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).



Is the Segre embedding projectively normal?
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.


