A Corollary on Estimation of Fiber Dimension from Kemper's Course on Commutative Algebra

I have a question about following proof from Kemper's Course in Commutative Algebra (page 142):

Corollary 10.6. Let $f: X \to Y$ be a morphism of equidimensional affine varieties over an algebraically closed field (!). For a point $y \in Y$ , every irreducible component $Z \subset f^{-1}(\{y\})$ of the fiber has dimension $$\dim(Z) \ge \dim(X) − \dim(Y) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10.8)$$

If $f$ is dominant and $Y$ is irreducible, there exists an open, dense subset $U \subset Y$ such that for every $y \in U$, the fiber $f^{-1}(\{y\})$ is nonempty and equidimensional of dimension $\dim(X) − \dim(Y )$.

Proof. Let $P \in \operatorname{Spec}_{max} (K[Y ])$ be the maximal ideal corresponding to a point $y \in Y$ . The fiber over y is an affine variety, so by Corollary 8.24, the inequality (10.8) follows if we can show that every maximal ideal in the coordinate ring of the fiber has height at least $d := \dim(X) − \dim(Y )$. By Proposition 7.11, this is the same as showing that every maximal ideal $Q$ in the fiber ring over $P$ has height at least $d$. Such a $Q$ corresponds to a maximal ideal $Q \in \operatorname{Spec}_{max} (K[X ])$, so $\operatorname{ht}(Q) = \dim(X)$ by Corollary 8.24.
Since $\operatorname{ht}(P) = \dim(Y)$ (again by Corollary 8.24), the inequality (7.7) in Theorem 7.12 guarantees $\operatorname{ht}(P) \ge d$, so (10.8) is proved.

Question: Where the proof uses that $k$ is algebraically closed? Everytime the statements of such kind with algebraically closedness assumption for base field occure, one thinks that somewhere the Nullstellensatz is involved giving a 1-1 identification between primes and irreducible subvarieties.

Here I nowhere found a reason to exploit it. Indeed it's true for every affine variety $Y \subset k^n$ that every point $y:=(y_1,..., y_n) \subset Y \subset k^n$ has an associated maximal ideal. It's the converse, that every maximal ideal arise as such geometric point, which indeed exploits the Nullstellensatz. But it is not used here as I see. So why we need here this algebraically closedness assumption on $k$?

I'm not going to spell out what the results 7.11, 7.12 and 8.24 (can be looked up in the book) but no one of them uses the algebraically closedness assumption of $k$.

I think what's going on here is that the algebraically closed hypothesis is being used implicitly when we compute the dimension of an algebraic variety by looking at its coordinate ring.

Kemper defines an algebraic variety $X$ over $K$ to be a subset of $K^n$ cut out by a set of polynomials in $K[x_1,...,x_n]$, and defines the dimension of $X$ to be its Krull dimension for the Zariski topology on $X$. This only coincides with the Krull dimension of the coordinate ring $K[X]$ when $K$ is algebraically closed. Over a general field, we lose the bijective correspondence between prime ideals in $K[X]$ and irreducible closed subsets of $X$. See Definition 5.1(c) on p. 62 of your book.

For an explicit example, let $X$ be the cone in $\mathbb{R}^3$ defined by $x^2+y^2-z^2=0$, let $Y$ be $\mathbb{R}^1$, and let $f\colon X\to Y$ be $f(x,y,z)=z$. Then $\dim(X)=2$ and $\dim(Y)=1$, so we expect each irreducible component of each fiber $f^{-1}(\{z\})$ to have dimension at least $1$. This is true when $z>0$ (the fibers are circles) and true vacuously when $z<0$ (the fibers are empty, so there are no irreducible components). But when $z=0$, the fiber is a point, which has dimension $0$.