Statements which feel like they shouldn't be first-order expressible, but are

Question

I recently had the opportunity to study some (very basic) model-theory, and that made some theorems in ring theory become immediately more interesting. For example, one characterization of the Jacobson radical of a ring $R$ makes it possible to express the formula "$x \in J(R)$" as a first-order formula in the language of rings:

$$\forall r \exists s (s(1-rx) = 1)$$

as it is known that $x \in J(R) \iff 1 - rx$ is invertible for all $r \in R$.

Similarily, the statement "every $R$-module is flat", which feels very far from first-order expressible, can be written as the sentence:

$$\forall a \exists x (a = axa)$$

as both of these statements are satisfied precisely by the von Neumann regular rings.

I am also aware of this question, which talks about a way of expressing the sentence $\dim(R) \leqslant k$ for a commutative ring $R$.

In that spirit, I was hoping for more examples (not necessarily from algebra) of seemingly "complex" statements which are equivalent to some first-order formula (or maybe a set of formulas). I would also like to know if a statement being expressible in first-order logic has any applications in situations like these.

Answer 1

A very important example of this is the incredible expressive power of first-order arithmetic. For instance, in the structure $(\mathbb{N},+,\cdot)$ it is possible to express the fundamental theorem of arithmetic as a first-order sentence. At first glance this seems very unlikely: to state that for all $n\in\mathbb{N}$ there exists a prime factorization of $n$, you need to state the existence of a finite sequence of arbitrary length of elements of $\mathbb{N}$ with a certain property. First order logic lets you quantify over a fixed finite number of elements of your structure, but how can you quantify over a finite sequence of arbitrary length? However, by some very clever trickery using the Chinese remainder theorem (Gödel's $\beta$ function), it actually is possible to encode finite sequences of natural numbers using single natural numbers in the first-order language of arithmetic, so that you can quantify over them. This makes it possible for first-order arithmetic to express much more than it would appear to at first glance.

(To be clear, this is a different sort of example than the ones you gave. The sentence which expresses the fundamental theorem of arithmetic in $(\mathbb{N},+,\cdot)$ would not necessarily express the existence of unique factorizations in the usual sense when interpreted in a different structure; indeed, unique factorization is not a first-order property of rings (or semirings). This is because the first-order encoding of finite sequences only works when it is interpreted in the actual true natural numbers. So the relevance of this example is not from the abstract algebra perspective of being able to classify different structures using first-order properties. Rather, it is relevant when thinking about things like what various first-order axiomatizations of the natural numbers are capable of proving (since the list of interesting things they might be able to prove is much longer than you would expect).)

Answer 2

Here's a possible example that occurred in my own research. If $(a(n))$ is an automatic sequence, then the first-order theory of $\langle \mathbb{N}, +, n \rightarrow a(n) \rangle$ is decidable, and there is a free theorem-prover called Walnut that implements a decision procedure for it.

One property of sequences that people are interested in is balance: we say a sequence over a finite alphabet $(a(n))$ is balanced if for all $\ell$ the number of occurrences of each letter $c$ in any two length-$\ell$ contiguous blocks of $(a(n))$ differs by at most $1$. However, there's no obvious way to express this property in first-order logic, because it seems to require counting the number of $c$'s occurring in length-$\ell$ blocks. However, in the case where $(a(n))$ takes only two distinct values, say $0$ and $1$, we can use an alternative characterization of the balance property, which says that a finite block is unbalanced if and only if it contains both $0w0$ and $1w1$ for some block $w$. And this assertion is first-order expressible.

For more details about this, see my paper here.

Answer 3

Many important classes of finite groups can be described by some family of sentences in the first-order language of group theory. See these slides of John Wilson.

1. Finite soluble groups of derived length at most $d$
2. Groups of odd order
3. Finite non-abelian simple groups can be defined by a single sentence [1]
4. Finite soluble groups can also be defined by a single sentence [2]: a finite group is soluble if and only if it satisfies the following first order sentence, which states that no non-trivial element $g$ is a product of $56$ commutators of pairs of conjugates of $g$: $$ \forall g\ \forall x_1 \dotsm \forall x_{56}\ \forall y_1 \dotsm \forall y_{56}\ \bigl(g = 1 \vee g \not= [g^{x_1}, g^{y_1}] \dotsm [g^{x_{56}}, g^{y_{56}}]\bigr) $$

[1] U. Felgner, ‘Pseudo-endliche Gruppen’, Proc. 8th Easter Conf. on Model Theory (Humboldt-Universit ̈at, Berlin, 1990) 82–96.

[2] J. S. Wilson, Finite axiomatization of finite soluble groups, J. London Math. Soc. (2) 74 (2006), 566–582.

Answer 4

Ordinal definability is a great example of this (and a very useful one).

A set is said to be definable if it is first-order definable in the class structure $\langle V; \in\rangle$ (with no parameters allowed). As is well-known, the notion of definability is not first-order definable. (One needs to be a bit careful in stating this: There are countable models of ZFC in which every set is definable, so for those particular models, definability happens to be definable. But this doesn't happen in "most" models of ZFC, and definability cannot be defined in general within ZFC.)

In contrast, consider the following:

A set is said to be ordinal definable if it is first-order definable in the class structure $\langle V; \in\rangle,$ allowing ordinal parameters.

At first glance, the notion of ordinal definability would appear also not to be first-order definable (the definition above can't be directly expressed in first-order logic). But, in fact, the reflection principle can be used to show that the above is equivalent to:

A set $x$ is ordinal definable if there is some ordinal $\eta$ such that $x$ is first-order definable in the structure $\langle V_\eta; \in\rangle$ (either without parameters or allowing ordinal parameters — that doesn't matter here).

And this yields a first-order definition of ordinal definability. (If I recall correctly, this is due to Myhill and Scott.)

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This is important in set theory, since the class of hereditarily ordinal definable sets (which is definable, since ordinal definability is definable) is a useful inner model of ZFC.

Answer 5

This may be a bit more straightforward, but I was once surprised to learn that the statement for (abelian) groups "$A$ is a vector space over some field" is equivalent to a countable set of formulas: An abelian group in question has this property if and only if each multiplication-by-$n$-map is either identically zero or an isomorphism, which (for any fixed $n$ at a time) is clearly expressible as a first order statement in the language of groups.