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I'm trying to study basic facts of field extensions from the model-theory point of view.

Consider the language $L$ of fields, and call $at$ the family of atomic formulae. A morphism of fields is by definition an $at$-morphism of $L$-structures, that is a function $f \colon F_1 \to F_2$ such that

  • for every $\phi(\vec{x}) \in at$, if $F_1 \models \phi(\vec{a})$ then $F_2 \models \phi(f \, \vec{a})$.

From the theory of fields, it is known that every morphism of fields is in fact an embedding.

Let $F \subseteq E_{i = 1, 2}$ be two field extensions.
An $F$-morphism is a field morphism $f \colon F_1 \to F_2$ such that $f(x) = x$ for all $x \in F$.
In this case $f$ is an $at(F)$-morphism, where $at(F)$ is the family of atomic formulae with parameters in $F$.

For example, if $\alpha \in E_1$ is solution to a polynomial equation with coefficients in $F$, then $f(\alpha) \in E_2$ is solution to the same equation.

It is also well known that

For any set of formulae $\Delta$, every $\Delta$-embedding is a $\{\exists\}\Delta$-morphism.

Let $\alpha \in E_1$ be a root of a polynomial $p(x) \in F[x]$. Then we can express that $\alpha$ has multiplicity at least $m$ as an $\{\exists\}\,at(F)$-formula as follows:

$$ \varphi(\alpha) \equiv \exists \, b_0, \ldots, b_s \, . \, p(x) = (x - \alpha)^m (b_s x^s + \ldots b_0)$$ where the equality of polynomials is replaced by the equality of the corresponding coefficients and $s$ is the appropriate integer.

This implies that if $\alpha \in E_1$ is a root of $p(x)$ with multiplicity $m$, then $f(\alpha) \in E_2$ is a root of $p(x)$ with multiplicity at least $m$. The intuition suggests that this multiplicity should be exactly $m$.

Whence we wish to express the property

  • $\alpha$ has multiplicity less than or equal to $m$

with a formula that is preserved by $F$-morphisms. This seems to be expressible with a $\{\neg \exists\} \, at$-formula, or a $\{\forall\} \, at^{\pm}$-formula. But those should not be preserved by $F$-morphisms, as the following example shows:

$$\mathbb{Q} \models (\forall x \, . \, x^2 - 2 \neq 0), \qquad \mathbb{R} \not\models (\forall x \, . \, x^2 - 2 \neq 0).$$

Questions:

  • is the intuition of the fact that the exact multiplicity is preserved wrong?
  • can this property be expressed as a formula that is preserved?
  • also, is there any reference which study field extensions using model theory?

Thank you in advance.

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The property that $a$ is a root of $p(x)$ of multiplicity $m$ can be expressed by the following $L(F)$ formula $\varphi(a)$: $$\exists b_0,\dots,b_s\, (p(x) = (x-a)^m(b_sx^s+\dots+b_1x + b_0)\land b_sa^s+\dots+b_1a+b_0\neq 0),$$ where, as in your question, equality of polynomials is expressed by the conjunction of atomic formulas expressing equality of their coefficients. This formula says that $p(x)$ can be factored into $(x-a)^m$ times a polynomial which does not have $a$ as a root.

$\varphi(a)$ is an existential $L(F)$ formula, and truth of existential formulas is preserved under embeddings. In general, homomorphisms only preserve the truth of positive existential formulas, and $\varphi(a)$ is not positive, because of the inequation $(b_sa^s+\dots+b_1a+b_0\neq 0)$. But as you pointed out, embeddings and homomorphisms coincide for fields. This corresponds to the fact that we can use the Rabinowitsch trick to turn inequations into equations. That is, $\varphi(a)$ is equivalent (modulo the theory of fields) to the following positive existential formula:

$$\exists b_0,\dots,b_s,c\, (p(x) = (x-a)^m(b_sx^s+\dots+b_1x + b_0)\land (b_sa^s+\dots+b_1a+b_0)c= 1).$$

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  • $\begingroup$ Thank you, this was very clear. I wanted to ask again, at the cost of sounding boring, have you ever seen someone using those methods to study field theory from "the basics"? I have seen applications of model theory to prove the existence of algebraic closure, proofs of Nullstellensatz, transfer principles, dimension axiomatizations and studying more specific topics. I find the logical point of view more clear, even if it means spelling out more things. But maybe it is not so interesting/relevant, the usual algebraic arguments are enough, and nobody had reason to report that. $\endgroup$
    – user574499
    Commented Jul 14, 2018 at 23:17
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    $\begingroup$ @mononote Hmm... I'm not aware of a development of the basics of field theory that uses the language of model theory. In any development of algebra, one could use this language, but my sense is that it would really be a cosmetic choice, i.e. the proofs would just be translated, with the main ideas remaining the same. So people typically stick with the traditional presentation. That said, it's probably a good exercise in both model theory and field theory, and it could be enlightening in some places, to try translating as much as possible into model-theoretic language. $\endgroup$ Commented Jul 15, 2018 at 0:14

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