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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

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

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


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  • $\begingroup$ 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 $V(\mathfrak{a})=\operatorname{im}(\psi)$? $\endgroup$ – Nils Matthes Apr 7 '13 at 14:37
  • $\begingroup$ @NilsMatthes I have posted an answer below. $\endgroup$ – user38268 Apr 8 '13 at 5:22
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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 $M=\left\{0,1,\ldots ,m\right\}$ 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

\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 $\kk\left[Z_{i,j}\mid \left(i,j\right)\in M\times N\right]$ 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:

\begin{equation} \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} \end{equation}

[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 $W$.)

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    $\begingroup$ 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, $\endgroup$ – user38268 Apr 8 '13 at 5:55
  • $\begingroup$ 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? $\endgroup$ – Arrow Nov 17 '18 at 14:05
  • $\begingroup$ @Arrow: how do you prove $I$ is prime? $\endgroup$ – darij grinberg Nov 17 '18 at 15:30
  • $\begingroup$ thanks for this. It's great! BTW, in your definition of $T$ don't you need to use strict inequalities? $\endgroup$ – user347489 Nov 19 '18 at 8:16
  • $\begingroup$ @user347489: No, I don't. Is anything wrong with the argument? $\endgroup$ – darij grinberg Nov 19 '18 at 8:19
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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.

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    $\begingroup$ Shouldn't you prove that $\mathfrak a$ is actually a homogeneous ideal? $\endgroup$ – 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 $V_i\neq V$ 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.

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    $\begingroup$ 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. $\endgroup$ – Ragib Zaman Jun 3 '13 at 10:11

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