 Solution

Factorization over $\mathbb{Q}$ givesFor $R_1$, the factors $x-1$ and $x+1$ are coprime. The nonzero elements $(1,0)$ and $(0,1)$ multiply to zero, so the ring is neither an integral domain nor a field. Squaring coordinatewise shows it has no nonzero square-zero element.

For $R_2$, the polynomial $x^2+1$ has no rational root and is irreducible of degree two over $\mathbb{Q}$. The quotient is consequently a field, hence an integral domain, and has no nonzero square-zero element.

For $R_3$, the class $\epsilon$ is nonzero by its degree, but ${\epsilon }^2=0$. Thus this ring is neither an integral domain nor a field, and does have a nonzero square-zero element.No two are isomorphic. Being a field separates $R_2$ from the other two; existence of a nonzero square-zero element separates $R_1$ from $R_3$. Both properties are preserved by ring isomorphisms.