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Krull Dimension and Zariski spectrum, Slides of Mathematics

Krull dimension of a lattice, Complementary sequence and Kronecker’s theorem.

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2020/2021

Uploaded on 06/21/2021

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Krull Dimension
Thierry Coquand
Trieste, August 2008
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Download Krull Dimension and Zariski spectrum and more Slides Mathematics in PDF only on Docsity!

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.