Fleshing out a proof that $\phi$ is dominant iff $\phi^*$ is injective.

Question

This is a solution-verification/proof-explanation question, since I have a proof that I need to flesh out; I am unsure of a couple of steps.

The Details:

Most of the terminology here is given in this question of mine.

A morphism $\phi:X\to Y$ of affine varieties is dominant if $\phi(X)$ is dense in $Y$, i.e., $\overline{\phi(X)}=Y$ in the Zariski topology.

The Proof in Question:

We have

$$\begin{align} \phi^*\text{ is injective}&\iff \ker(\phi^*)=\{0\}\\ &\iff \forall f\in k[Y](\phi^*(f)=0\iff f=0)\\ &\iff \forall f\in k[Y](f\circ\phi=0\iff f=0)\\ &\stackrel{(1)}{\iff} \mathcal I(\phi(X))=\{0\}\\ &\stackrel{(2)}{\iff }\overline{\phi(X)}=Y\\ &\iff \phi\text{ is dominant.} \end{align}$$

The Problem:

How do $(1)$ and $(2)$ hold?

Thoughts:

I think $(2)$ holds by taking $\mathcal V$ of both sides in the forward direction, recognising that $\mathcal V(\mathcal I(Z))=\overline Z$; conversely, I think we use Hilbert's Nullstellensatz.

I have no clue about $(1)$.

Further Context:

Since I have no clue for $(1)$, to add context, let me answer some of the questions here:

• What are you studying?

A postgraduate research degree in linear algebraic groups. I need to get comfortable with algebraic geometry.

• What text is this drawn from, if any? If not, how did the question arise?

My supervisor set me this question a few weeks ago. The proof above is largely my own, although the things I am not sure about are based on Springer's book, "Linear Algebraic Groups (Second Edition)".

• What kind of approaches (to similar problems) are you familiar with?

See the following questions:

Understanding $\mathcal V(I)$, $\mathcal I(X)$, and their relationship to each other.

A morphism of affine varieties $\phi: X\to Y$ is an isomorphism iff the algebra homomorphism $\phi^*$ is an isomorphism.

• What kind of answer are you looking for? Basic approach, hint, explanation, something else?

I would prefer an explanation, please, of $(1)$ and $(2)$.

• Is this question something you think you should be able to answer? Why or why not?

Yes, but I would need much more time than I have.

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Please help :)

Answer 1

Perhaps it is better to write things down a bit more verbosely.

Suppose that $\phi : X\longrightarrow Y$ is dominant so that $\phi(X)$ is dense in $Y$. The map $k[Y] \longrightarrow k[X]$ is given by the pullback along $\phi$, and if $f \in k[Y]$ and vanishes on $\phi(X)$, then it vanishes on its closure, which is $Y$, since $V(f)$ is closed and $\phi(X)\subseteq V(f)$. Hence, $f$ is the zero element in $k[Y]$ and $\phi^*$ is injective.

Conversely, suppose that $\phi^*$ is injective, so that if $f\in k[Y]$ vanishes on $\phi(X)$ then $f$ vanishes on all of $Y$. This means that $I(\phi(X))$, the ideal of polynomials that vanish on $\phi(X)$, is just the zero ideal $0$. But since $V(I(-))$ is the closure operator, we have that $$ \overline{\phi(X)} = V(I(\phi(X))) = V(0) = Y.$$

Answer 2

About (2): Actually there's a version of the Hilbert Nullstellensatz that holds for ideals in the coordinate ring of an affine variety, see this answer for details.

About (1): $\mathcal I(\phi(X))$ is the ideal of all functions $f \in k[Y]$ that vanish on $\phi(X)$, i.e. by definition we have $$\forall f \in k[Y]: f\in \mathcal I (\phi(X)) \Leftrightarrow f(\phi(X))=\{0\}$$

But $f(\phi(X))=0$ means that $f(\phi(x))=0$ for all $x \in X$ which is equivalent to saying that $f \circ \phi$ is the zero function in $k[X]$. That's because in the coordinate ring $k[X]$ we take a quotient of the polynomial ring by the functions that vanish identically on $X$, which is exactly the case for $f \circ \phi$ here. Thus we have, without any condition on $\phi$

$$\forall f \in k[Y]: f \in \mathcal I(\phi(X)) \Leftrightarrow f \circ \phi=0$$

Now to say that $\mathcal I(\phi(X))=\{0\}$ is precisely to say that $\forall f \in k[Y]: f \in \mathcal I(\phi(X)) \Leftrightarrow f=0$. If we combine this with the equivalence above, we get (1).