= Solution
Factor $X^3+X=X(X-i)(X+i)$. These factors generate pairwise comaximal <ideals>, so the <Chinese remainder theorem> gives
$$
\Phi([p])=(p(0),p(i),p(-i)).
$$
The three copies of $\mathbb C$ have different <module> structures: $X$ acts by $0,i,-i$, respectively. Thus $q(X)$ acts on $(a,b,c)$ as $(q(0)a,q(i)b,q(-i)c)$, making $\Phi$ a <module homomorphism>.
The inverse, obtained by <Lagrange interpolation>, is
$$
\boxed{\Phi^{-1}(a,b,c)=\left[a(1+X^2)-\frac b2(X^2+iX)-\frac c2(X^2-iX)\right].}
$$
Evaluating at $0,i,-i$ recovers $(a,b,c)$. Conversely, a polynomial whose three evaluations vanish is divisible by $X(X-i)(X+i)$, so these maps are mutual inverses on the <quotient module>. Three copies with identical trivial $X$-action would not give the required isomorphism.
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