Segre map is an embedding

Question

I'd like to show that there exists an embedding $P^m \times P^n \to P^{(m+1)(n+1)-1}$, where $P^i$ denotes the real projective space.

I found the Segre map

$$\Sigma_{m,n}:P^m \times P^n \to P^{(m+1)(n+1)-1}$$ $$[x_0:\cdots: x_m] \times [y_0:\cdots:y_n] \to [x_0y_0: x_0y_1: \cdots: x_iy_j:\cdots: x_my_n].$$ We define $[z_{00}:z_{01}:\cdots:z_{ij}:\cdots:z_{mn}]:= [x_0y_0: x_0y_1: \cdots: x_iy_j:\cdots: x_my_n]$ for $[x_0:\cdots: x_m] \times [y_0:\cdots:y_n]\in P^m \times P^n$. It is easy to see the image in matrix form \begin{equation*} \begin{bmatrix} z_{00} & \cdots & z_{0n} \\ \vdots & \vdots & \vdots \\ z_{m0} & \cdots & z_{mn} \end {bmatrix} = \begin{bmatrix} x_{0} \\ \vdots \\ x_{m} \end {bmatrix} \begin{bmatrix} y_{0}: \cdots : y_{n} \end {bmatrix}. \end{equation*}

Let $z=[z_{00}:z_{01}:\cdots:z_{ij}:\cdots:z_{mn}]$ be an element of the image of $\Sigma_{m,n}$ and let $(a,b)\in P^m \times P^n$ such that $\Sigma_{m,n}(a,b)=z$. WLOG we can assume $a_0=b_0=z_{00}=1$. then $b_j=z_{0j}$ for all $0\leq j \leq n$ and $a_i=z_{i0}$ so $a,b$ are uniquely determined and this map is bijective onto the image.

Is this enough to conclude that we have an embedding? Is $(m+1)(n+1)-1$ the smallest number for which we can have an embedding of $P^m\times P^n$?

Answer 1

Being bijective is not enough to be an embedding, but what you've done is: You have constructed an inverse morphism from $\Sigma_{m,n}$ to $\mathbb{P}^m \times \mathbb{P}^n$, showing that the Segre map is an isomorphism onto its image, which is what it means to be an embedding.

No, it is definitely not the smallest number. It is a general fact that you can embed any smooth $n$-dimensional projective variety $X$ into $\mathbb{P}^{2n+1}$. The idea is you take an embedding into some high dimensional projective space and then repeatedly project down to hyperplanes, one dimension at a time. As long as the point you project from doesn't lie on any secant or tangent line of $X$, this will continue to be an embedding.