Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2008 / B / 7

Prime elements are first-order definable in the ring first-order language. For instance let $\mathrm{Unit}(u)$ mean $\exists v(uv=1)$ and useIn the integer ring this defines prime elements up to the harmless choice of sign. Expand the first-order language by names for every integer and a new constant $c$. Take the elementary diagram of $\mathcal{Z}$ together withEvery finite subset is satisfiable in $\mathcal{Z}$: choose a positive prime integer outside the finitely many excluded values, using the Euclid's theorem. Compactness theorem gives a first-order model of the entire set. The complete-diagram argument embeds $\mathcal{Z}$ elementarily into its reduct; identify that copy with the original integer ring.

The resulting elementary extension contains $c$, a prime element unequal to every named standard integer. ThereforeThe construction rules out all standard integers, not merely the finite list appearing in any one approximation.

Apply downward Löwenheim-Skolem to the extension from part (i), keeping all standard integers and the new prime $c$. The first-order language and this prescribed subset are countable, so there is a countable elementary substructure $N$ containing them. In particular $N\equiv \mathcal{Z}$ and contains an element outside the standard copy of $\mathbb{Z}$.

No ring isomorphism $N\to \mathcal{Z}$ can exist. Such an isomorphism fixes $0$ and $1$, hence every nonnegative integer numeral. It also fixes each negative integer, which is the unique solution of $x+\bar{n}=0$ for the corresponding positive $n$. Consequently it fixes the whole standard copy. The image of $c$ would be some standard integer $a$, already the image of that same named standard element, violating injectivity.

Thus $N$ and $\mathcal{Z}$ are two nonisomorphic countably infinite first-order models of $T$, so

Every integer is definable without parameters in the ring first-order language. For $a\ge 0$, use $x=\bar{a}$; for $a=-m<0$, use $x+\bar{m}=0$. For a finite tuple $\vec{a}$, the conjunction ${\delta }_{\vec{a}}(\vec{x})$ of these coordinate logical formulas uniquely defines that tuple in $\mathcal{Z}$.

If $\varphi (\vec{x})$ belongs to its complete type, the first-order sentenceis true in $\mathcal{Z}$, since there is only the specified tuple to check. It therefore belongs to $T=\mathrm{Th}(\mathcal{Z})$. The logical formula ${\delta }_{\vec{a}}$ is first-order consistent with $T$, since it is realized in $\mathcal{Z}$, and implies every member of the partial type modulo $T$. Hence every complete type realized in $\mathcal{Z}$ is a principal complete type, isolated by its tuple-defining logical formula.

Take the parameter-free complete typeof the nonstandard prime constructed in part (i). It is first-order consistent and complete because it consists of exactly the logical formulas true of an actual element in a first-order model of $T$. It contains $x\ne \bar{a}$ for every integer $a$, and consequently no element of $\mathcal{Z}$ realizes it. ThusIt is necessarily a nonprincipal complete type. If a first-order consistent logical formula isolated it, completeness of $T$ would force the existential closure of that logical formula to hold in every first-order model of $T$, including $\mathcal{Z}$; a witness there would realize the partial type, a contradiction. This connects the explicit omission with the hypothesis of the omitting types theorem.

Enumerate the standard positive primes as $(p_n{)}_{n<\omega }$. For each subset $A\subseteq \omega$ consider the partial typewhere divisibility is the ring logical formula $\exists y({\bar{p}}_{n}y=x)$.

Every finite subset is realized in $\mathbb{Z}$: take the product of the finitely many primes required to divide $x$, or $1$ if there are none. This is nonzero and is not divisible by any of the distinct finitely many primes required not to divide it. Compactness theorem supplies a complete type extending $q_A$.

Distinct subsets $A,B$ yield distinct complete extensions, because at an index in their symmetric difference they require opposite divisibility statements. There are $2^{{\aleph }_0}$ such subsets, so there are at least that many complete one-types. There are at most that many because the first-order language has only countably many logical formulas and a partial type is a set of them. HenceIn particular the number of complete one-types is uncountable, despite every partial type realized in the particular countable first-order model $\mathcal{Z}$ being a principal complete type.