Preimage of algebraic set under Segre embedding is algebraic

Question

Let $S:\mathbb{P}^n\times\mathbb{P}^m\rightarrow\mathbb{P}^N$ be the Segre embedding, given by $S([x_0:\dots:x_n],[y_0:\dots:y_m])=[x_0y_0:\dots:x_ny_m]$, where $N=mn+m+n$. Let $W$ be an algebraic subset of $\mathbb{P}^N$. Can anyone help me show that $S^{-1}(W)$ is an algebraic set? I'm having trouble explicitly finding bihomogeneous polynomials whose simultaneous solutions correspond to $S^{-1}(W)$. Any help is appreciated.

Answer 1

I change the notation: let \begin{equation} \sigma_{m,n}:([x_0:\dots:x_m],[y_0:\dots:y_n])\in\mathbb{P}^m\times\mathbb{P}^n\to[x_0y_0:\dots:x_my_n]\in\mathbb{P}^N \end{equation} be the Segre embedding, where $N=(m+1)(n+1)-1$, I denote as $\mathbb{K}$ the underground field; one can prove easily that $\sigma_{m,n}$ is a well defined map; via simply but long computation, one can prove that $\sigma_{m,n}$ is an injective map.

Let $z_{ij}$'s be the coordinates of $\mathbb{P}^N$, by definition we can state that $\Sigma_{m,n}$, the image of $\sigma_{m,n}$ in $\mathbb{P}^N$, is \begin{equation} V_p(z_{ij}-z_{ji}\mid i\in\{0,\dots,m\},j\in\{0,\dots,n\},i\neq j); \end{equation} therefore $\mathbb{P}^m\times\mathbb{P}^n$ is in bijection with $\Sigma_{m,n}$; let \begin{equation} \forall i\in\{0,\dots,m\},j\in\{0,\dots,n\},\,T_{ij}=\{[z_{00}:\dots:z_{mn}]\in\mathbb{P}^N\mid z_{ij}\neq0\}, \end{equation} these are open subset of $\mathbb{P}^N$ and therefore: \begin{equation} \Sigma_{m,n}^{ij}=\Sigma_{m,n}\cap T_{ij}\neq\emptyset. \end{equation} Without loss of generality, let $X=\Sigma_{m,n}^{00}$, then: \begin{gather} \sigma_{m,n}^{-1}(X)=\sigma_{m,n}^{-1}\left(\left\{\left[1:\frac{z_{01}}{z_{00}}:\dots:\frac{z_{mn}}{z_{00}}\right]\in\mathbb{P}^N\mid z_{ij}=z_{ji}\right\}\right)=U_0\times V_0\,\text{where:}\\ i\in\{0,\dots,m\},\,U_i=\{[x_0:\dots:x_m]\in\mathbb{P}^m\mid x_i\neq0\},\\ j\in\{0,\dots,n\},\,V_j=\{[y_0:\dots:y_n]\in\mathbb{P}^n\mid y_j\neq0\}, \end{gather} or more in general: \begin{equation} \sigma_{m,n}^{-1}\left(\Sigma_{m,n}^{ij}\right)=U_i\times V_j; \end{equation} from all this, one can state that $\sigma_{m,n}^{ij}=\sigma_{m,n|U_i\times V_j}$ is a bijection between $U_i\times V_j$ and $\Sigma_{m,n}^{ij}$.

One remember that $U_i\times V_j$ is regular isomorphic to $\mathbb{A}^{m+n}$ and $\Sigma_{m,n}^{ij}$ is regular isomorphic to a closed subset of $\mathbb{A}^{mn}$; without loss of generality, let $i=j=0$ and: \begin{equation} \varphi=\sigma_{m,n}^{00}:(x_1,\dots,x_m,y_1\dots,y_n)\in\mathbb{A}^{m+n}\to(x_1y_1,\dots,x_my_n)\in\mathbb{A}^{mn}; \end{equation} by previous reasoning: $\varphi$ is an injective regular map!

Let $W$ be a closed subset of $\mathbb{P}^N$ and let $Y=W\cap X$, then $Y$ is regular isomorphic to a closed subset of $\mathbb{A}^{mn}$; by definition: \begin{equation} \exists f_1,\dots,f_r\in\mathbb{K}[z_{11},\dots,z_{mn}]:Y=V(f_1,\dots,f_r) \end{equation} and in particular: \begin{equation} \varphi^{-1}(Y)=V(\varphi^{*}(f_1),\dots,\varphi^{*}(f_r)), \end{equation} where $\varphi^{*}$ is the pull back morphism associated to $\varphi$.

Because $\mathbb{A}^{m+n}$ is regular isomorphic to $U_0\times V_0$: \begin{equation} \exists d_1,e_1,\dots,d_r,e_r\in\mathbb{N_0}\mid\sigma_{m,n}^{-1}(W)=V_p\left(x_0^{d_1}y_0^{e_1}\varphi^{*}(f_1),\dots,x_0^{d_r}y_0^{e_r}\varphi^{*}(f_r)\right) \end{equation} that is the inverse image of $W$ via $\sigma_{m,n}$ is the zero set of a (finite) family of bihomogeneous polynomial with coefficients in $\mathbb{K}$.

Is it all clear?