An affine $K$-algebra factored by a maximal ideal is contained in a finitely generated $K$-domain

Question

I am trying to understand the proof of Proposition 1.2 from Gregor Kemper's "A Course in Commutative Algebra", which says that if $\varphi: A \rightarrow B$ is a homomorphism of the algebras $A,B$ over the field $K$ and if $B$ is an affine $K$-algebra and $\mathfrak{m}$ is a maximal ideal in $B$, then $\varphi^{-1}(\mathfrak{m})$ is a maximal ideal in $A$.

The step I'm having trouble with is when he uses Lemma 1.1,(b), which he proves previously and which states that an if an algebra $A$ over a field $K$ is itself a field and also $A$ is contained in an affine $K$-domain, then $A$ must be algebraic over $K$. This lemma is used for the $K$-algebra $B/\mathfrak{m}$ to conclude that $B/\mathfrak{m}$ is algebraic over $K$. I know that $B/\mathfrak{m}$ is a field because $\mathfrak{m}$ is maximal. But why is $B/\mathfrak{m}$ contained in an affine $K$-domain? Which is this finitely generated $K$-domain or how do we know $B$ is contained in one?

(A similar question was asked here for prime ideals instead of maximals, but got no answer.)

Answer 1

$B/\mathfrak{m}$ itself is an affine $K$-domain. It is finitely generated as a $K$-algebra since $B$ is finitely generated as a $K$-algebra. It is a domain since it is a field.

Answer 2

I do not have a copy of your book and cannot comment on "affine $K$-algebras", but the following seems to hold: Let $A,B$ be commutative unital $k$-algebras with $k$ a field, where $B$ is finitely generated over $k$.

Lemma. Let $\phi:A \rightarrow B$ be a map of rings and let $\mathfrak{m}\subseteq B$ be a maximal ideal. It follows $\phi^{-1}(\mathfrak{m})\subseteq A$ is a maximal ideal.

Proof. Assume $k$ is a field and let $\mathfrak{n}:=\phi^{-1}(\mathfrak{m})$. We get an inclusion of rings

$$ k \subseteq \kappa(\mathfrak{n}) \subseteq \kappa(\mathfrak{m})$$

and $\kappa(\mathfrak{n})$ is an integral domain. Since $\kappa(\mathfrak{m})$ is a finite extension of $k$ (this follows from the HNS) the following holds: Let $0\neq x\in \kappa(\mathfrak{n})$ be an element. There is a canonical surjective map

$$ \psi: k[t]\rightarrow k(x)$$

defined by $\psi(t):=x$, with $ker(\psi):=(p(t))$ for a non-zero ideal $p(t)\in k[t]$. Since $k(x)$ is an integral domain it follows $(p(t))$ is a maximal ideal, hence $x$ is a unit in $\kappa(\mathfrak{n})$. Since $x$ was an arbitrary non-zero element, it follows $\kappa(\mathfrak{m})$ is a field and hence $\mathfrak{n}$ is a maximal ideal. QED

I believe a similar proof holds when $k$ is the ring of integers.

There is a class of rings - Hilbert-Jacobson rings - where the following result holds:

Theorem. Let $A$ be a Hilbert-Jacobson ring and let $\phi: A\rightarrow B$ be a map of commutative unital rings where $B$ is a finitely generated $A$-algebra. It follows $B$ is a Hilbert-Jacobson ring, and for any maximal ideal $I\subseteq B$ it follows $J:=\phi^{-1}(I)\subseteq A$ is a maximal ideal and the canonical map $A/J \rightarrow B/I$ is a finite extension of fields.

A ring $A$ is a Hilbert-Jacobson ring iff every prime ideal in $A$ is the intersection of the maximal ideals containing it. You may find this result proved in Bourbakis "Algebre Commutative" V5.