Formulas in a Field and in a Field Extension

Question

Let $\mathbb F$ be a field and let $a, b, c, d$ be fixed elements in the field $\mathbb F$. Consider the formulas

1) $\exists\;x\;\;:\;\;x^2=-1.$

2) $\exists\;x\;\;:\;\;(xa=c\land xb=d).$

Formula $(1)$ can be false in $\mathbb F$ but true in a field extension of $\mathbb F$. For exemple, formula $(1)$ is false in the reals $\mathbb R$ but true in the complex numbers $\mathbb C$.

The same does not happen with the formula $(2)$ since formula $(2)$ is true if and only if the matrix $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right),$$ is singular, and it is well known that the singularity of a matrix is not changed by an extension of the field that contains the elements of the matrix.

My question is:

Is there any characterization of the formulas, in a field, whose validity does not change by an extension of the field?

Remark: I put this question yesterday in https://math.stackexchange.com/questions/561178/formulas-in-a-field-and-in-a-field-extension. But was not answered.

Answer 1

This is a major clean-up of my previous argument, thanks to the very helpful comments of Emil Jerabek.

I interpret the question as follows: Characterize all sentences in the language of fields that are preserved under field extensions.

Claim: A formula $\phi$ is preserved under field extensions if and only if the field axioms prove that $\phi$ is equivalent to an existential formula.

For definiteness, we take the language of fields to have operations for addition and multiplication, and distinguished elements for 0 and 1.

The main tool to establish the claim is the following

Lemma: Suppose $M$ is a structure in any old language, and $T$ is a theory in that language. Let $\forall(T)$ denote the set of universal consequences of $T$. If $M\models\forall(T)$ then $M$ extends to a model of $T$.

Proof-sketch: The diagram of $M$ cannot be inconsistent with $T$, because otherwise $T$ would imply the negation of some formula in the diagram of $M$, and therefore would imply the universal closure of the negation of that formula, contrary to the assumption $M\models\forall(T)$. But then any model of the diagram of $M$ plus $T$ will do the job we require.

Now we prove the Claim. The non-trivial part is to show that if the theory of fields (which we will henceforth call $T$) cannot prove the equivalence between $\phi$ and any existential sentence, then $\phi$ is not preserved under extensions.

But our premiss is equivalent to the statement that $T$ cannot prove the equivalence between $\neg\phi$ and any universal sentence. Let $S$ be the set of universal consequences of $T+\neg\phi$. Then $S+T\nvdash\neg\phi$. For otherwise, using the fact that $S$ is closed under conjunction (at least up to logical equivalence) we could argue that $\sigma+T\vdash\neg\phi$, for some sentence $\sigma\in S$. But this, together with $T+\neg\phi\vdash\sigma$ would imply that $T\vdash\sigma\longleftrightarrow \neg\phi$, contrary to the premiss of our argument.

We have shown that $S+T\nvdash\neg\phi$. Therefore there is a structure $k$ satisfying $S+T+\phi$. Evidently $k$ is a field. Recalling that $S$ is the set of universal consequences of $T+\neg\phi$, we can now use the Lemma to conclude that $k$ extends to a field $K$ satisfying $\neg\phi$. This proves that $\phi$ is not preserved under extensions.

I'd love some feedback on this!

Answer 2

These are exactly the formulas that are provably (in the theory of fields) equivalent to a formula without quantifiers.

The proof is that, in the theory of algebraically closed fields, they must be equivalent to a formula without quantifiers, because quantifier elimination holds in the theory of algebraically closed fields.

I claim the two formulas are also equivalent in the theory of fields. If not, there must be some field where one is true and the other is false, by Godel's completeness theorem. But if we pass to the algebraic closure of that field, each formula retains its truth value, which contradicts the fact that the two formulas are equivalent for algebraically closed fields.

Answer 3

This was intended to be a comment to SJR's post, but was too long. Here is a proof that any formula that is preserved downwards in models of $T$ is equivalent in $T$ to a universal formula, from which we deduce that formulas preserved upwards are equivalent to existential formulas.

Let $T$ be an $\cal L$-theory. Suppose $\phi(\bar v)$ is preserved downwards in models of $T$. Let $\Gamma(\bar v)=\{\psi(\bar v): \psi$ is universal and $T\models \phi(\bar v)\rightarrow \psi(\bar v)\}$. We claim that $T+\Gamma(\bar v)\models \psi(\bar v)$. If that happens then there are $\psi_1,\dots,\psi_n\in\Gamma(\bar v)$ such that $T\models \phi(\bar v)\leftrightarrow (\psi_1(\bar v)\land\dots\land\psi_n(\bar v)$.

Suppose not. Then there is ${\cal M}\models T$ and $\bar a\in \cal M$ such that ${\cal M}\models \psi(\bar a)$ for $\psi\in\Gamma$ and ${\cal M}\models \neg\phi(\bar a)$.

Let $T_1= T+ \phi(\bar a) + $ atomic diagram of $\cal M$. If $T_1$ is satisfiable there is ${\cal N}\models T$ with ${\cal M}\subset {\cal N}$ and ${\cal N}\models \phi(\bar a)$, contradicting the fact that $\phi$ is preserved downward. Thus $T_1$ must be unsatisfiable. Thus there is a quantifier free formula $\theta(\bar v,\bar w)$ and $\bar b\in \cal M$ such that $$T\models \theta(\bar a,\bar b)\rightarrow \neg\phi(\bar a).$$ Since the parameters $\bar a$ and $\bar b$ don't occur in $T$, $$T\models \forall \bar v\ (\phi(\bar v)\rightarrow\forall \bar w\ \neg\theta(\bar v,\bar w))$$ But then $\forall \bar w\ \neg\theta(\bar v,\bar w)\in \Gamma$ contradicting the fact that ${\cal M}\models \theta(\bar a,\bar b)$.