# Segre map is an embedding

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 $$P^m\times P^n$$?

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.

• So for example $P^m \times P^n$ could be embedded to $P^{2n+1}$ by what embedding? Commented May 20, 2020 at 8:52
• No, $\mathbb{P}^m \times \mathbb{P}^n$ could be embedded into $\mathbb{P}^{2n+2m+1}$, by picking any $2n+2m+2$ generic bihomogeneous degree 1,1 polynomials in the $a_i$ and $b_j$.
– nkm
Commented May 20, 2020 at 8:54
• Ok, thank you. Is there any literature you know of about this topic? I would like to see a bit more explanations. Commented May 20, 2020 at 8:56
• This should be in most introductory algebraic geometry textbooks. For one, it's in Shafarevich's Basic Algebraic Geometry 1, chapter II, section 5.4.
– nkm
Commented May 20, 2020 at 9:07
• +1, this is a nice explanation. One nitpick: your statement about a variety of dimension $n$ embedding in to $\Bbb P^{2n+1}$ is not quite correct. You should require that the variety be smooth, otherwise there are counterexamples (pick a projective variety of dimension $n$ with a tangent space at some point of dimension at least $2n+2$, for instance). Commented May 21, 2020 at 22:40