= Solution
Factorization over $\mathbb Q$ gives
$$
R_1\cong\mathbb Q\times\mathbb Q,\qquad
R_2\cong\mathbb Q(i),\qquad
R_3\cong\mathbb Q[\varepsilon]/(\varepsilon^2),\quad\varepsilon=x-1.
$$
For $R_1$, 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 $\varepsilon$ is nonzero by its <degree>, but $\varepsilon^2=0$. Thus this ring is neither an <integral domain> nor a <field>, and does have a nonzero square-zero element.
$$
\boxed{\begin{array}{c|ccc}&\text{integral domain}&\text{field}&\exists u\neq0:\ u^2=0\\\hline
R_1&\text{no}&\text{no}&\text{no}\\
R_2&\text{yes}&\text{yes}&\text{no}\\
R_3&\text{no}&\text{no}&\text{yes}
\end{array}}
$$
\b[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>.
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