A ring $R$ (commutative, with identity) is said to be solid if the unique homomorphism $\mathbb{Z} \to R$ is an epimorphism in the category of rings, or equivalently if $r \otimes 1 = 1 \otimes r$ in $R \otimes R$ for all $r \in R$.

The rings $\mathbb{Q}$ and $\mathbb{Z}/p$ (for primes $p$) are each solid. But is the ring $\mathbb{Q} \times \prod_{\text{primes } p} \mathbb{Z}/p$ solid?

No, it's not. Here's a proof that uses an ultraproduct.

Admittedly this proof is rather elaborate, and there may well be simpler proofs (perhaps already published decades ago) that don't use ultraproducts or even any form of choice. You can take that as a challenge. Nevertheless, it's an instance where ultraproducts can be used to answer a question in the mainstream of algebra.

Interestingly, the proof has some points in common with Asaf's wonderful proof of the irrationality of $\sqrt{2}$.