MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question is in the title exactly as I want to ask it, but let me provide some background and motivation.

Many of the properties of fields studied in the algebraic theory of quadratic forms are manifestly elementary properties in the sense of model theory: that is, if one field has this property, then any other field which has the same first-order theory in the language of fields has that same property. Examples:

being quadratically closed, being formally real, being real-closed, being Pythagorean (sum of two squares is always a square), for any fixed positive integer n, having I^n = 0 (follows from the Milnor conjectures!), the u-invariant, the level, the Pythagoras number...

These properties imply that at least for some fields $K$, if $L$ is any field elementarily equivalent ot $K$, then $W(L) \cong W(K)$: e.g. $K$ is quadratically closed, $K$ is real-closed, $K = \mathbb{C}((t))$. Is it always the case that $K \equiv L$ implies $W(K) \cong W(L)$? I am pretty sure the answer is no because for instance if $\operatorname{dim}_{\mathbb{F}_2} K^{\times}/K^{\times 2}$ is infinite, I think it is not an elementary invariant. And if you take a field with vanishing Brauer group, then $W(K)$ is, additively, an elementary $2$-group of dimension $\operatorname{dim}_{\mathbb{F}_2} K^{\times}/K^{\times 2} + 1$.

But are there known positive results in this direction?

share|cite|improve this question
up vote 6 down vote accepted

As you point out, one cannot hope that the Witt ring, up to isomorphism, be an elementary invariant of a field. The strongest statement which I might conjecture would be that if $K \preceq L$ is an elementary extension of fields, then $W(K) \to W(L)$ is an elementary extension of rings. If this statement were true, then the theory of the Witt ring would be an elementary invariant as any two elementarily equivalent fields have a common elementary extension.

It is true that if $K \preceq L$ is an elementary extension of fields, then map $W(K) \hookrightarrow W(L)$ is an inclusion [Why? Being zero in the Witt ring is defined by an existential condition.] One might try to prove that $W(K) \hookrightarrow W(L)$ is elementary by induction where the key step would be to show that if $W(L) \models (\exists x) \phi(x;a)$ where $a$ is a tuple from $W(K)$, $x$ is a single variable, and $\phi$ is a formula in the language of rings, then $W(K) \models (\exists x) \phi(x;a)$. The witness in $W(L)$ would be represented by a quadratic space of some finite dimension $n$. One would like to argue that the set defined by $\phi(x;a)$ in the space of $n$-dimensional quadratic forms is definable in the field language in $K$ in which case a witness could be found in $K$ via elementarity. This last part of the argument is delicate as it would require knowing bounds for checking equalities in the Witt ring.

The Witt ring construction is an example of an ind-definable set modulo an ind-definable equivalence relation. These are discussed in some detail in Hrushovski's paper on approximate groups ( arXiv:0909.2190 ). With Krajiceck, I considered similar issues (how does the Grothendieck ring of a first-order structure depend on its theory) in Combinatorics with definable sets: Euler characteristics and Grothendieck rings. Bull. Symbolic Logic 6 (2000), no. 3, 311--330.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.