Is the Segre embedding projectively normal?

Question

Consider the Segre embedding $$ \mathbb P^n\times \mathbb P^m \hookrightarrow \mathbb P^N. $$

It seems to me that from the definition it is clear that

$$ H^0(\mathbb P^N, \mathscr O_{\mathbb P^N}(1)) = H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(1))\otimes H^0(\mathbb P^m, \mathscr O_{\mathbb P^m}(1)), $$

but is it obvious that

$$ H^0(\mathbb P^n\times \mathbb P^m, {\mathscr O_{\mathbb P^{N}}(1)}|_{\mathbb P^n\times \mathbb P^m}) = H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(1))\otimes H^0(\mathbb P^m, \mathscr O_{\mathbb P^m}(1))? $$

More generally:

Question: Is the Segre embedding projectively normal?

EDIT Note that strictly speaking the above condition is only linearly normal, I did mean to ask projective normality, which in this case follows from linear normality. (See J.C. Ottem's answer).

Answer 1

Yes. The Segre embedding $i:P:=\mathbb P^n \times \mathbb P^m\to \mathbb P^N$ is defined by the sections of the line bundle $O_P(1,1):=pr_1^*O_{\mathbb P^n}(1)\otimes pr_2 O_{\mathbb P^m}(1)$ on $\mathbb P^n \times \mathbb P^m$ and by definition of $i$ we have $i^*O_{\mathbb P^N}(1)=O(1,1)$. In other words, the equality $$H^0(\mathbb P^n \times \mathbb P^m,i^*O_{\mathbb P^N}(k))=H^0(\mathbb P^n \times \mathbb P^m,O(k,k))$$ holds for any $k\in \mathbb Z$, and so $H^0(\mathbb P^n \times \mathbb P^m,i^*O(k))$ decomposes as the tensor product $H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(k))\otimes H^0(\mathbb P^m, \mathscr O_{\mathbb P^m}(k))$ by Kunneth's theorem. Note however, that the definition of projectively normal requires that the map $$H^0(\mathbb P^N,O(k))\to H^0(P,O(k,k))$$is surjective for all $k\ge 0$ not just for $k=1$ (in which case the embedding is called linearly normal). This is true in our case, since global sections of $O(k,k)$ are polynomials in sections of $O(1,1)$. Of course, when $k=2$, the kernel of the above map is generated by the quadrics defined as the $2\times 2$ minors defining the Segre embedding.