Isomorphic ring $\mathbb{Q}[x]/(x^3+x) \cong \mathbb{Q} \times \mathbb{Q}[x]/(x^2+1)$ I want to prove an isomorphism of the form $\mathbb{Q}[x]/(x^3+x) \cong \mathbb{Q} \times \mathbb{Q}[x]/(x^2+1)$. I want to use the Chinese Remainder Theorem
$\mathbb{Q}[x]/(x^3+x) = \mathbb{Q}[x]/(x(x^2+1))$
So I choose $I = (x), J = (x^2 + 1)$
What I have to show first is that $I$, $J$ are coprime. (How can I do this? Because it feels trivial)
Secondly I have to show that $\mathbb{Q}[x]/(x) = \mathbb{Q}$. My question is: How can I prove this?
 A: To find an explicit isomorphism, we note that
$$1=(-x)\cdot x+1\cdot(x^2+1).$$
This proves also that $I=(x)$ and $J=(x^2+1)$ are coprime ideals.  So, for all $p(x)\in\mathbb{Q}[x]$, we can write
$$p(x)=\big(-xp(x)\big)\cdot x+p(x)\cdot (x^2+1)=(x^2+1)p(x)-x^2p(x).$$
Therefore, a good isomorphism $\varphi:\Bbb{Q}/\big(x(x^2+1)\big)\to \big(\mathbb{Q}[x]/(x)\big)\oplus \big(\mathbb{Q}/(x^2+1)\big)$ is given by
$$\varphi\big(p(x)\operatorname{mod}x(x^2+1)\big)=\big(p(x)\operatorname{mod} x,p(x)\operatorname{mod} (x^2+1)\big).$$
The inverse of $\varphi$ is $\psi: \big(\mathbb{Q}[x]/(x)\big)\oplus \big(\mathbb{Q}/(x^2+1)\big)\to \Bbb{Q}/\big(x(x^2+1)\big)$ given by
$$\psi\big(a(x)\operatorname{mod} x,b(x)\operatorname{mod}(x^2+1)\big)=\Big((x^2+1)a(x)-x^2b(x) \Big)\operatorname{mod} x(x^2+1).$$
Note that there exists an isomorphism $\mathbb{Q}[x]/(x)\to\mathbb{Q}$ sending $\big(f(x)\operatorname{mod} x\big)\mapsto f(0)$.  So, you can rewrite $\varphi$ and $\psi$ as $$\Phi:\Bbb{Q}/\big(x(x^2+1)\big)\to \mathbb{Q}\oplus \big(\mathbb{Q}/(x^2+1)\big)$$ and $$\Psi:\mathbb{Q}\oplus \big(\mathbb{Q}/(x^2+1)\big)\to \Bbb{Q}/\big(x(x^2+1)\big),$$ which are given by
$$\Phi\big(p(x)\operatorname{mod}x(x^2+1)\big)=\big(p(0),p(x)\operatorname{mod} (x^2+1)\big)$$
and
$$\Psi\big(t,b(x)\operatorname{mod}(x^2+1)\big)=\Big((x^2+1)t-x^2b(x) \Big)\operatorname{mod} x(x^2+1).$$
