Algebraic extensions are isomorphic if the same polynomials have roots

Question

I want to solve the following excercise.

Let $K$ be a (perfect) field and $L_1, L_2$ be two algebraic extensions of $K$. Then $L_1$ and $L_2$ are isomorphic if every polynomial $f\in K[X]$ which has a root in $L_1$ has a root in $L_2$ and vice versa.

The hint is to use either the Compactness Theorem to reduce to the case of finitely generated subfields of $L_1, L_2$ or to use König's Lemma from Graph Theory.

For the finitely generated case let $F=K(a_1,\dots,a_n)$ for some $a_1,\dots,a_n\in L_1$. By the Primitive Element Theorem we have $F=K(a)$ for some $a\in L_1$ and by assumption the minimal polynomial $\mathfrak{m}_a$ of $a$ over $K$ has a root $b\in L_2$. Then clearly $K(a)\cong K(b)$.

My problem is that I'm not really sure how I can use the compactness theorem to reduce to this case. It is probably easy, but I think my problem is that I am used to applying the Compactness Theorem to build structures with certain properties and not maps between structures.

Question: How can this proof be finished off by the compactness theorem? Are there any general techniques how compactness can be used to build isomorphisms between structures rather than structures?

I would also like to see a proof using the König's Lemma approach (because afaik König's Lemma is strictly weaker than the Compactenss Theorem over $\mathrm{ZF}$(?)) but I dont know how to build the tree in this case. (Maybe we have to assume that $K$ is countable?)

Question: Can we prove the above statement (under some reasonable additional assumptions) in $\mathrm{ZF}+$ König's Lemma?

Answer 1

Question: How can this proof be finished off by the compactness theorem? Are there any general techniques how compactness can be used to build isomorphisms between structures rather than structures?

You can prove the following in general.

Assume that $A$ and $B$ are structures which have the same finitely generated substructures up to isomorphisms. Assume that under all these isomorphisms every $a$ in $A$ has only finitely many possible images in $B$, and that every similarly every $b$ in $A$ has only finitely many possible preimages in $A$. Then $A$ and $B$ are isomorphic.

The proof uses the compactness theorem of propositional calculus---introduce for every pair $(a, b)$ in $A \times B$ a variable $R_{ab}$. That$$f = \{(a, b) : R_{ab} \text{ is true}\}$$is a partial isomorphism can easily be expressed by a set of formulas. To express that $f$ is defined everywhere and is surjective needs the above finiteness condition.

Question: Can we prove the above statement (under some reasonable additional assumptions) in $\mathrm{ZF}+$ König's Lemma?

I do not think so. König's Lemma is about countable graphs, so it is of use only in a countable situation. The right combinatorial lemma is Rado's Selection Principle.

https://caicedoteaching.wordpress.com/2009/03/21/580-iii-partition-calculus/#more-1825

https://de.wikipedia.org/wiki/Auswahlprinzip_von_Rado

$L_i$ can be realized as ascending chain of union of finite algebraic extensions. In particular, you can identify the direct limit system. The induced map is isomorphism as well.

This is not true. Only if $L_i/K$ is countably generated. It is then conceivable to define a tree of partial isomorphism for which König applies.

Läuchli constructed a model where $\mathbb{Q}$ has two non-isomorphic algebraic closures. I don't know how badly choice fails there (in particular, if König's lemma fails there).

I would guess that one cannot uniformly choose a zero of a polynomial. The algebraic closure is then not well-ordered of course.

You need to require that the extensions are normal too, no?

No. Possibly, in $L_i$, not every finite extension is contained in a normal extension---take for $L_i$ a finite, not normal extension.

I found the statemant for example formulatet as "Algebraic field extensions of K are isomorphic over K if and only if the same polynomials in K[X] have a zero in these extensions." in the model theory book by Tent and Ziegler.

Well. I must be missing something algebraic about this, then.

No. For me the algebraic difficulty occurs, when $K$ is not perfect.