In algebraic geometry, thedimension of a scheme is a generalization of adimension of an algebraic variety.Scheme theory emphasizes therelative point of view and, accordingly, therelative dimension of amorphism of schemes is also important.

By definition, the dimension of a schemeX is the dimension of the underlyingtopological space: the supremum of the lengthsℓ of chains ofirreducible closed subsets:

${\displaystyle \mathrm{\varnothing }\ne V_0⊊V_1⊊\cdots ⊊V_{\ell }\subset X.}$^[1]

In particular, if${\displaystyle X=\mathrm{Spec}A}$ is anaffine scheme, then such chains correspond to chains ofprime ideals (inclusion reversed) and so the dimension ofX is precisely theKrull dimension ofA.

IfY is an irreducible closed subset of a schemeX, then the codimension ofY inX is the supremum of the lengthsℓ of chains of irreducible closed subsets:

${\displaystyle Y=V_0⊊V_1⊊\cdots ⊊V_{\ell }\subset X.}$^[2]

An irreducible subset ofX is anirreducible component ofX if and only if the codimension of it inX is zero. If${\displaystyle X=\mathrm{Spec}A}$ is affine, then the codimension ofY inX is precisely the height of the prime ideal definingY inX.

• If a finite-dimensionalvector spaceV over a field is viewed as a scheme over the field,^{[note 1]} then the dimension of the schemeV is the same as the vector-space dimension ofV.
• Let${\displaystyle X=\mathrm{Spec}k[x,y,z]/(xy,xz)}$,k a field. Then it has dimension 2 (since it contains thehyperplane${\displaystyle H=\{x=0\}\subset {\mathbb{A}}^3}$ as an irreducible component). Ifx is a closed point ofX, then${\displaystyle \mathrm{codim}(x,X)}$ is 2 ifx lies inH and is 1 if it is in${\displaystyle X-H}$. Thus,${\displaystyle \mathrm{codim}(x,X)}$ for closed pointsx can vary.
• Let${\displaystyle X}$ be an algebraic pre-variety; i.e., an integral schemeof finite type over a field${\displaystyle k}$. Then the dimension of${\displaystyle X}$ is thetranscendence degree of thefunction 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]
• LetR be adiscrete 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 themaximal 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 ofR 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${\displaystyle R[t]/({\omega }_{R}t-1)=R[{\omega }_R^{-1}]=}$ thefield of fractions ofR. Also,${\displaystyle {\mathfrak{p}}_1}$ has height one byKrull's principal ideal theorem and${\displaystyle {\mathfrak{p}}_2}$ has height two since${\displaystyle {\mathfrak{m}}_R[t]⊊{\mathfrak{p}}_2}$. Consequently,

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

whileX is irreducible.

Anequidimensional scheme (or,pure dimensional scheme) is ascheme all of whoseirreducible components are of the samedimension (implicitly assuming the dimensions are all well-defined).

All irreducible schemes are equidimensional.^[5]

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

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

Let${\displaystyle f:X\to Y}$ be amorphism locally offinite type between twoschemes${\displaystyle X}$ and${\displaystyle Y}$. The relative dimension of${\displaystyle f}$ at a point${\displaystyle y\in Y}$ is thedimension of thefiber${\displaystyle f^{-1}(y)}$. 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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