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Let $f$ be an endomorphism of projective space $\mathbb{P}^n$. From this answer I know that $f$ is proper. But how does one actually determine the image of a given arbitrary Zariski closed subset?

For instance, if $f(x:y:z) = (x^2 - y^2 : x^2 + y^2 : z^2)$ is a random morphism and $X = V(x^2 + y^2 - z^2)$ is a circle then I have found, purely by trial and error, that $f(X) = V(y - z)$ is a line. If instead $f(x:y:z) = (x^3 : y^3 : z^3)$ is the cubing map then more trial and error shows that $f(X) = V((z^2 - x^2 - y^2)^3 - 27(xyz)^2)$ is a sextic.

How would one find the defining equation(s) of $f(X)$ in general? In particular, what's the relationship between the degrees of $f$, $X$, and $f(X)$? Is there some sort of elimination theory/intersection theory involved?

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  • There is a lovely book which covers many topics like this one, Ideals, Varieties, and Algorithms by Cox, Little and O'Shea. I don't think this is the place to give a good overview, by the general answer is, do something like the answer below to reduce to a calculation with a homomorphism of rings - then hit it hard with Grobner basis machinery. This is for example, what Magma will do if you ask it to compute the image of a rational map. – Mummy the turkey Jun 03 '21 at 17:44
  • You ask for the degrees, I this is a complicated question in general, even for curves. For example, take the cuspidal cubic $y^2z = x^3 + x^2z$. Then (almost all) maps to $\mathbb{P}^2$ given by degree $2$ polynomials will be birational on the image (they are birational automorphisms of $\mathbb{P}^2$). hence of degree $1$. Almost always the degree of the image is going to be higher. However, if you choose $[xy, yz, x^2 + xz]$ then the image is $xy + y^2 - z^2$. For curves with uglier equations (especially highly singular ones) the behaviour will be even more extreme. – Mummy the turkey Jun 03 '21 at 17:55
  • What you can do is bound the degree of $f(X)$ given $\deg(X)$ and a bound on the degrees of the polynomials defining $f$. See https://mathoverflow.net/questions/63451/degree-of-image-of-a-polynomial-map – H A Helfgott Oct 08 '23 at 09:20

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Question: "How would one find the defining equation(s) of f(X) in general?"

Answer: Locally if $f:X:=Spec(B) \rightarrow S:=Spec(A)$ there is a map

$$\phi: A \rightarrow B$$

and let $I:=ker(\phi)$. It follows $\phi$ factors as $A\rightarrow A/I \rightarrow B$ and you get a canonical map

$$g: X \rightarrow Spec(A/I):=V(I) \subseteq S.$$

By definition $V(I):=\overline{f(X)}$ is the closure of the image. If the map $f$ is closed it follows the image $f(X)$ is defined by a set of generators $g_i$ of the ideal $I$. Hence if you want equations defining the image you must calculate such a set of generators.

For your map defined for the projective space, you may cover projective space with open affine subsets and do this locally. You will end up with a sheaf of ideals. Let $f: \mathbb{P}^n \rightarrow \mathbb{P}^n$ be your map and $f^{\#}: \mathcal{O}_{\mathbb{P}^n} \rightarrow f_*\mathcal{O}_{\mathbb{P}^n}$ the map of structure sheaves. It follows $ker(f^{\#})$ will define your image $f(X)$ and you must calculate this ideal locally. Any closed subscheme of projective space comes from a homogeneous ideal $I$ in $k[x_0,..,x_n]$ and $I$ is finitely generated, hence you must calculate a set of generators of this ideal. This is true in general: If $X \subseteq \mathbb{P}^n_A$ where $A$ is any ring, it follows $X$ is defined by a homogeneous ideal $I \subseteq A[x_0,..,x_n]$. Hence if $f: \mathbb{P}^n_A \rightarrow \mathbb{P}^n_A$ you must calculate $ker(f^{\#})$.

Question: "In particular, what's the relationship between the degrees of f, X, and f(X)? Is there some sort of elimination theory/intersection theory involved?"

Answer: For some special cases there are formulas for the degree of the image $f(X)$. You will find this topic discussed on this site.

In general you must calculate generators of the ideal $ker(f^{\#})$ using an affine open cover. The relationship between endomorphisms of projective space and maps of graded rings is not as simple as for affine schemes (see Ex II.2.14 and Theorem II.7.11 in Hartshorne).

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