I am reading Harris' Algebraic Geometry, A First Course and could not understand a part of Theorem 3.16.

The setting is this: $X\subset\mathbb{P}^m$ is a quasi-projective variety, $f:X\to\mathbb{P}^n$ is a regular map, and $U\subset X$ is a constructible set.

The part of the proof needs to establish the claim $f(U)$ contains a nonempty open subset $V\subset\overline{f(U)}$ of the closure of $f(U)$.

I am copying the part I don't understand from Harris:

To begin with, we may replace $U$ by an open subset, and so may assume that it is affine; restricting to a smaller affine open, we may assume the target space is also affine space. After replacing $U$ by the graph of $f$ we may realize the map $f$ as the restriction to a closed subset $U\subset\mathbb{A}^n$ of a linear projection $\mathbb{A}^n\to\mathbb{A}^m$, so that it is enough to prove the claim for a locally closed subset $U\subset\mathbb{A}^n$ under the projection

$$\pi:\mathbb{A}^n\to \mathbb{A}^{n-1}$$


Finally, we can replace $\mathbb{A}^{n-1}$ by the Zariski closure Y = $\overline{\pi(U)}$ of the image of $U$ and $\mathbb{A}^{n}$ by the inverse image $\pi^{-1}(Y) = Y \times \mathbb{A}^{1}$. It will thus suffice to establish the claim for a locally closed subset $U$ of a product $Y\times\mathbb{A}^{1}$ (or, equivalently, a locally closed subset $U\subset Y\times\mathbb{P}^{1}$) and the projection map $\pi: Y \times\mathbb{P}^{1}\to Y $on the first factor, with the further assumption that $\pi(U)$ is dense in $Y$.

Basically I have trouble understanding what exactly he is trying to say in "we may realize the map $f$ as the restriction to a closed subset $U\subset\mathbb{A}^n$ of a linear projection $\mathbb{A}^n\to\mathbb{A}^m$", I have no idea what he is trying to do, particularly, is the new closed subset $U$ related to the old constructible $U$ in some way? What is this linear projection from $\mathbb{A}^n$ to $\mathbb{A}^m$, explicitly?

I am a beginner in algebraic geometry and could speak the language of scheme yet. Any help is really appreciated.


1 Answer 1


At that stage, you have a regular map from an affine to another affine $f:U\subset \mathbb{A}^n\to f(U)\subset \mathbb{A}^m$. The graph of $f$ is defined to be the closed subset $$G:=\{(x,f(x))\mid x\in U\}\subset \mathbb{A}^n\times\mathbb{A}^m=\mathbb{A}^{n+m}.$$ It is also canonically isomorphic to $U$ by projecting to the first factor $(x,f(x))\mapsto x$. Under this isomorphism, the map $f$ becomes the projection to the second factor, i.e. it now maps a closed subset $G\subset \mathbb{A}^{n+m}$ under this linear projection $\mathbb{A}^{n+m}\to\mathbb{A}^m$.


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