Krull Dimension and Zariski spectrum, Slides of Mathematics

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Thierry Coquand

Trieste, August 2008

Zariski spectrum

Any element of the Zariski lattice is of the form D(a 1 ,... , an) = D(a 1 ) ∨ · · · ∨ D(an). We have seen that D(a, b) = D(a + b) if D(ab) = 0

In general we cannot write D(a 1 ,... , an) as D(a) for one element a

We can ask: what is the least number m such that any element of Zar(R) can be written on the form D(a 1 ,... , am). An answer is given by the following version of Kronecker’s Theorem: this holds if Kdim R < m

Krull dimension of a lattice

If L is a lattice, we say that u 1 ,... , un and v 1 ,... , vn are (n-)complementary iff

u 1 ∨ v 1 = 1, u 1 ∧ v 1 6 u 2 ∨ v 2 ,... , un− 1 ∧ vn− 1 6 un ∨ vn, un ∧ vn = 0

For n = 1: we get that u 1 and v 1 are complement

Proposition: Kdim L < n iff any n-sequence of elements has a complementary sequence

Krull dimension of a lattice

What is important here is the logical complexity

Distributive lattice: equational theory

The notion of complementary sequence is a (first-order) coherent notion

Krull dimension of a ring

Kdim R < n is defined as Kdim (Zar(R)) < n

Proposition: Kdim R < n iff for any sequence a 1 ,... , an in R there exists a sequence b 1 ,... , bn in R such that, in Zar(R), we have

D(a 1 , b 1 ) = 1, D(a 1 b 1 ) 6 D(a 2 , b 2 ),... , D(an− 1 bn− 1 ) 6 D(an, bn), D(anbn) = 0

This is a first-order condition in the multi-sorted language of rings and lattices

Example: Kronecker’s theorem

Kronecker in section 10 of

Grundz¨uge einer arithmetischen Theorie der algebraischen Gr¨ossen. J. reine angew. Math. 92, 1-123 (1882)

proves a theorem which is now stated in the following way

An algebraic variety in Cn^ is the intersection of n + 1 hypersurfaces

Kronecker’s Theorem

In particular if R is a polynomial ring k[X 1 ,... , Xm] with m < n then this says that given n + 1 polynomials we can find n polynomials that have the same set of zeros in an arbitrary algebraic closure of k

Kronecker’s Theorem

This concrete proof/algorithm, is extracted from R. Heitmann “Generating non-Noetherian modules efficiently” Michigan Math. J. 31 (1984), 167-

Though seeemingly unfeasible (use of prime ideals, topological arguments on the Zariski spectrum) this paper contains implicitely a clever and simple algorithm which can be instantiated for polynomial rings

Forster’s Theorem

We say that a sequence s 1 ,... , sl of elements of a commutative ring R is unimodular iff D(s 1 ,... , sl) = 1 iff R = <s 1 ,... , sl>

If M is a matrix over R we let ∆n(M ) be the ideal generated by all the n × n minors of M

Theorem: Let M be a matrix over a commutative ring R. If ∆n(M ) = 1 and Kdim R < n then there exists an unimodular combination of the column vectors of M

This is a non Noetherian version of Forster’s 1964 Theorem

Forster’s Theorem

We get a first-order (constructive) proof.

It can be interpreted as an algorithm which produces the unimodular combination.

The motivation for this Theorem comes from differential geometry

If we have a vector bundle over a space of dimension d and all the fibers are of dimension r then we can find d + r generators for the module of global sections

Serre’s Spliting-Off Theorem

This is the special case where the matrix is idempotent

The existence of a unimodular combination of the column in this case has the following geometrical intuition.

We have countinuous family of vector spaces over a base space. If the dimension of each fibers of a fibre bundle is > the dimension of the base space, one can find a non vanishing section

This is not the case in general: Moebius strip, tangent bundle of S^2

Vector bundles are represented as finitely generated projective modules

Elimination of noetherian hypotheses

Kronecker’s Theorem, Forster’s Theorem were first proved with the hypothesis that the ring R is noetherian

The fact that we can eliminate this hypothesis is remarkable

An example of a first-order statement for which we cannot eliminate this hypothesis is the Regular Element Theorem which says that if I = <a 1 ,... , an> is regular (that is uI = 0 implies u = 0) then we can find a regular element in I.