# 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.

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The first paragraph becomes logical when you exchange sentences #2 and #3 with #4. –  Martin Brandenburg Mar 10 '13 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. –  Jesko Hüttenhain Mar 10 '13 at 11:03
@user32240: Is $k$ algebraically closed ? –  user18119 Mar 11 '13 at 8:08
@QiL'8: Yes, I forgot to write it. –  Bombyx mori Mar 11 '13 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. –  Norbert Pintye Mar 10 at 18:07