# How to show $\phi(X)$ contains a non-empty open set of its closure $\overline{\phi(X)}$?

From T.A.Springer, Linear Algebraic Groups, the end of Chapter 1.

Assume $X\rightarrow Y$ is a morphism of varieties. Using a covering of $Y$ by affine open sets, we reduce the proof to the case that $Y$ is affine. Similarly we can also assume $X$ is affine. Now I need to prove $\phi(X)$ contains a non-empty open set of its closure $\overline{\phi(X)}$. Here $k$ is algebraically closed.

Let $A=k[\overline{\phi(X)}]$, $B=k[X]$, What I know is there exist some $a\in A$ such that for any homomorphism from $A$ to $k$ such that $f(a)\not=0$, there is an extension $f:B\rightarrow k$ with $f(1)\not=0$.

It seems to me that any homomorphism $A\rightarrow k$ is a morphism of varieties $A^{1}\rightarrow \overline{\phi(X)}$ via the duality. So the fact that any such map non-vanishing on $a$ can be ''extended'' to $A^{1}\rightarrow X$ should give us what we wants. But this is far away from having a principle open set of the type $D_{f}=f(x)\not=0,x\in \overline{\phi(X)}$. I feel the logic at here is a bit inverted and I need some help to straighten it back.

If anyone can give a hint how to prove this via Noether's normalization lemma I would be grateful.

• The first paragraph becomes logical when you exchange sentences #2 and #3 with #4. Commented Mar 10, 2013 at 10:51
• Not an answer at all, but this is Proposition 15.4.2 in the sweet book Lie Algebras and algebraic groups by Patrice Tauvel and Rupert Yu. Commented Mar 10, 2013 at 11:03
• @user32240: Is $k$ algebraically closed ?
– user18119
Commented Mar 11, 2013 at 8:08
• @QiL'8: Yes, I forgot to write it. Commented Mar 11, 2013 at 8:11

We can replace $Y$ by $\overline{\phi(X)}$. Let $a$ be as in the OP. Then any point $y$ of the principal open subset $D_a$ of $Y$ correspond to a homomorphism $f: A\to k$ such that $f(a)\ne 0$, it is $A\to A/y=k$ when $y$ is viewed as a maximal ideal. By what you already known, this $f$ extends to $g : B\to k$ and $g$ defines a point $x\in X$ lying over $Y$: indeed, $x$ is the maximal ideal $\mathrm{ker}(g)$. As $g$ extends $f$, we have $\mathrm{ker}(g)\cap A\supseteq \mathrm{ker}(f)$. As both ideals are maximal, they are equal, so $\phi(x)=y$.
Edit I forgot to mention the answer on Noether's normalization lemma. A generalized form (Nagata, Local rings, I, § 14) is : if $A\subseteq B$ are integral domains and $B$ is finitely generated $A$-algebra, then there exists $a\in A$ non-zero, $d\ge 0$ such that $A\to B$ decomposes into $$A_a\to A_a[T_1, \dots, T_d] \to B_a$$ with the last map being injective and finite. This implies that $D_X(a)\to \mathbb A^d\times {D_Y(a)}$ is surjective. On the other hands, $\mathbb A^d\times {D_Y(a)}\to D_Y(a)$ is obviously surjective, so $D_X(a)\to D_Y(a)$ is surjective. This proof doesn't need $k$ algebraically closed and works even for any domain $A$ (not necessarily a $k$-algebra).
• 'This proof doesn't need $k$ algebraically closed'. It does. Take for example $k=\mathbb{Q}$, $X=Y=\mathbb{A}^1_k$ and $\varphi(x)=x^2$. Then $\varphi$ is dominant, yet its image does not contain any non-empty open.