= Solution
An element $t\in T$ is an <integral element> over $R$ when it satisfies a monic equation
$$
t^n+r_{n-1}t^{n-1}+\cdots+r_0=0,
\qquad r_i\in R.
$$
The extension $R\subseteq T$ is an <integral extension> when every $t\in T$ is integral over $R$.
Let $P\in\operatorname{Spec}R$ 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 $PR_P$, so $\mathfrak n\cap R_P=PR_P$. Contracting $\mathfrak n$ back to $T$ produces $Q\in\operatorname{Spec}T$ with $Q\cap R=P$. This proves the <Lying-over theorem> and hence the surjectivity of
$$
\pi:\operatorname{Spec}T\longrightarrow\operatorname{Spec}R.
$$
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 $\operatorname{Frac}(R/P)$, 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
$$
\boxed{Q_1=Q_2.}
$$
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 $\dim T\leq\dim R$. 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
$$
\boxed{\dim T=\dim R.}
$$
For the concrete surface, write $x,y,z$ for the residue classes of $X,Y,Z$ and put
$$
u=x-z,\qquad v=y-z.
$$
Then
$$
xy+yz+zx=uv+2(u+v)z+3z^2,
$$
and characteristic zero allows division by three. Therefore
$$
T\cong k[u,v][z]\big/\left(z^2+\frac23(u+v)z+\frac13uv\right).
$$
Division 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
$$
\boxed{R=k[x-z,y-z]\cong k[U,V].}
$$
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