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For an integral extension $A\subseteq B$ and $\mathfrak{p}\in \mathrm{Spec}A$, localization gives a bijectionIt combines the prime ideal correspondence for localization with the contraction of a maximal ideal under an integral extension.

An element $t\in T$ is an integral element over $R$ when it satisfies a monic equationThe extension $R\subseteq T$ is an integral extension when every $t\in T$ is integral over $R$.

Let $P\in \mathrm{Spec}R$ and localize both rings at $S=R\setminus P$. The extension $R_P\subseteq S^{-1}T$ remains integral. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}T$. The contraction of a maximal ideal under an integral extension is maximal, and the local ring $R_P$ has unique maximal ideal $P{R}_P$, so $\mathfrak{n}\cap R_P=P{R}_P$. Contracting $\mathfrak{n}$ back to $T$ produces $Q\in \mathrm{Spec}T$ with $Q\cap R=P$. This proves the Lying-over theorem and hence the surjectivity of

Suppose $Q_1\subseteq Q_2$ and both contract to $P$. Quotient by $Q_1$ and localize the resulting integral domain at the nonzero elements of $R/P$. The localized ring is an integral domain integral over the field $\mathrm{Frac}(R/P)$, and is therefore itself a field. The localization of $Q_2/Q_1$ must consequently be zero; since the localized ring is a domain, $Q_2/Q_1=0$. Thus the incomparability theorem for integral extensions gives

The Krull dimension of a ring is the supremum of the lengths of its strict chains of prime ideals. Incomparability makes the contraction of every strict prime chain in $T$ strict, so $\dim T\le \dim R$. Conversely, start over the bottom prime of any chain in $R$ using lying over and lift each subsequent inclusion using the going-up theorem. Hence

For the concrete surface, write $x,y,z$ for the residue classes of $X,Y,Z$ and putThenand characteristic zero allows division by three. ThereforeDivision by the monic quadratic shows that $T$ is free over $k[u,v]$ with basis $1,z$. In particular, $u,v$ are algebraically independent and the requested Noether normalization of the quadratic surface xy plus yz plus zx is

Put $S=A\setminus \mathfrak{p}$ and definewhere $A$ acts on $B$ through the given inclusion. By the prime ideal correspondence for localization, primes of $B_{\mathfrak{p}}$ correspond to primes $\mathfrak{q}$ of $B$ satisfying $\mathfrak{q}\cap S=\varnothing$, equivalently $\mathfrak{q}\cap A\subseteq \mathfrak{p}$.

The localized extension $A_{\mathfrak{p}}\subseteq B_{\mathfrak{p}}$ remains integral. If $\mathfrak{q}\cap A=\mathfrak{p}$, thenis an integral domain integral over the field $A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$. An integral domain integral over a field is a field, so $S^{-1}\mathfrak{q}$ is maximal.

Conversely, if $\mathfrak{n}$ is maximal in $B_{\mathfrak{p}}$, the contraction of a maximal ideal under an integral extension is maximal in the local ring $A_{\mathfrak{p}}$, hence equals $\mathfrak{p}{A}_{\mathfrak{p}}$. Contracting once more to $A$ gives $\mathfrak{p}$. Extension and contraction are inverse under localization, proving the required fiber primes of an integral extension bijection.