Definition A topological space is called noetherian if every descending chain of closed subsets terminates.

Remark If R is anoetherian ring, then X = Spec(R) is a noetherian topological space, since the closed subsets correspond to ideals.

Proposition Every closed subset in a noetherian topological space can be uniquely decomposed as an irredundant union of irreducible subspaces.

Proof: Use the same proof that showed the corresponding result for algebraic subsets of affine space over a field.

Definition Let X be a noetherian topological space. The dimension of X is defined to be
dim(X) = sup { n : there exists a chain of nonempty irreducible distinct
closed subsets Z0 Z1 Zn in X},

provided that this supremum exists.


  1. dim(*) = 0.
  2. If k is a field, then dim(A1k) = 1, since the irreducible sets are either points or all of A1.
  3. If k is a field, then dim(A2k) = 2, since the irreducible sets are either all of A2, an irreducible curve, or a point.
  4. With the usual topology on the real line R, one has dim(R)=0. For R is not irreducible, and the only irreducible closed subsets of R are single points.
  5. dim(Spec(Z)) = 1, since every prime ideal is either zero or maximal.

Definition Let P be a prime ideal in a ring R (as always, commutative with 1). The height of P is defined to be
ht(P) = sup { n : there exists a chain of distinct prime ideals
P0 P1 Pn = P R}

Definition The Krull dimension of a ring R is defined to be

dim(R) = sup{ht(P) : P R is prime}.

Lemma Let X = Spec(R) be an affine scheme. Then

dim(X) = dim(R).

Proof: There is a one-to-one correspondence between the irreducible closed subsets of X and the prime ideals of R.

Proposition Let X be a noetherian topological space. Then
  1. dim(X) = sup(dim(Xi)), where the Xi range through the irreducible components of X.
  2. If U X is a dense open set and if dim(U) is finite, then dim(X) = dim(U).
  3. If Y- X, then dim(Y) dim(X). Moreover, if X is irreducible and the inclusion is proper, then the inequality is strict.

Proof: (i) Let Z0 Zn be a proper chain of irreducible closed subsets of X. Since

Zn = i(Xi Zn)

and Zn is irreducible, there exists an i such that Zn Xi. But then the entire chain is contained inside Xi.

(ii) By (i), we may assume that X is irreducible. If Z0 Zn is a proper chain of closed irreducible subsets of U, then their closures Z-0Z-n form a proper chain of closed irreducibles of X. (Properness follows because Zi = UZ-i, and closures always stay irreducible.) So dim(U) dim(X).

Since dim(U) is finite, we can choose a chain Z0 Zn in U of maximal length. Now let W be any irreducible closed subset of X such that Z-i WZ-i+1. Intersecting back with U yields

Zi W U Zi+1.

Since the original chain had maximal length, one of these inclusions must be an equality. But W U is dense in the irreducible space W, hence one of the earlier inclusions in X was already an equality. It follows that Z-0Z-n is a maximal chain in X, and the dimensions are equal.

(iii) Without loss of generality, we may assume that Y=Y-. Let Z0 Zn be a proper chain in Y. Then it is also a proper chain in X. If the containment is proper, then we can add X to the top of the chain to lengthen it.

Proposition Let X be a scheme that is a noetherian topological space. Then dim(X)=0 if and only if X is a finite set with the discrete topology.

Proof: It is clear that any set with the discrete topology has dimension zero, since the only irreducible subsets are singeltons. On the other hand, we know that every scheme contains at least one closed point (coming from a maximal ideal inside some ring defining an open affine subset). If X contained a non-closed point, then its closure would be irreducible, forcing the dimension of X to be greater than zero. So, any zero-dimensional scheme has the property that every point is closed. Thus, in any affine open subset Spec(R) of X, the ring R must have the property that every prime ideal is maximal. This forces Spec(R), and hence X to have the discrete topology. Now use the finite decomposition into irreducibles for the noetherian topological space X to obtain the desired conclusion.

Theorem (Dimension Theorem) Let X be an algebraic variety over a field k. Then

dim(X) = tr.degk(k(X)).

Proof: Since every variety contains an open, dense, affine variety, the result will follow if we can establish the following three facts.
  1. The coordinate ring A(X) is an integral extension of a polynomial ring.
  2. If B A is an integral extension of noetherian rings, then dim(B)=dim(A) and tr.deg(B)=tr.deg(A).
  3. If k is a field, then dim(Ank) = n.

Let's start with the easiest one first.

Proposition If k is a field, then dim(Ank) = n.

Proof: Since (0) (x1) (x1, x2) (x1,, xn) is a proper chain of prime ideals, we have dim(An) n.

On the other hand, we know that n = tr.deg(k[x1,,xn]). Moreover, if A is an integral domain and if P is any nonzero prime ideal in A, then tr.deg(A/P)<tr.deg(A). For, take any algebraically independent set x-1, , x-r A/P. Lift them arbitrarily to elements x1, , xr A, and choose a nonzero element x P. Then {x1, , xr, x} is an algebraically independent set in A. To see this, suppose F(t1,, tr, tr+1) k[t1,,tr+1] is any polynomial such that F(x1,,xr,x) = 0 in A. Since A is an integral domain, we can assume that F is an irreducible polynomial. Then F(x-1,,x-r,0) = 0 in A/P. Since there are no nonzero algebraic relations between these elements in A/P, the polynomial F(t1,,tr,0) must be identically zero. But then F is divisible by tr+1; by irreducibility, F=tr+1. Since x0 in A; this is a contradiction.

Now let P0 Pt be a chain of prime ideals in k[x1, , xn]. Then

n > tr.deg(A/P0) > tr.deg(A/P1) >> tr.deg(A/Pt),

so n t. Thus, dim(An) n, and the result follows.

In order to establish the other two parts, we need some preliminaries from commutative algebra; in particular, we need to know how prime ideals behave in integral extensions of rings. To understand that, we introduce one of the essential tools of commutative algebra: localization.

Definition If A is a ring and if P is a prime ideal in A, we define the localization of A at P to be the ring AP obtained by inverting all elements of A\ P.

Remark The localization AP is a local ring, with maximal ideal PAP.

Definition Let f:B A be an extension of rings. If P A is a prime ideal, then Q=f-1(P) = P B is a prime ideal in B. In this circumstance, we say that P lies over Q.

Lemma Let B be a local ring with maximal ideal Q. Let B A be an integral extension. Then the set of prime ideals of A lying over Q is just the set of maximal ideals of A.

Proof: We show first that every maximal ideal M of A lies over Q. Define N=M B. Then A- = A/M is a field that is integral over the subring B- = B/N. Now, let 0 x B-. Since 1/xA-, we have a monic polynomial equation

(1/x)n + b1(1/x)n-1++bn = 0

with coefficients in B-. Now multiply by xn-1 to get

(1/x) = - (b1 + b2 x ++ bnxn-1) B-.

So, B- is a field and N=Q is the unique maximal ideal in B.

Next, we must show that every ideal P in A that lies over Q is maximal. This time, A- = A/P is an integral domain that is integral (thus algebraic) over the field B-=B/Q. Take 0 y A-. There is an algebraic dependence relation

b0yn + b1yn-1 ++bn = 0

of minimal degree with coefficients in B-. By minimality, bn0. Since B- is a field, we can divide the polynomial by bn, and assume that bn=1. Now we have

[1 / y] = -(b0yn-1 + b1yn-2 ++bn-1) A-.

Hence, A- is a field and P is maximal.

Proposition (The Lying-Over Theorem) Let B A be an integral extension of rings. If Q B is any prime ideal, then there exists a prime ideal P A lying over Q.

Proof: Let BQ be the localization of B at Q. Then AQ = (B\ Q)-1A is an integral extension of BQ, and contains it as a subring. The prime ideals of A lying over Q correspond to the prime ideals of AQ lying over QBQ, which are necessarily the maximal ideals of AQ. Since BQ0, we know that AQ is nonzero, so it has maximal ideals.

Proposition (The Going-Up Theorem) Let B A be an integral extension of rings. Let Q Q' B be prime ideals and let P A be a prime ideal lying over Q. Then there exists an ideal P' lying over Q' such that P P'.

Proof: The quotient A/P contains and is integral over B/Q. By the Lying Over Theorem, there exists a prime P'/P in A/P lying over the prime ideal Q'/Q. By the isomorphism theorems, P' is a prime ideal of A lying over Q'.

Lemma Let B be an integral domain that is integrally closed in its field of fractions L. Let K/L be a normal extension with Galois group G, and let A be the integral closure of B in K. If Q is any prime ideal of B, then G acts transitively on the set of prime ideals lying over Q.

Proof: One can assume (with some work that I'm omitting) that K/L is a finite Galois extension. Suppose now that P' and P are two primes lying over Q, and that P' is not contained in any of the conjugates Pi= σi(P) of P for σi G. Then there is an element x P' that is contained in no Pi. But y=NK/L(x) B is not in Q (because all σi(x) P) and is in P' A. This is a contradiction.

Proposition (The Going-Down Theorem) Let B A be an integral extension of integral domains, and assume that B is integrally closed in its field of fractions. Let Q Q' B be prime ideals and let P' A be a prime ideal lying over Q'. Then there exists an ideal P lying over Q such that P P'.

Proof: Let K be the field of fractions of A, let L be the field of fractions of B, and let F be the Galois closure of the field extension K/L. Inside F, let C denote the integral closure of B (and hence also of A). By the Lying Over Theorem, there exists a prime ideal Q0 in C lying over Q. By the Going Up Theorem, there exists a prime ideal Q'0 Q0 lying over Q'. Also by the Lying Over Theorem, there exists a prime ideal Q'1C lying over P'. Now the ideals Q'0 and Q'1 lie over the same ideal Q' in B. Because the extension is Galois, there exists an automorphism σ Gal(F/L) such that σ(Q'0) = Q'1. Now the result follows by taking P= σ(Q0) A.

Now we can carry out part (ii) of the proof of the dimension theorem.

Proposition Let B A be an integral extension of noetherian rings. Then dim(B)=dim(A) and tr.deg(B) = tr.deg(A).

Proof: Since integral extensions are algebraic, the equality for transcendence degrees is trivial. The Lying Over and Going Up Theorems show that any chain of prime ideals in B can be lifted to a chain in A of the same length; thus, dim(A) dim(B). Conversely, if P P' are distinct prime ideals in A, then the ideals P B P' B must also be distinct. (Localize at the prime ideal P B; since the ideals lying over its maximal ideal are precisely the maximal ideals of the localization of A, there cannot be any proper containments between them.) Thus, any chain of primes in A produces a chain of the same length in B, and dim(A) dim(B).

Finally, we can carry out step (i).

Theorem Let A=k[x1, , xn] be a polynomial ring over a field k, and let P be an ideal of A of height h. Then there exist elements v1, , vn A such that
  1. A is integral over k[v], and
  2. P k[v] = (v1, , vh).

Proof: The proof is by induction on the height h=ht(P). The case h=0 is trivial, for then P=0 and we can take vi=xi. A better base for the induction, however, is provided by the case h=1. Choose a nonzero polynomial v1=f(x) P. Write f(x) = i=1s ci Mi(x) where ci k and each Mi(x) is a monomial. Given any positive integers d1,,dn, we define the d-weight of a monomial M(x)= xiai to be aidi. Now choose weights d1=1, d2, , dn so that all the monomials appearing in f(x) have distinct weights. (We can achieve this by using large prime powers for the weights.) Put vi=xi-x1di for i=2,,n. Then
v1 = f(x) = f(x1, v2+x1d2, , vn + x1dn)
= ajx1e + g(x1, v2, , vn)

where g is a polynomial whose degree in x1 is strictly less than e, and aj is the coefficient of the term of largest weight in f. It follows that x1 is integral over the ring k[v1, , vn], and hence so are the elements xi= vi+x1di.

We're not done yet, however, since we only know that k[v] k[x] is an integral extension. We still need to mod out the prime ideal P and its pullback P k[v]. We clearly have a containment (v1) P k[v]. In fact, however, both these ideals are primes of height 1, so they must be equal. (Note: that P k[v] is height 1 uses the fact that k[v] is integrally closed, so the Going Down Theorem applies.)

Now suppose h>1. Choose an ideal Q P of height h-1. By induction, there exist w1, , wn such that k[x] is integral over k[w] and Q k[w] = (w1, , wh-1). Define P'=P k[w]. Then P' also has height h, so it contains (w1,,wh-1) properly. Choose a nonzero polynomial of the form f(wh, , wn) P' and repeat the weight argument of the case h=1.

Theorem (Noether Normalization Theorem) Let A be an integral domain that is finitely generated as an algebra over a field k. Then there exist elements y1, , yr A that are algebraically independent over k such that A is integral over k[y1, , yr].

Proof: Write A=k[x1, , xn]/P where P is a prime ideal of height n-r. By the previous theorem, there are elements v1, , vn in k[x1, , xn] such that k[v] k[x] is integral and such that P k[v] = (vr+1, , vn). The result follows by taking yi vi mod P.

Corollary Let A be an integral domain that is finitely generated as a k-algebra. Then for any prime P in A, we have

ht(P) + dim(A/P) = dim(A).

Proof: Reduce to the case A=k[x1,,xn] and use the proof of the Noether Normalization Theorem.

Corollary If X and Y are algebraic varieties, then dim(X× Y) = dim(X) + dim(Y).

Proof: The transcendence degree of a tensor product satisfies the corresponding relation.

Proposition Let X be a hypersurface in Ank. Then every irreducible component of X has dimension n-1.

Proof: We may assume that X is irreducible, and hence of the form Z(f) for a nonconstant irreducible polynomial f k[x1,,xn]. By renumbering the variables, we can assume that xn actually occurs in f. Now let ti A(X) be the image of xi. We claim that {t1, , tn-1} is an algebraically independent set, and therefore dim(X) n-1. For, suppose that G(t1, , tn-1) =0 is an algebraic relation. Then G I(X)= (f), so f divides G. But this is impossible, since xn occurs in f but not in G. On the other hand, the proof that An has dimension n shows that dim(X) n-1.

Theorem Let XAnk be an algebraic set all of whose irreducible components have dimension n-1. Then X is a hypersurface.

Proof: Without loss of generality, we may assume that X is irreducible. Let f I(X) be a nonzero polynomial. By irreducibility, some irreducible factor h of f must vanish on X. So, X Z(h). But this is an inclusion of irreducible sets of the same dimension, so it must be an equality.

Comments on this web site should be addressed to the author:

Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210