Show that $\mathbb{Q}[X]/\langle X^3+X\rangle\cong \mathbb{Q} \times \mathbb{Q}[i]$

Question

I've seen that $\mathbb{R}[X]/\langle X^2+1\rangle\cong \mathbb{C}$ with the evaluation in i. I tried to replicate the idea, but I don't end up having that $\ker(f)=\langle X^3+X\rangle$, but $\langle X^2+1\rangle$ instead. I thought of putting $f(p)=\frac{1}{3}(p(i)+p(-i)+p(0))$, but this is not multiplicative, so it's not a ring morphism.

Answer 1

By Chinese Remainder Theorem, and specifically the coprimality of ideals $(x)$ and $(x^2+1)$ in $\mathbb Q[x]$, $$\mathbb Q/\left((x)\cdot(x^2+1)\right)\cong \mathbb Q/(x)\times\mathbb Q/(x^2+1)\cong \mathbb Q\times\mathbb Q[i]$$