+W$$ is a submodule of this $$\kk$$-module, we obtain $$\kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]=T+W$$ from \eqref{darij.pf.thm1.2}.

Applying Lemma 2 to $$C=\kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$$, $$A=T$$, $$B=W$$ 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 $$\left(d\right)$$ contains at least one row of length $$\geq 2$$, which is easily seen to place it inside the ideal $$W$$.)

• 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 Nov 17 '18 at 14:05
• @Arrow: how do you prove $I$ is prime? – darij grinberg Nov 17 '18 at 15:30
• thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities? – user347489 Nov 19 '18 at 8:16
• @user347489: No, I don't. Is anything wrong with the argument? – darij grinberg Nov 19 '18 at 8:19

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.

• Shouldn't you prove that $\mathfrak a$ is actually a homogeneous ideal? – Andrea Gagna May 15 '14 at 15:46

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 $V_i\neq V$ 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 $x_i \in V\setminus V_i$ 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.

• 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