Confused about Algebraic Geometry Theorem

**slevvio** · October 13th 2011, 09:14 AM

Let $\phi: X \rightarrow Y$ be a regular map between algebraic sets.
This induces a $k$ -algebra homomorphism $\phi^*: k[Y] \rightarrow k[X]$ given by $f \mapsto f\phi$

I have a theorem here that is really confusing me, and help would be appreciated.

It says " $\phi^*$ is injective $\iff \phi(X) \subseteq Y$ is dense "

While proving this, the author takes density to mean $\phi(X) \subseteq V_Y(I) \implies V_Y(I) = Y$ . What has this got to do with density in the usual sense?

(where $V_Y(I) = \{ y \in Y : f(y) = 0 \forall f \in I \}$ )

**Drexel28** · October 13th 2011, 10:00 AM

Originally Posted by slevvio

Let $\phi: X \rightarrow Y$ be a regular map between algebraic sets.
This induces a $k$ -algebra homomorphism $\phi^*: k[Y] \rightarrow k[X]$ given by $f \mapsto f\phi$

I have a theorem here that is really confusing me, and help would be appreciated.

It says " $\phi^*$ is injective $\iff \phi(X) \subseteq Y$ is dense "

While proving this, the author takes density to mean $\phi(X) \subseteq V_Y(I) \implies V_Y(I) = Y$ . What has this got to do with density in the usual sense?

(where $V_Y(I) = \{ y \in Y : f(y) = 0 \forall f \in I \}$ )

Because $V_Y(I)$ is closed in the Zariski topology, and so $\phi(X)$ being dense means the only closed set which contains it is the full space.

**slevvio** · October 13th 2011, 10:09 AM

thanks drexel.

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