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The question is:

Show that the pullback $\mathbb{C}[W] \rightarrow \mathbb{C}[V]$ is injective if and only if $F$ is dominant, that is, the image set $F(V)$ is dense in $W$.

The $W, V$ are algebraic varieties. $F$ represents the pullback function. This is Exercise 2.5.1 in An Invitation to Algebraic Geometry. I'm not sure how to deal with the image set being dense in $W$. Are we using the same "dense" concept as in set theory: every point in $W$ is a limit point of $F(V)$? Then I'm not sure how to make connections between the image of an algebraic variety in another variety. Thanks for any help!

nekodesu
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  • I'm assuming that $V$ and $W$ are algebraic varieties but it would help if you provided more context. – Brian Fitzpatrick Jan 11 '16 at 20:34
  • I assume that $\mathbb{C}[V]$ stands for regular functions on $V$. Then you have to assume more, for example they are affine. This is false for projective varieties. – Mohan Jan 11 '16 at 21:59
  • @BrianFitzpatrick Hi! I just edited the problem statements. Hope this helps. – nekodesu Jan 11 '16 at 22:52
  • Hint: consider the closed subset defined by the kernel of the ring homomorphism. See also Atiyah–MacDonald, exercise 1.21(iii). – Remy Jan 12 '16 at 06:16

1 Answers1

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I am also reading this book. And I think this problem is a little weird. Here is my proof.

The pullback is injective is equivalent to $$\forall g \in \mathbb{C}[W] \;g(F(x))\equiv 0 \Leftrightarrow g(x)\equiv0.\tag{1} $$

In my opinion, the dense means dense in Zariski topology i.e. any open set $S\subset W$, $Im(F(x))\cap S \neq \emptyset$.

If $g\neq 0$, consider the open set $M:=\{x|g(x)\neq 0\}$. However, $\exists x\in M\cap Im(F), g(x)\neq 0$. We get a contradictory.

Now we need to prove that $(1)\Rightarrow $ dense.

If $Im(F)$ is not dense，there exists an open set $M$, $M\cap Im(F) = \emptyset$.

Consider the closed set $M^c$, and the corresponding polynomial $g$ (just take out one of them). Because $Im(F)\subset M^c$, $g(F(x))\equiv 0$. But $g\neq0$. We get a contradicgory too.

Will Cai
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