Dimension of a scheme

In algebraic geometry, the dimension of a scheme is a generalization of the dimension of an algebraic variety. Scheme theory emphasizes the relative point of view and, accordingly, the relative dimension of a morphism of schemes is also important.

By definition, the dimension of a scheme X is the dimension of the underlying topological space: the supremum of the lengths ℓ of chains of irreducible closed subsets:

${\displaystyle \emptyset \neq V_{0}\subsetneq V_{1}\subsetneq \cdots \subsetneq V_{\ell }\subset X.}$^[1]

In particular, if ${\displaystyle X=\operatorname {Spec} A}$ is an affine scheme, then such chains correspond to chains of prime ideals (inclusion reversed), so the dimension of X is precisely the Krull dimension of A.

If Y is an irreducible closed subset of a scheme X, then the codimension of Y in X is the supremum of the lengths ℓ of chains of irreducible closed subsets:

${\displaystyle Y=V_{0}\subsetneq V_{1}\subsetneq \cdots \subsetneq V_{\ell }\subset X.}$^[2]

An irreducible subset of X is an irreducible component of X if and only if its codimension in X is zero. If ${\displaystyle X=\operatorname {Spec} A}$ is affine, then the codimension of Y in X is precisely the height of the prime ideal defining Y in X.

• If a finite-dimensional vector space V over a field is viewed as a scheme over the field,^{[note 1]} then the dimension of the scheme V is the same as the vector-space dimension of V.
• Let ${\displaystyle X=\operatorname {Spec} k[x,y,z]/(xy,xz)}$, 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 ${\displaystyle \operatorname {codim} (x,X)}$ is 2 if x lies in H and is 1 if it is in ${\displaystyle X-H}$. Thus, ${\displaystyle \operatorname {codim} (x,X)}$ 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 ${\displaystyle k}$.^[3] Also, if ${\displaystyle U}$ is a nonempty open subset of ${\displaystyle X}$, then ${\displaystyle \dim U=\dim X}$.^[4]
• Let R be a discrete valuation ring and ${\displaystyle X=\mathbb {A} _{R}^{1}=\operatorname {Spec} (R[t])}$ the affine line over it. Let ${\displaystyle \pi :X\to \operatorname {Spec} R}$ be the projection. ${\displaystyle \operatorname {Spec} (R)=\{s,\eta \}}$ 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,^{[note 2]} 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 ${\displaystyle R[t]\to R/{\mathfrak {m}}_{R},f\mapsto f(0){\bmod {\mathfrak {m}}}_{R}}$. 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 ${\displaystyle {\mathfrak {m}}_{R}[t]\subsetneq {\mathfrak {p}}_{2}}$. Consequently,

${\displaystyle \operatorname {codim} ({\mathfrak {p}}_{1},X)=1,\,\operatorname {codim} ({\mathfrak {p}}_{2},X)=2,}$

while X is irreducible.

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).

All irreducible schemes are equidimensional.^[5]

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.

Let ${\displaystyle f:X\rightarrow Y}$ 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 ^{[clarification needed]} are purely of the same dimension ${\displaystyle n}$, then one says that ${\displaystyle f}$ is of relative dimension ${\displaystyle n}$.^[6]

• Kleiman's theorem
• Glossary of scheme theory
• Equidimensional ring

• William Fulton. (1998), Intersection theory, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge., vol. 2 (2nd ed.), Berlin, New York: Springer-Verlag, ISBN 978-3-540-62046-4, MR 1644323
• Hartshorne, Robin (1977), Algebraic Geometry, Graduate Texts in Mathematics, vol. 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9, MR 0463157

• The Stacks Project authors. "28 Properties of Schemes/28.10 Dimension".
• The Stacks Project authors. "29.29 Morphisms of given relative dimension".

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