Krull Dimension and Height Theorems — Examples

The only strict prime chain from $\mathfrak{p}$ to $\mathfrak{q}$ is $(x)⊊(x,y)$, so [L1] gives $\mathrm{ht}(\mathfrak{q}\mathrm{/}\mathfrak{p})=1$.

By [L2], $A$ has dimension $3$, $A\mathrm{/}\mathfrak{q}\cong k[z]$ has dimension $1$, and $A\mathrm{/}\mathfrak{p}\cong k[y,z]$ has dimension $2$. Thus $\mathrm{ht}(\mathfrak{q})=2$ and $\mathrm{ht}(\mathfrak{p})=1$, so the same relative height is $1$.

So the quotient-chain computation and the affine-dimension computation agree.

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.

By [L2], $\mathrm{ht}(I)+\dim (A\mathrm{/}I)=\dim A=n$. Since $\dim (A\mathrm{/}I)=0$, we get $\mathrm{ht}(I)=n$. This matches the upper bound from [L1].

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

The quotient $R\mathrm{/}(\bar{x})\cong k[y{]}_{(y)}\mathrm{/}(y^2)$ is Artinian local, so $(\bar{x})$ has radical $\mathfrak{m}$. Thus there exists a one-generated ideal with maximal radical, and [L2] gives $\dim R\le 1$.

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.

Since $R$ is a local domain and not a field, [L3] shows that $\dim R\mathrm{\ne }0$. Together with 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], $\dim R=2$. Since $R\mathrm{/}(x)\cong k[y]$ has dimension $1$, [L3] gives $\mathrm{ht}((x))=1$. Therefore [L4] yields $\dim (S^{-1}R)=\dim (R_{(x)})=1$.

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 $\dim k=0$, so [L1] gives $\dim k[x]=1$.

The ring $A=k[t]\mathrm{/}(t^2)$ is Artinian local with the single prime $(\bar{t})$, so $\dim A=0$. Applying [L1] again gives $\dim A[x]=1$. The nilpotent element $\bar{t}$ 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.

The ring $A\cong k[x]$ has dimension $1$, and [L1] at the zero prime reads $0+{\mathrm{trdeg}}_k\mathrm{Frac}(A)=1$, which is true because $\mathrm{Frac}(A)\cong k(x)$.

For the maximal ideal $\mathfrak{m}$, the quotient is $A\mathrm{/}\mathfrak{m}\cong k$, so its quotient dimension is $0$. Then [L2] gives $\mathrm{ht}(\mathfrak{m})=1$, and the formula becomes $1+0=1$.

Thus the affine dimension formula is verified term by term on this plane curve domain.