There exists a field with u-invariant 9

Algebra, Number theory 1 Minute

You need to know: The concept of a Field.

Background: A number u(F) is called u-invariant of a field F, if (a) equation \sum_{i=1}^n \sum_{j=1}^n a_{ij} x_i x_j = 0 (where x_1, x_2, \dots, x_n are variables and a_{ij} are some coefficients from F) in n=u(F) variables may have no solutions, except of x_1 = x_2 = \dots = x_n = 0, and (b) the same equation in n=u(F)+1 variables always have a non-zero solution.

The Theorem: On 4th January 2000 Oleg Izhboldin submitted to Annals of Mathematics a paper in which he proved that there exists a field F with u-invariant 9.

Short context: In 1953, Kaplansky conjectured that u-invariant of any field is always a power of 2. In 1989, Merkurev disproved this, and showed that u-invariant can be any even number, but left open whether it can be any odd number grater than 1. It is known that u-invariant can never be 3, 5, or 7. However, the Theorem proves that it can be 9.

Links: The original paper is available here. See also Section 1.7 of this book for an accessible description of the Theorem.

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Published by Bogdan Grechuk

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  1. Pingback: The u-invariant of any field of transcendence degree 2 over R is at most 4 – Theorems of the 21st century

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