# Zariski closure and affine transformations commute

Let $$\mathbb K$$ be a field, $$X\subseteq \mathbb K^n$$ a subset, and $$\ f:\mathbb K^n\rightarrow \mathbb K^m$$ an affine map. Using the standard notation $$V$$, $$I$$ for algebraic geometry, it should hold that $$V(I(f(X)))=f(V(I(X))),$$ i.e. that the Zariski closure $$VI$$ commutes with the affine transformation $$f$$. Why is that?

(To specify the notation more: for a set $$Y$$, we mean by $$I(Y)$$ the set of all polynomials vanishing on $$Y$$; and for a set of polynomials $$J$$, we mean by $$V(J)$$ the set of all zeros common to all polynomials in $$J$$.)

Note that I do NOT assume $$f$$ to be bijective, and that $$X$$ does not need to be an algebraic set.

The equality obviously doesn't hold for every polynomial mapping $$\ f$$, as that would mean that an image of Zariski closed set under polynomial mapping is always Zariski closed, which is not true (e.g. $$f(x)=x^2:\mathbb R\rightarrow \mathbb R$$ and $$X=\mathbb R$$).

EDIT: As pointed out by reuns, using the general topology fact that $$\overline{f(X)}\supseteq f(\overline{X})$$ for continuous maps, it is enough to show that $$f(V(I(X)))$$ is Zariski closed (i.e. an algebraic set).

• The reformulation of the question would be "if $f$ is $\textbf{affine}$ then $\overline{f(X)}=f(\overline{X})$". The equality does not hold for $f$ continuous in general, as my example shows. If $f(x)=x^2$, we have $\overline{f(R)}=\overline{[0,\inf)}=R \neq [0,\inf)=f(R)=f(\overline{R})$. But you are right that $\supseteq$ holds in general, so that shows one direction. – OnDragi Jan 21 at 13:26
• In what sense is $f : \mathbb{R} \to \mathbb{R}, x \to x^2$ continuous, is it continuous for the topology induced by $V(I(X))$ ? – reuns Jan 21 at 13:40
• Yes, it is a polynomial map, and polynomial maps are continuous w.r.t. Zariski topology. – OnDragi Jan 21 at 13:42
• $\overline{f(X)} = f(\overline{X})$ is true whenever $f$ maps closed sets to closed sets (that's part of your definition of affine right ?). Here you are saying $f^{-1}(U)$ is open whenever $U$ is open (this would make $f$ continuous) but $f(X)$ isn't closed whenever $X$ ? Isn't the situation different for polynomial maps $f : \mathbb{C} \to \mathbb{C}$. – reuns Jan 21 at 13:53
• Indeed, that is the case. I am not sure whether the situation is different for algebraically closed fields. – OnDragi Jan 21 at 14:00