Let $Y \subseteq \Bbb{P}^n$ and $Z \subseteq \Bbb{P}^m$ be two projective varieties. By $Y \times Z$, we really mean the image of $Y \times Z$ via the Segre embedding $\psi$ in $\Bbb{P}^N$ with $N = (n+1)(m+1) - 1$. We want to determine the $t$ -th graded piece of the coordinate ring of a product of two projective varieties $Y$ and $Z$. Now I have looked in many texts (Hartshorne, Liu, Shafarevich, Fulton, Gortz - Wedhorn, Milne's notes, etc) but no one seems to say what it is!

My guess is: $$S(Y\times Z)_t \cong S(Y)_t \otimes_k S(Z)_t$$

Now I have tried to prove this as follows. First let $x_{00},\ldots,x_{nm}$ be coordinates on $\Bbb{P}^N$, $y_0,\ldots,y_n$ coordinates on $\Bbb{P}^n$ and $z_0,\ldots,z_n$ coordinates on $\Bbb{P}^m$. Then the induced map $\psi^\ast$ that sends $x_{ij} \mapsto y_i \otimes z_j$ is actually a surjection when we restrict to graded pieces

$$\psi^\ast : k[x_{ij}]_t \to k[y_0,\ldots,y_n]_t \otimes_k k[z_0,\ldots,z_m]_t.$$

Thus we get a surjection $\psi^\ast : k[x_{ij}]_t \to S(X)_t \otimes_k S(Y)_t $.

My question is: Why is the kernel of this map $I(\psi(X \times Y))$? It seems almost impossible to prove this because in general to say what elements are actually zero in a tensor product, difficult it is to tell. How can I prove my claim above via an alternative method?


1 Answer 1


First assume that $Y = \Bbb{P}^n$ and $Z = \Bbb{P}^m$

The map $\psi^\ast : k[X_{ij}] \to k[Y_0,\ldots,Y_n] \otimes_k k[Z_0,\ldots,Z_m]$ defined by $X_{ij} \rightarrow Y_i \otimes Z_j$ will induce a well defined map $\overline{\psi^\ast} : k[x_{ij}] \to k[Y_0,\ldots,Y_n] \otimes_k k[Z_0,\ldots,Z_m]$. where $k[x_{ij}] \equiv \frac{k[X_{ij}]}{(X_{ij}X_{pq} - X_{iq}X_{pj})}$. Clearly as you have observed the map $\overline{\psi^\ast}_t : k[x_{ij}]_t \to k[Y_0,\ldots,Y_n]_t \otimes_k k[Z_0,\ldots,Z_m]_t$ is onto for all $t$. Now we shall claim that it is also one one.

From the relations we can prove that $x_{i_1j_1}x_{i_2j_2} \ldots x_{i_tj_t} = x_{i_1 \sigma(j_1)}x_{i_2 \sigma(j_2)} \ldots x_{i_t \sigma(j_t)}$ and $x_{i_1j_1}x_{i_2j_2} \ldots x_{i_tj_t} = x_{\sigma(i_1)j_1}x_{\sigma(i_2)j_2} \ldots x_{\sigma(i_t)j_t}$ in $k[x_{ij}]$ for any permutation $\sigma$ of $t$ symbols.

This will give $x_{i_1j_1}x_{i_2j_2} \ldots x_{i_tj_t} = x_{p_1q_1}x_{p_2q_2} \ldots x_{p_tq_t}$ in $k[x_{ij}]_t$ if $\{i_1, i_2, \ldots, i_t\} = \{p_1, p_2, \ldots, p_t\}$ and $\{j_1, j_2, \ldots, j_t\} = \{q_1, q_2, \ldots, q_t\}$.

So from these conclude that $k[x_{ij}]_t$ can be generated by elements as many as in the basis of $k[y_0, y_1, \ldots y_n]_t \otimes k[z_0, z_1, \ldots z_m]_t$. So the map $\overline{\psi^\ast}_t : k[x_{ij}]_t \to k[Y_0,\ldots,Y_n]_t \otimes_k k[Z_0,\ldots,Z_m]_t $ is injective for all $t$. Therefore $\overline{\psi^\ast}$ is injective.

For two positively graded rings $R = \oplus_{i =0}^\infty R_i$ and $S = \oplus_{j =0}^\infty S_j$ define $V(R, S) = \oplus _{t =0}^\infty R_t \otimes S_t$.

The map $\psi^\ast: k[X_{ij}] \to k[Y_0,\ldots,Y_n] \otimes_k k[Z_0,\ldots,Z_m]$ has image $V(k[Y_0,\ldots,Y_n], k[Z_0,\ldots,Z_m])$.

Let $Y$ and $Z$ be arbitrary projective variety with $I(Y) = (f_1, f_2, \ldots, f_u)$ and $I(Z) = (g_1, g_2, \ldots, g_v)$. Then $I(Y \times Z) = {\psi^\ast}^{-1}(f_1 \otimes 1, f_2 \otimes 1, \ldots, f_u \otimes 1, 1 \otimes g_1, 1 \otimes g_2, \ldots, 1 \otimes g_v)$. So $\psi^\ast$ will induce a one one map $\overline{\psi^\ast} : \frac{k[X_{ij}]}{I(Y \times Z)} \rightarrow \frac{[Y_0,\ldots,Y_n]}{I(Y)} \otimes_k \frac{k[Z_0,\ldots,Z_m]}{I(Z)} = S(Y) \otimes_k S(Z)$ with image $V(S(Y), S(Z))$

  • $\begingroup$ Dear @A.G. thanks for your answer. I am now trying to prove this for arbitrary $Y$ and $Z$ but how do I know what is in $I(Y \times Z)$? $\endgroup$
    – user38268
    Commented Jun 22, 2013 at 0:07
  • $\begingroup$ I have edited my answer. $\endgroup$
    – A.G
    Commented Jun 23, 2013 at 3:32

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