# Height of maximal ideals in a polynomial ring over a field

How to show that the height of maximal ideals in a polynomial ring of several variables over a field (not necessarily algebraically closed) are all the same?

It seems quite complex to make sure that all maximal ideals have the same length. E.g., in $$\mathbb{Q}[X]$$ maximal ideals are not always linear terms, $$(X^2-2)$$ is also maximal.

Lemma: Let $$k$$ be a field and $$A$$ be an affine domain of transcendence degree $$d$$ over $$k$$. Let $$P\subset A$$ be a height $$1$$ prime. Then $$A/P$$ has transcendence degree $$d-1$$ over $$k$$.
Proof: By Noether normalization, there exist algebraically independent elements $$x_1,\dots,x_d\in A$$ such that $$A$$ is integral over $$B=k[x_1,\dots,x_d]$$. Then $$P\cap B$$ is a height $$1$$ prime of $$B$$ ($$P\cap B\neq 0$$ by incomparability of primes lying over the same prime in an integral extension, and there can be no nonzero prime strictly contained in $$P\cap B$$ by going down). But $$B$$ is just a polynomial ring and in particular a UFD, so this means $$P\cap B$$ is generated by some irreducible polynomial $$f\in B$$. It follows that $$B/(P\cap B)$$ has transcendence degree $$d-1$$ over $$k$$ (explicitly, if $$x_i$$ is a variable that appears in $$f$$, then the other $$d-1$$ variables are still algebraically independent mod $$f$$ but $$x_i$$ is algebraically dependent over them via $$f$$). Since $$A/P$$ is integral and thus algebraic over $$B/(P\cap B)$$, $$A/P$$ also has transcendence degree $$d-1$$ over $$k$$.
Theorem: Let $$k$$ be a field and $$A$$ be an affine domain over $$k$$. Then every maximal chain of prime ideals in $$A$$ has length equal to the transcendence degree of $$A$$ over $$k$$.
Proof: Let $$d$$ be the transcendence degree of $$A$$ over $$k$$ and let $$0=P_0\subset P_1\subset\dots\subset P_n$$ be a maximal chain of prime ideals in $$A$$. By induction on $$i$$ using the Lemma, $$A/P_i$$ has transcendence degree $$d-i$$ for each $$i$$, and in particular $$A/P_n$$ has transcendence degree $$d-n$$. But by Zariski's lemma, $$A/P_n$$ is algebraic over $$k$$ (since $$P_n$$ is a maximal ideal so $$A/P_n$$ is a finitely generated algebra over $$k$$ which is a field). Thus $$d-n=0$$ so $$d=n$$, as desired.