# Confused about Algebraic Geometry Theorem

• Oct 13th 2011, 09:14 AM
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 $V_Y(I) = \{ y \in Y : f(y) = 0 \forall f \in I \}$ )
• Oct 13th 2011, 10:00 AM
Drexel28
Re: Confused about Algebraic Geometry Theorem
Quote:

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.
• Oct 13th 2011, 10:09 AM
slevvio
Re: Confused about Algebraic Geometry Theorem
thanks drexel.