Prove that none of the following theories is finitely axiomatizable:

group-theorylogicmodel-theory

prove that none of the following theories is finitely axiomatizable:

$(a)$ infinite models of sets with only equality;
$(b)$ fields of characteristic zero;
$(c)$ divisible Abelian groups;
$(d)$ torsion-free Abelian groups
Using the following therom:

Let K be an arbitrary class of models. Then:
$(1)$ K is an elementary class if and only if K is closed under ultraproducts and elementary equivalence.
$(2)$ K is a finitely axiomatizable class if and only if both K and the complement of K are closed under ultraproducts and elementary equivalence.

I managed to proof that $(b)$ is not finitely axiomatizable by proof by negation I assumed that if there was such finite set of sentences I can lead to a contradiction (essentially saying that theory consist of the sentences 2$\neq$0,3$\neq$0,…..,p$\neq$0 is consistit for all primes p then by compactness I get an infinite axiomatizable theory), but it doesn't use that theorm (I couldn't make the connection…).

and for (d) I claimed: lets conclude the sentensces $\varphi_n$ that say the only element raised to the $n$th power that equals $e$ is $e$.If I take the ultraproudct of groups that have torsion elements I will get an torison free group hence K is not close uneder ultraproducts hence by the theorm is not not finitly axiomtazed (but I think its a bit off ;\ my knowledge in group theory is lacking).

For the others I tried to either proof they are not closed under elementary equivalence or under ultraproducts but I didn't find a way.
Any help would be appericated !

Best Answer

For each prime $p$, $\Bbb{Z}/p\Bbb{Z}$ has the following properties.

  • It is a finite field of characteristic $p$ with $p$ elements.
  • In the additive group, all non-zero elements have order $p$ so are torsion.
  • It is not divisible by $p$, because $p=0$.

Now consider the ultraproduct $\text{UP}$. The filter used is free and contains all cofinite subsets of primes.

The axioms of a field are a finite collection of statements of first-order logic, and since each factor in the ultraproduct satisfies them, the ultraproduct satisfies them and is a field.

Any positive integer $n>0$ is embedded in $\text{UP}$ as the constant sequence $(n, n, n, \dots)$. Since $n$ is divisible by only finitely many primes, it is invertible in $\Bbb{Z}/p\Bbb{Z}$ for all but finitely many primes. Since the ultrafilter contains the complement of the set of primes dividing $n$, $n$ is invertible in $\text{UP}$ on a large set. This implies it is invertible as an element of the field $\text{UP}$. (In the primes dividing $n$ you put anything you want for an inverse, because that's a negligible set.)

This implies all positive integers are invertible in the field $\text{UP}$. So it contains a copy of $\Bbb{N}, \Bbb{Z}$ and $\Bbb{Q}$. Its prime sub-field is $\Bbb{Q}$, so it has characteristic zero. It is infinite. Since it's a field extending $\Bbb{Q}$, it is a vector space over $\Bbb{Q}$. The additive group of any vector space over $\Bbb{Q}$ is a torsion-free divisible Abelian group, so the additive group of $\text{UP}$ is torsion-free and divisible.

Related Question