This is exercise 2.16.10 in Kenneth's Foundation of Mathematics. How to make use of the hypothesis the group is of exponent $p$ to prove that the theory of infinite abelian group of exponent $p$ is model-complete?

The following lemma is given as a hint in the book:

Let $\mathfrak{A} \subseteq \mathfrak{B}$ and if for all $a_1, …, a_n \in A$ (universe of $\mathfrak{A}$) and $b \in B$, there is an automorphism $\phi$ of $\mathfrak{B}$ such that $\phi(b) \in A$ and $\phi(a_i)=a_i$, then $\mathfrak{A}$ is an elementary submodel of $\mathfrak{B}$.

  • $\begingroup$ What have you tried? What tools do you know for proving a theory is model complete? $\endgroup$ – Alex Kruckman Jul 31 '18 at 21:01
  • $\begingroup$ I am asked to use the automorphism lemma. $\endgroup$ – YuiTo Cheng Aug 1 '18 at 0:13
  • $\begingroup$ I recommend adding this information, along with the statement of the lemma, to the question. $\endgroup$ – Alex Kruckman Aug 1 '18 at 0:29
  • $\begingroup$ The more context, the better! $\endgroup$ – Alex Kruckman Aug 1 '18 at 1:34
  • 2
    $\begingroup$ I don't know much about model theory, but if the groups is of exponent $p,$ then it is a vector space over the field $\mathbb{Z}/p\mathbb{Z}.$ I don't know if this is of any help. $\endgroup$ – Chris Leary Aug 1 '18 at 2:14

There are many ways to prove this, but I'll give a proof that follows the hint.

Let $T$ be the theory of infinite abelian groups of exponent $p$. To show $T$ is model complete, we assume $\mathfrak{A}\subseteq \mathfrak{B}$ are models of $T$, and we will show that $\mathfrak{A}\preceq \mathfrak{B}$. By the lemma, it suffices to take $a_1,\dots,a_n\in A$ and $b\in B$ and find an automorphism $\phi$ of $\mathfrak{B}$ such that $\phi(a_i) = a_i$ for all $i$ and $\phi(b)\in A$. We may assume $b\notin A$, since otherwise the identity automorphism works.

Abelian groups of exponent $p$ are vector spaces over $\mathbb{F}_p$, so we can use some linear algebra. A finite dimensional vector space over $\mathbb{F}_p$ is finite, so $\dim(\mathfrak{A})$ is infinite. Let $V$ be the subspace of $\mathfrak{A}$ spanned by $a_1,\dots,a_n$, and let $a_1',\dots,a_k'$ be a basis for $V$ (we can take the $a_i'$ to be elements of the tuple $a_1,\dots,a_n$, but this isn't necessary). Since $\dim(V)\leq n$, while $\dim(\mathfrak{A})$ is infinite, $V$ is a proper subset of $A$. Picking some $a^*\in A\setminus V$, the set $\{a_1',\dots,a_k',a^*\}$ is linearly independent. Let $V' = \text{Span}(a_1',\dots,a_k',a^*)$.

Now $V'\subseteq A$ and $b\notin A$, so the set $\{a_1',\dots,a_k',a^*,b\}$ is also linearly independent. Extend this set to a basis for $\mathfrak{B}$. The permutation of the basis obtained by swapping $a^*$ and $b$ and fixing all other basis elements determines an automorphism $\phi$ of $\mathfrak{B}$. We have $\phi(b) = a^*\in A$, and $\phi(a_i) = a_i$ for all $i$, since each $a_i$ is a linear combination of the basis elements $a_1',\dots,a_k'$, each of which is fixed by $\phi$.

  • $\begingroup$ I've actually come up with the same solution. $\endgroup$ – YuiTo Cheng Aug 2 '18 at 0:43

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