Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 144 / 3

In a real closed field, the formula $\exists z(z^2=x)$ defines the nonnegative elements. A quantifier-free formula in one variable over the prime field is a Boolean combination of polynomial equations, so in an infinite field it defines a finite or cofinite set. The nonnegative cone is neither finite nor cofinite. Hence the Theory of real closed fields does not eliminate quantifiers in the pure ring language.

Expand RCF by a constant $c$ and the sentences $c>n$ for all natural numbers $n$. Every finite subset is realized in $\mathbb{R}$. The compactness theorem gives a model realizing all of them, hence a Non-Archimedean real closed field.

RCF is not aleph-zero-categorical: the real algebraic numbers and a countable real closure of $\mathbb{Q}(t)$ are countable models with different transcendence degrees. It is not categorical in any uncountable cardinal either. At cardinality $2^{{\aleph }_0}$, for example, $\mathbb{R}$ is Archimedean whereas the real closure of $\mathbb{R}(t)$ with $t$ infinitely large is non-Archimedean. If RCF were categorical in any uncountable cardinal, the Morley categoricity theorem would make it categorical in every uncountable cardinal, contradicting this pair.

Let $f\in \mathbb{R}(X_1,\dots ,X_n)$ be positive semidefinite. The assertion that $f(x)\ge 0$ wherever its denominator is nonzero is first-order in ordered fields and holds in $\mathbb{R}$. Since RCF is model-complete, it holds in every real closed extension of $\mathbb{R}$.

If $f$ were not a sum of squares, the Artin-Schreier ordering criterion would give an ordering of $\mathbb{R}(X_1,\dots ,X_n)$ in which $f<0$. Its real closure is a real closed extension of $\mathbb{R}$, contradicting the transferred assertion at the generic tuple $(X_1,ldots,X_n)$. Hence $f$ is a sum of squares. This is the Model-theoretic proof of Hilbert's seventeenth problem.