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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

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1 Answer 1

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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 $X$.

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