# Dominant and Injective Morphisms reverse implication

As mentioned before, I am self-studying some commutative algebra out of "A Course in Commutative Algebra" by Kemper. In the text, Kemper has the following problem:

Let $$X$$ and $$Y$$ be affine varieties over a field $$K$$, and let $$f: X → Y$$ be a morphism with induced homomorphism $$ϕ: K[Y ] → K[X].$$ We say that $$f$$ is dominant if the image $$f(X)$$ is dense in $$Y$$ , i.e., $$\overline{f(X)} = Y$$ .

Show that $$f$$ is dominant if and only if ϕ is injective.

He provides a proof of the forward implication and the reverse implication. I have come terms in understanding the forwards implication, but I have two questions regarding the reverse implication. Here is the proof given:

Assume that $$ϕ$$ is injective. Write $$K[X] = K[x_1, . . . , x_m]/I, K[Y ] = K[y_1, . . . , y_n]/J,$$ and let $$f$$ be given by $$f_1, . . . , f_n ∈ K[x_1, . . . , x_m].$$ Take $$g ∈ \mathcal{I}_{K[y_1,...,y_n]} (f(X)).$$ Then the polynomial $$g(f_1, . . . , f_n) ∈ K[x_1, . . . , x_m]$$ vanishes on $$X$$, so $$ϕ(g + J) = 0.$$ This implies $$g ∈ J.$$ We have shown that $$\mathcal{I}_{K[y_1,...,y_n]} (f(X)) ⊆ J$$, so

$$Y ⊆ \mathcal{V}_{K^n} (J) ⊆ \mathcal{V}_{K^n} (\mathcal{I}_{K[y_1,...,y_n]}(f(X))) = \overline{f(X)} ⊆ Y$$$$\square$$

Now form here, I can see why $$f(X)$$ is dominant.

Before I ask my questions let me clarify some of the author's notations:

$$\mathcal{I}_{K[y_1,...,y_n]} (f(X))$$ is the vanishing ideal of $$f(X)$$.

$$\mathcal{K^n}(J)$$ is the affine variety given by $$J$$.

For the most part, I am okay with the proof, my questions are as follows:

1. What justifies the containment $$Y ⊆ \mathcal{V}_{K^n} (J)$$?

2. What justifies the equality $$\mathcal{V}_{K^n} (\mathcal{I}_{K[y_1,...,y_n]}(f(X))) = \overline{f(X)}$$?

Any help would be appreciated.

1. We have $$k[Y]=K[y_1,\ldots,y_n]/J$$, this means that $$Y\subset K^n$$ is exactly the vanishing set of $$J$$. Or we can say that the class of any function $$a\in J$$ is equal to zero in $$K[y_1,\ldots,y_n]/J$$, hence for any maximal (or prime) ideal $$\mathfrak m\subset K[y_1,\ldots,y_n]/J$$ the element $$a$$ is equal to zero in $$(K[y_1,\ldots,y_n]/J)/\mathfrak m$$, i.e. it vanishes on the corresponding point.
2. $$\mathcal V(\mathcal I_{K[y_1,\ldots,y_n]}(f(X)))$$ is a closed set containing $$f(X)$$, hence $$\overline{f(X)}\subset \mathcal V(\mathcal I_{K[y_1,\ldots,y_n]}(f(X)))$$. Conversely, if $$p\in K^n$$ does not lie in the closure, then we can choose a distinguished open set $$D(b)$$ containing $$p$$ and not intersecting with $$\overline{f(X)}$$. This $$b$$ vanishes on $$f(X)$$, and does not vanish on $$p$$, hence $$p\notin\mathcal V(\mathcal I_{K[y_1,\ldots,y_n]}(f(X)))$$.