# Morphisms between two projective varieties

I know similar questions have been asked, but I don't think this specific question has been asked.

I wish to prove the following

Let $$X\subset \mathbb{P}^n$$ and $$Y\subset \mathbb{P}^m$$ be two projective varieties. Show that $$f:X\rightarrow Y$$ is a morphism if and only if there exists a collection of $$m+1$$ homogeneous polynomials $$F_{0},...,F_{m}\in k[X_{0},...,X_{n}]$$ of the same degree such that $$f([x_{0}:...:x_{n}])=[F_{0}(x):...:F_{m}(x)]\in Y$$, and such that for each $$x\in X$$, at least one of $$F_{i}(x)\neq 0$$

The "only if" part is giving me a lot of grief. The definition of morphism I am using is from Hartshorne which is the following

Definition: Let $$k$$ be a fixed algebraically closed field. A variety over $$k$$ (or simply variety) is any affine, quasi-affine, projective or quasi-projective variety as defined above. If $$X$$ and $$Y$$ are two varieties, a morphism $$\phi: X\rightarrow Y$$ is a continuous map such that for every open set $$V\subset Y$$, and for every regular function $$f:V\rightarrow k$$, the function $$f\circ \phi: \phi^{-1}(V)\rightarrow k$$ is regular.

My attempt:

Describe the co-ordinates of $$\mathbb{P}^n$$ by $$[x_{0},...,x_{n}]$$ and describe the co-ordinates of $$\mathbb{P}^m$$ by $$[y_{0},...,y_{m}]$$.

Assume $$\phi$$ is a morphism. Let $$V_{0}\subset \mathbb{P}^m$$ be the open subset of $$\mathbb{P}^m$$ corresponding to the subset $$y_{0}\neq 0$$. Then we have that $$V_{0}\cap Y$$ is an open subset of $$Y$$ which corresponds to an affine variety in $$\mathbb{A}^m$$. Now since $$\phi$$ is a morphism it follows that for $$f:\phi^{-1}(V_{0}\cap Y)\rightarrow k$$ a regular function that $$f\circ \phi$$ is regular.

This doesn't give me a lot to work with. So now consider $$f\circ \phi$$ restricted to $$U_{0}\cap\phi^{-1}(V_{0}\cap Y)$$, here $$U_{0}$$ is the open subset of $$\mathbb{P}^{n}$$ corresponding to $$x_{0}\neq 0$$. This looks promising, but the map $$f\circ \phi: U_{0}\cap\phi^{-1}(Y\cap V_{0})\rightarrow Y\cap V_{0}$$ is a map from a QUASI affine variety to an affine variety. Had $$U_{0}\cap\phi^{-1}(Y\cap V_{0})$$ been an affine variety I could have used the fact that morphisms between affine varieties are made up of polynomials in each co-ordinate and then used the construction in Hartshorne proposition 2.2 to "projectify (apply $$\beta$$" to these polynomials to obtain a map of the form stated in the question. However, even if I could do this, this would only be from $$U_{0}\cap\phi^{-1}(Y\cap V_{0})\rightarrow Y\cap V_{0}$$ and I could repeat this process for $$U_{i}\cap\phi^{-1}(Y\cap V_{j})\rightarrow Y\cap V_{j}$$ for various $$i$$ and $$j$$ but then how would I "stick them together" to get one polynomial map as stated in the question.

Any hints or solutions are appreciated

If $$X = \mathbb P^n$$ and $$Y = \mathbb P^m$$, then you can proceed as follows.

Think of your morphism $$f : \mathbb P^n \to \mathbb P^m$$ as a rational map $$f : \mathbb P^n \dashrightarrow \mathbb A^m$$, and let $$f_1, \dots, f_m$$ be its coordinates. Then,

$$f([x_0 : \dots : x_n]) = [1 : f_1 : \dots : f_n]$$

Rescaling by the lowest common denominator of the $$f_i$$, we have a polynomial expression

$$f([x_0 : \dots : x_n]) = [g_0 : \dots : g_m]$$

where $$\gcd(g_0, \dots, g_m) = 1$$, valid on a dense open subset of $$\mathbb P^n$$.

This expression is essentially unique. If there were another, say,

$$f([x_0 : \dots : x_n]) = [h_0 : \dots : h_m]$$

then $$p/q = g_i / h_i$$ would be a well defined rational function, independent of $$i$$, defined on an open subset of $$\mathbb P^n$$. Since the homogeneous coordinate ring of $$\mathbb P^n$$ is a unique factorization domain, then $$p$$ divides all the $$g_i$$, and $$q$$ divides all the $$h_i$$. Hence $$p$$ and $$q$$ are both units, i.e., nonzero constants.

Finally, since the $$g_i$$ provide a representation of $$f$$ that is valid everywhere on $$\mathbb P^n$$, they cannot simultaneously vanish anywhere on $$\mathbb P^n$$, so we are done.

However, I do not think what you are trying to prove is actually true if $$X$$ and $$Y$$ are arbitrary. For example, let $$X \subset \mathbb P^2$$ be the conic $$xz = y^2$$. Define $$\varphi : X \to Y = \mathbb P^1$$ by

$$\varphi([x:y:z]) = \begin{cases} [x:y], & \text{if } x \ne 0 \\ [y:z], & \text{if } z \ne 0 \end{cases}$$

(You need to check that this is actually well defined: (a) the point $$[0:1:0]$$ is not on $$X$$, and (b) $$[x:y] = [y:z]$$ for any point on $$X$$ where $$x, z$$ are both nonzero.)

Then sadly $$\varphi$$ does not admit a single representation of the form you want that is defined everywhere on $$X$$.