Dimension of a scheme

Examples

• If a finite-dimensional vector space V over a field is viewed as a scheme over the field, then the dimension of the scheme V is the same as the vector-space dimension of V.
• Let ${\displaystyle X=\mathrm{Spec}k[x,y,z]/(xy,xz)}$, k a field. Then it has dimension 2 (since it contains the hyperplane ${\displaystyle H=\{x=0\}\subset {\mathbb{A}}^3}$ as an irreducible component). If x is a closed point of X, then ${\displaystyle \mathrm{codim}(x,X)}$ is 2 if x lies in H and is 1 if it is in ${\displaystyle X-H}$. Thus, ${\displaystyle \mathrm{codim}(x,X)}$ for closed points x can vary.
• Let ${\displaystyle X}$ be an algebraic pre-variety; i.e., an integral scheme of finite type over a field ${\displaystyle k}$. Then the dimension of ${\displaystyle X}$ is the transcendence degree of the function field ${\displaystyle k(X)}$ of ${\displaystyle X}$ over ${\displaystyle k}$. Also, if ${\displaystyle U}$ is a nonempty open subset of ${\displaystyle X}$, then ${\displaystyle \dim U=\dim X}$.
• Let R be a discrete valuation ring and ${\displaystyle X={\mathbb{A}}_R^1=\mathrm{Spec}(R[t])}$ the affine line over it. Let ${\displaystyle \pi :X\to \mathrm{Spec}R}$ be the projection. ${\displaystyle \mathrm{Spec}(R)=\{s,\eta \}}$ consists of 2 points, ${\displaystyle s}$ corresponding to the maximal ideal and closed and ${\displaystyle \eta }$ the zero ideal and open. Then the fibers ${\displaystyle {\pi }^{-1}(s),{\pi }^{-1}(\eta )}$ are closed and open, respectively. We note that ${\displaystyle {\pi }^{-1}(\eta )}$ has dimension one, while ${\displaystyle X}$ has dimension ${\displaystyle 2=1+\dim R}$ and ${\displaystyle {\pi }^{-1}(\eta )}$ is dense in ${\displaystyle X}$. Thus, the dimension of the closure of an open subset can be strictly bigger than that of the open set.
• Continuing the same example, let ${\displaystyle {\mathfrak{m}}_R}$ be the maximal ideal of R and ${\displaystyle {\omega }_R}$ a generator. We note that ${\displaystyle R[t]}$ has height-two and height-one maximal ideals; namely, ${\displaystyle {\mathfrak{p}}_1=({\omega }_{R}t-1)}$ and ${\displaystyle {\mathfrak{p}}_2=}$ the kernel of ${\displaystyle R[t]\to R/{\mathfrak{m}}_R,f\mapsto f(0){mod\mathfrak{m}}_R}$. The first ideal ${\displaystyle {\mathfrak{p}}_1}$ is maximal since ${\displaystyle R[t]/({\omega }_{R}t-1)=R[{\omega }_R^{-1}]=}$ the field of fractions of R. Also, ${\displaystyle {\mathfrak{p}}_1}$ has height one by Krull's principal ideal theorem and ${\displaystyle {\mathfrak{p}}_2}$ has height two since ${\displaystyle {\mathfrak{m}}_R[t]⊊{\mathfrak{p}}_2}$. Consequently,

${\displaystyle \mathrm{codim}({\mathfrak{p}}_1,X)=1,\mathrm{codim}({\mathfrak{p}}_2,X)=2,}$

while X is irreducible.

Equidimensional scheme

An equidimensional scheme (or, pure dimensional scheme) is a scheme whose irreducible components are of the same dimension (implicitly assuming the dimensions are all well-defined).

Examples

All irreducible schemes are equidimensional.

In an affine space, the union of a line and a point not on the line is not equidimensional. Generally, if two closed subschemes of some scheme, neither containing the other, have unequal dimensions, then their union is not equidimensional.

If a scheme is smooth (for instance, étale) over Spec k for some field k, then every connected component (which is then, in fact, an irreducible component) is equidimensional.

Relative dimension

Let ${\displaystyle f:X\to Y}$ be a morphism locally of finite type between two schemes ${\displaystyle X}$ and ${\displaystyle Y}$. The relative dimension of ${\displaystyle f}$ at a point ${\displaystyle y\in Y}$ is the dimension of the fiber ${\displaystyle f^{-1}(y)}$. If all the nonempty fibers are purely of the same dimension ${\displaystyle n}$, then one says that ${\displaystyle f}$ is of relative dimension ${\displaystyle n}$.