# Application of the Löwenheim-Skolem Theorem

In the lecture notes I'm currently reading, the professor wrote that an application of the Löwenheim-Skolem theorem (stated below) is to show that the theory of infinite vector spaces over a field $K$ in the language $\mathcal{L} = (0, +, (f_q)_{q \in K})$ is complete. Here $f_q$ are the unary functions to be understood like scalar-multiplication by $q$ in the vector space.

My confusion is: The Löwenheim-Skolem theorem makes a statement about the existence of models of a certain size. Why would that help? To apply Löwenheim-Skolem I need the theory to be consistent anyway - and can't I then just directly apply Gödel's completeness theorem?

Theorem (Löwenheim-Skolem (ZFC)): Let $T$ be a consistent theory with infinite models in a language $\mathcal{L}$ of cardinality $\kappa$. Then for any infinite cardinal $\lambda \geq \kappa$ the theory $T$ admits a model of size $\lambda$.

• You probably mean Löwenheim-Skolem – Primo Petri Jun 14 '17 at 17:25
• Oh boy, thanks! I've been studying the topic for a while now and I always misread it... – Steven Jun 14 '17 at 17:25
• There is also some confusion between "completeness" and "concistency". The L-S thm assume concistency and does not say anything about completeness. – Primo Petri Jun 14 '17 at 17:33
• Ah, I see! I confused completeness with coherency of a theory, so Gödel's theorem isn't going anywhere. – Steven Jun 14 '17 at 17:37
• If remember well, Gödel completeness theorem says that some syntatic calculus is complete: if you cannot derive inconsistencies, then there is a model. Completeness of a theory is a different notion. – Primo Petri Jun 14 '17 at 17:41

First let me address your confusion over why Lowenheim-Skolem is necessary. The theory of a class $\mathbb{K}$ of structures is the set of sentences satisfied by all structures in that class. Unlike the theory of a single structure, this theory need not be complete - for instance, if $\mathbb{K}$ is the class of groups, then the commutativity axiom "$\forall x, y(x*y=y*x)$" is not in $Th(\mathbb{K})$ (since there are non-abelian groups), but neither is its negation (since there are abelian groups).
So we can't just automatically conclude that the theory in your question is complete; we need to argue that any two structures in your class are "basically the same" (= elementarily equivalent). A standard trick for doing this is: given $\mathcal{A}$ and $\mathcal{B}$ which we want to show are elementarily equivalent, build "bigger" structures $\mathcal{A}'$ and $\mathcal{B}'$, such that we know $\mathcal{A}\equiv\mathcal{A}'$ and $\mathcal{B}\equiv\mathcal{B}'$, and show that $\mathcal{A}'\cong\mathcal{B}'$. Do you see a way to use Lowenheim-Skolem to do this, here? HINT: what determines the isomorphism type of a vector space? How is this related to its cardinality?
Incidentally, re: the strategy above, a neat theorem of Keisler and Shelah gives a way to do this that always works: if $\mathcal{A}\equiv\mathcal{B}$, then they have isomorphic ultrapowers (and obviously conversely)! This is Theorem 10.7 of this survey article on ultraproducts by Keisler, which is valuable reading if you're interested in the more set-theoretic side of model theory.
• The fact that "if $A$ and $B$ are elementary equivalent then they have isomorphic ultrapowers" has some hypotheses, doesn't it ? I vaguely remember seeing that what's always true is that they have isomorphic ultralimits, or that one embeds in an ultrapower of the other; and that in some specific cases they have isomorphic ultrapowers – Maxime Ramzi Jun 14 '17 at 18:11