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I am using Karen E. Smith et al.'s Invitation to Algebraic Geometry, and was wondering the following: if $$\phi : V \subseteq \mathbb{P}^n \to W \subseteq \mathbb{P}^m $$ is a morphism of quasiprojective varieties (as they define it in the textbook, so without using any commutative algebra!), and $V$ is Zariski closed in $\mathbb{P}^n$, in other words, it is a projective variety, is $\phi(V)$ then a Zariski closed subset of $W$?

I have looked through this site and found a few questions that look very similar, like Image of morphism of projective varieties is projective variety and Is the image of a projective variety projective? . However, it appears that they contradict each other. Which is probably because they use different definitions of what projective means?

I was wondering if the statement is true in the context of the definitions handled by Karen E. Smith et al. and how to prove it.

What I have tried:

If $V \subseteq \mathbb{P}^n$ is Zariski closed, then I know it is given by the vanishing of finitely many homogeneous polynomials in the variables $x_0, \ldots, x_n$, say, $p_1, \ldots, p_K$.

The definition of morphism given in Smith et al. is very local: in each $p \in V$ we find an open subset $U_p \subseteq V$ such that there exist homogeneous polynomials $F_0^{(p)}, \ldots, F_m^{(p)}$ in the variables $x_0, \ldots, x_n$ such that $\phi$ agrees on $U^{(p)}$ with the map $$q \mapsto [F_0^{(p)}(q) : \cdots : F_m^{(p)}(q)].$$

Because $V$ is compact in the Zariski topology, we can pick finitely many $U^{(p_1)}, \ldots, U^{(p_L)}$ that cover $V$. I suppose I should try to find equations for $\phi(V)$ now. But I do not see an obvious way to combine the $F_j^{(p_l)}(x_0, \ldots, x_n)$'s and the $p_k(x_0, \ldots, x_n)$'s to find a set of homogeneous equations for $\phi(V)$.

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There's no contradiction between the two linked questions: in the first link, the target of the morphism is a subvariety (or subscheme) of $\Bbb P^n$, while in the second link, the target is not assumed to live anywhere in particular. This matters to the end result - the underlying claim here which is true is the following:

Theorem. Let $X\subset \Bbb P^n$ be a closed subvariety. Then any map of varieties $f:X\to Y$ is closed.

If $Y\subset\Bbb P^m$, you can consider the composition $X\to Y \to \Bbb P^m$ and so the image of $X$ is closed in $\Bbb P^m$ and therefore a subvariety. If $Y\not\subset\Bbb P^m$, then you can't make this conclusion.

See here on MSE for a proof of the theorem at approximately the level of the text you're referencing.

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  • $\begingroup$ Thank you for the answer, this helps a lot. Is this one of the cases were this would be a lot easier to show using the tools of commutative algebra (which I have not learned yet)? $\endgroup$ Commented Nov 19, 2023 at 15:37
  • $\begingroup$ Please read the linked proof (that's why I linked it). You'll need to either black-box the elimination theorem there or prove it yourself, which will take some algebra. $\endgroup$
    – KReiser
    Commented Nov 19, 2023 at 15:44

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