Confused about Algebraic Geometry Theorem

Let $\displaystyle \phi: X \rightarrow Y$ be a regular map between algebraic sets.

This induces a $\displaystyle k$-algebra homomorphism $\displaystyle \phi^*: k[Y] \rightarrow k[X]$ given by $\displaystyle f \mapsto f\phi$

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

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

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

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

Re: Confused about Algebraic Geometry Theorem

Quote:

Originally Posted by

**slevvio** Let $\displaystyle \phi: X \rightarrow Y$ be a regular map between algebraic sets.

This induces a $\displaystyle k$-algebra homomorphism $\displaystyle \phi^*: k[Y] \rightarrow k[X]$ given by $\displaystyle f \mapsto f\phi$

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

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

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

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

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

Re: Confused about Algebraic Geometry Theorem