Dominant and Injective Morphisms reverse implication

Question

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.

Answer 1

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 $p\notin\mathcal V(\mathcal I_{K[y_1,\ldots,y_n]}(f(X)))$.