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.

Definition

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:

$\emptyset \neq V_{0}\subsetneq V_{1}\subsetneq \cdots \subsetneq V_{\ell }\subset X.$

In particular, if $X=\operatorname {Spec} A$ 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:

$Y=V_{0}\subsetneq V_{1}\subsetneq \cdots \subsetneq V_{\ell }\subset X.$

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

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

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

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

See also

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

References

• 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

External links

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