Why does undecidability of arithmetic not follow from that of first-order logic?

Question

As far as I understand, first-order arithmetic incorporates first-order logic. It is a fact that a first-order logic with at least two binary predicates is undecidable. Doesn't this imply immediately the undecidability of arithmetic?

Answer 1

Suppose you have two binary predicates $R$ and $S$. If you add the axiom $(\forall x)(\forall y)[R(x,y) \land S(x,y)]$, the resulting theory is complete and decidable. Basically, with that axiom $R$ and $S$ are always true, so you can replace them with "True" and ignore all quantifiers.

So, adding axioms to an undecidable system can result in a decidable system.

Answer 2

One can make an argument along those lines, by making "incorporates" precise enough. However, note that the first-order theory of algebraically closed fields of characteristic $0$ is decidable, as is the theory of real-closed fields.