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The question is the Subject. This comes from Barwise: Intro to First Order Logic, it is prop 2.1 and it is answered 2 pages later. I however don't understand the proof well.

As Barwise says, the notion of divisible groups is not finitely axiomatizable in FOL. I can see this in that an abelian group G is divisible if $\forall n \geq 1 \space \forall x \exists y [ny = x]$. So we would need an infinite $n \in \mathbb{N}$ sentences (call it the set M):

$$\forall x \exists y [2y = x]$$ $$\forall x \exists y [3y = x]$$ $$\vdots$$ $$\forall x \exists y [ny = x]$$ $$\vdots$$

So I see that we are ranging over x,y, and n and so we don't have a FOL.

Now we let ${\psi_{1}, \psi_{2}, ... \psi_{k}}$ be a set of first-order sentences true in all divisible abelian groups. Now letting $\psi$ be the conjunction $(\psi_{1} \wedge \psi_{2} \wedge ... \wedge \psi_{k})$ we want to prove that $\psi$ is true in some non-divisible abelian group.

Now if $T$ is the set of all axioms of abelian groups as well as those that are non-divisible for every n (which there are an infinite amount), then we would have that $T \models \psi$.

I think a major part of the difficulty that I have is that I don't understand what $\psi_{r}, r \in {1,2,...,k}$ is; in that while I know it is a set of first order sentences true in abelian groups as well as in the set M above, what does it look like? Is it just those sentances axiomatizable in FOL. I can see this in that an abelian group G is divisible if $\forall n \geq 1 \forall x \exists y [ny = x]$

Now letting the set of axioms for abelian groups be L, and the set of sentences M above being axioms, we could make a set of axioms $T = {L \cup M}$. Now by the compactness Theorem we have that we could have some finite subset $T_{0} \subset T$. Now before we employed the compactness theorem we could model the sentences $\psi \text{ by } T \text{ thus } T \models \psi$ after the employment of the Compactness Theorem we could have $T_{0} \models \psi$.

Now that we have some finite subset $T_{0}$ of size N we could just find (Question) some subset of $T$ which is bigger than $T_{0}$ and this somehow produces a non-divisible abelian group?

Barwise chose $\mathbb{Z}_{p}$ and if we take some number $p \in \mathbb{N}, \space p > N = 3$ we get some element that is true in some non-divisible abelian group. So for $p = 5 \text{ where } k \geq 1$ we would have:

$$\forall x \exists y [y = x], \text{ where } y = x \text{ works } \forall x, $$ $$\forall x \exists y [2y = x], \text{ where } y = x/2 \text{ works } \forall x, $$ $$\forall x \exists y [3y = x], \text{ where } y = x/3 \text{ works } \forall x. $$ Now since $p > N$ say $p = 5$, where $N = 3$ we would have an element such that ??? I am not sure how to proceed here... since $\mathbb{Z}_5 = {0,1,2,3,4}$ and $4 > N$ this means???

Help is much appreciated.

Thanks,

Brian

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1 Answer 1

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I'll use your proposed notation, so $L$ is the set of axioms for abelian groups and $M$ is the set of divisibility statements that you exhibited. Now suppose $F$ is a finite set of first-order sentences that are true in all divisible abelian groups. I need to find a non-dvisible abelian group in which all the sentences in $F$ are true.

Each sentence $\psi$ in $F$, being true in all divisible abelian groups, i.e., in all models of $L\cup M$, is, by compactness, a consequence of a finite subset of $L\cup M$, say $L\cup M_\psi$. Since $F$ is finite and each $M_\psi$ is finite, we have that $\bigcup_{\psi\in F}M_\psi$ is a finite set of sentences; call it $M_*$. It's a subset of $M$, so it consists of finitely many of the divisibility sentences you exhibited in the question. That is, it consists of sentences $\forall x\,\exists y\,[ny=x]$ for finitely many numbers $n$.

Now if $p$ is a prime number bigger than all the finitely many $n$'s involved in $M_*$, then $\mathbb Z/p$ satisfies all the sentences in $M_*$ --- indeed, it satisfies the divisibility statements $\forall x\,\exists y\,[ny=x]$ for all $n$ that are not divisible by $p$. So it satisfies $M_\psi$ for each $\psi \in F$ (as these $M_\psi$ are subsets of $M_*$), so it satisfies all the sentences in $F$ (because each $\psi\in F$ is a consequence of $L\cup M_\psi$). But $Z_p$ is not divisible; it doesn't satisfy $\forall x\,\exists y\,[py=x]$.

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