Integral Extensions and Going Up - Examples

Every polynomial $f(X)\in k[X]$ can be written uniquely as $a(X^2)+Xb(X^2)$ with $a(T),b(T)\in k[T]$, by separating the even and odd powers of $X$. Therefore $k[X]=k[X^2]\oplus Xk[X^2]$, so $1$ and $X$ form a basis of $k[X]$ over $k[X^2]$.

Fix $f(X)\in k[X]$. Because $k[X]$ is a faithful module over the subring $k[X^2][f(X)]$ and step 1.1 shows that it is finitely generated over $k[X^2]$, [L1] implies that $f(X)$ is integral over $k[X^2]$.

Thus $k[X]$ is finite free of rank $2$ over $k[X^2]$, and every element of $k[X]$ is integral over that subring.

The elements $\sqrt{2}$ and $\sqrt{3}$ satisfy the monic equations $T^2-2=0$ and $T^2-3=0$ over $\mathbb{Z}$, so they are integral over $\mathbb{Z}$. Therefore [L1] implies that $R$ is a finitely generated $\mathbb{Z}$-module.

Every element of $R$ is a polynomial in $\sqrt{2}$ and $\sqrt{3}$. Replacing $(\sqrt{2}{)}^2$ by $2$ and $(\sqrt{3}{)}^2$ by $3$ reduces every monomial to a $\mathbb{Z}$-linear combination of $1,\sqrt{2},\sqrt{3},\sqrt{6}$, so these four elements span $R$ over $\mathbb{Z}$.

Since $\sqrt{2}$ and $\sqrt{3}$ are integral over $\mathbb{Z}$, [L2] implies that their sum and product are integral over $\mathbb{Z}$. Here the product is exactly $\sqrt{6}$, and the spanning result of step 2.1 places both elements inside one explicit finite $\mathbb{Z}$-module.

The element $1\mathrm{/}p$ already belongs to the base ring $\mathbb{Z}[1\mathrm{/}p]$, so it satisfies the monic equation $T-1\mathrm{/}p=0$ there. Hence it is integral over $\mathbb{Z}[1\mathrm{/}p]$.

Suppose $1\mathrm{/}p$ were integral over $\mathbb{Z}$. Then there would be a monic equation $(1\mathrm{/}p{)}^n+a_{n-1}(1\mathrm{/}p{)}^{n-1}+\cdots +a_0=0$ with $a_i\in \mathbb{Z}$. Multiplying by $p^n$ gives $1+a_{n-1}p+\cdots +a_0{p}^n=0$, impossible because the left side is congruent to $1$ modulo $p$.

Thus localisation makes $1\mathrm{/}p$ integral only after $p$ has become invertible in the base ring.

The element $t\in B$ satisfies the monic equation $T^2-t^2=0$ with coefficient $t^2\in A$, so $B$ is integral over $A$. Since both $A$ and $B$ are subrings of the domain $k[t]$, the zero prime of $B$ contracts to the zero prime of $A$.

The quotient $B\mathrm{/}(t)$ is $k$, and the image of $A$ in it is also $k$ because both $t^2$ and $t^3$ vanish. Hence $(t)\cap A=(t^2,t^3)$. In particular, the cusp maximal ideal of $A$ has a prime above it in $B$, namely $(t)$.

This is the concrete cusp-ring instance of lying over: the two natural base primes $(0)$ and $(t^2,t^3)$ are realised upstairs by $(0)$ and $(t)$.

Because $k$ is algebraically closed, the prime ideals of $k[X]$ are $(0)$ and the maximal ideals $(X-a)$ for $a\in k$. The inclusion is integral because $X$ satisfies the monic equation $T^2-Y=0$ over $k[Y]$.

The contractions are easy to compute: $(0)\cap k[Y]=(0)$, while $(X-a)\cap k[Y]=(Y-a^2)$ because a polynomial in $Y$ vanishes at $X=a$ exactly when it vanishes at $Y=a^2$. Thus a proper inclusion of primes in $k[X]$ can only be $(0)⊊(X-a)$, and its contractions are $(0)⊊(Y-a^2)$.

Hence every distinct comparable pair of primes in $k[X]$ has distinct contraction, exactly as [L1] predicts.

The ring $S$ is a domain, and $R$ is a subring of it, so $R$ is also a domain. Moreover $x$ satisfies the monic equation $T^2-T+x(1-x)=0$ with coefficient $x(1-x)\in R$, and $S=R[x]$ because $y\in R$ already. Hence $S$ is integral over $R$.

The ideals $P_0$ and $P$ are prime because they are contractions of the prime ideals $(x)$ and $(x,y)$ of $S$, and they are distinct because $y\in P\setminus P_0$. The ideal $Q=(1-x,y)$ is prime in $S$, and its contraction to $R$ is $P$, since mod $Q$ one has $x=1$ and $y=0$, so $x(1-x)$, $y$, and $xy$ all vanish.

The base ring is not integrally closed: the element $x\in S$ is integral over $R$ by step 1.1, but $x\mathrm{\notin }R$. Indeed, if $x\in R$, then setting $y=0$ would express $x$ as a polynomial in $x(1-x)$; evaluating at $x=0$ and $x=1$ would then give the same value on both inputs, impossible because $x$ takes the values $0$ and $1$.

Suppose $Q_0\subseteq Q$ were a prime ideal of $S$ with $Q_0\cap R=P_0$. Because $Q=(1-x,y)$ does not contain $x$, neither does $Q_0$. But $x(1-x)\in P_0\subseteq Q_0$ and $xy\in P_0\subseteq Q_0$, so primality of $Q_0$ and $x\mathrm{\notin }{Q}_0$ force $1-x\in Q_0$ and $y\in Q_0$. Hence $Q=(1-x,y)\subseteq Q_0$, and therefore $Q_0=Q$. This contradicts $Q_0\cap R=P_0$ because step 2.1 showed $Q\cap R=P\mathrm{\ne }{P}_0$.

Thus the integral extension of domains $R\subseteq S$ has a prime chain $P_0⊊P$ and a prime $Q$ over $P$ with no prime below $Q$ lying over $P_0$. So the normality hypothesis in [L1] is essential.