Krull Dimension and Height Theorems — Examples

In $k[y,z]$, let $Q⊊(y)$ be prime. If $Q$ contained a nonzero polynomial $f$, write $f=y^{r}g$ with $r\ge 1$ and $y\nmid g$. Primality would give $y\in Q$ or $g\in Q$; the latter is impossible because $g\mathrm{\notin }(y)$, while the former contradicts $Q⊊(y)$. Hence $Q=(0)$. By [L1], the only strict chain from $\mathfrak{p}$ to $\mathfrak{q}$ is $(x)⊊(x,y)$ and $\mathrm{ht}(\mathfrak{q}\mathrm{/}\mathfrak{p})=1$.

The quotient correspondence identifies the chain from step 1.1 with $(0)⊊(y)$ in $A\mathrm{/}\mathfrak{p}$, confirming its length is one.

Therefore the relative height in the stated quotient is one.

In $R$, one has $\bar{x}\bar{y}=0$ with both factors nonzero, so $\bar{x}$ is a zero divisor. Also $R\mathrm{/}(\bar{x})\cong k[y]$ is a domain, hence $(\bar{x})$ is a prime ideal minimal over the principal ideal $(\bar{x})$.

The ideal $(\bar{x})$ is one of the two minimal primes of $k[x,y]\mathrm{/}(xy)$, so [L1] gives $\mathrm{ht}((\bar{x}))=0$. Therefore the exact-height-one conclusion fails for a zerodivisor generator.

So a principal ideal generated by a zero divisor can indeed have a minimal prime of height zero.

The quotient $A\mathrm{/}I\cong k$ is a field, so $I$ is maximal and therefore prime.

For $n\ge 1$, the chain $(0)⊊(x_1)⊊\cdots ⊊(x_1,\dots ,x_n)=I$ consists of primes, since each quotient is a polynomial domain over $k$; it gives $\mathrm{ht}(I)\ge n$. Fact [L1] gives the opposite bound. When $n=0$, $A=k$ and $I=(0)$ has height zero. Thus $\mathrm{ht}(I)=n$ in all cases.

Therefore the height bound in Krull's height theorem is sharp.

Put $T=k[x,y]\mathrm{/}(y^2-x^3)$. Division by the monic polynomial in $y$ writes every element uniquely as $a(x)+yb(x)$; substituting $x=t^2$, $y=t^3$ maps this to $a(t^2)+t^{3}b(t^2)$, whose even and odd powers cannot cancel. Thus $T$ embeds in $k[t]$ and is a domain. Its fraction field is algebraic over $k(x)$ because $y^2=x^3$, so [L2] gives $\dim T=1$; localizing at $(x,y)$ gives $\dim R\le 1$. Also $R\mathrm{/}(\bar{x})\cong k[y{]}_{(y)}\mathrm{/}(y^2)$ is Artinian local, so $(\bar{x})$ has radical $\mathfrak{m}$.

If $\mathfrak{m}$ were principal, then $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ would be one-dimensional over the residue field. But the classes of $\bar{x}$ and $\bar{y}$ are linearly independent modulo ${\mathfrak{m}}^2$, so $\mathfrak{m}$ is not principal. In particular $\mathfrak{m}\mathrm{\ne }(0)$, so $R$ is not a field.

The ring $R$ is a domain by step 1.1, so $(0)$ is prime. Its nonzero maximal ideal $\mathfrak{m}$ from step 1.2 gives a strict prime chain $(0)⊊\mathfrak{m}$, hence $\dim R\ge 1$ directly. Together with $\dim R\le 1$ from step 1.1, this forces $\dim R=1$. Then [L1] makes $(\bar{x})$ a parameter ideal.

Therefore $(\bar{x})$ is a parameter ideal while $\mathfrak{m}$ is not principal, so a system of parameters need not minimally generate the maximal ideal.

By [L2], $\dim R=2$. If a prime $Q⊊(x)$ contained a nonzero $f=x^{r}g$ with $x\nmid g$, primality would imply $x\in Q$ or $g\in Q$. The former contradicts strict containment, and the latter contradicts $g\mathrm{\notin }(x)$. Thus $Q=(0)$, and the chain $(0)⊊(x)$ proves $\mathrm{ht}((x))=1$. By [L4], $\dim (S^{-1}R)=\dim (R_{(x)})=1$.

Fact [L1] independently gives $\dim (S^{-1}R)\le \dim R=2$, so the computed value $1$ is compatible with the general one-sided inequality. Since $1<2=\dim R$, this localization strictly lowers dimension.

So localization can strictly lower Krull dimension.

For the field $k$, one has $\dim k=0$ and [L1] gives $\dim k[x]=1$.

The ring $A=k[t]\mathrm{/}(t^2)$ is Artinian local with the single prime $(\bar{t})$, so $\dim A=0$. Every prime of $A[x]=k[t,x]\mathrm{/}(t^2)$ contains the nilpotent $\bar{t}$, and [L3] identifies its prime chains with those of $A[x]\mathrm{/}(\bar{t})\cong k[x]$. Hence [L1] gives $\dim A[x]=1$. The nilpotent element does not change the one-step dimension jump.

The zero ring is excluded from the statement because it has no prime ideals at all. Its polynomial ring is again the zero ring, so the expression $\dim 0+1$ does not describe its behavior. Thus the theorem is intentionally stated only for ordinary Noetherian rings with the established zero-ring convention left separate.

Substituting $y=x^2$ gives $A\cong k[x]$. By [L1], any nonzero prime of this PID is maximal, and the chain $(0)⊊(x)$ shows that its dimension is exactly one. The zero prime has height zero and quotient $A$ of dimension one.

By [L2], $A\mathrm{/}\mathfrak{m}\cong k$ has dimension zero. The chain $(0)⊊\mathfrak{m}$ and [L1] give $\mathrm{ht}(\mathfrak{m})=1$. Thus $\mathrm{ht}(\mathfrak{m})+\dim (A\mathrm{/}\mathfrak{m})=1+0=1$; at $(0)$ the corresponding sum is $0+1=1$.

Both displayed instances hold by direct computation in $k[x]$, without invoking the general affine dimension formula.