Ytukyg

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?

To find an explicit isomorphism, we note that
$$1=(-x)cdot x+1cdot(x^2+1).$$ This proves also that $I=(x)$ and $J=(x^2+1)$ are coprime ideals. So, for all $p(x)inmathbb{Q}[x]$, 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
$$varphibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(x)operatorname{mod} x,p(x)operatorname{mod} (x^2+1)big).$$
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
$$psibig(a(x)operatorname{mod} x,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)a(x)-x^2b(x) Big)operatorname{mod} x(x^2+1).$$
Note that there exists an isomorphism $mathbb{Q}[x]/(x)tomathbb{Q}$ sending $big(f(x)operatorname{mod} xbig)mapsto f(0)$. 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
$$Phibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(0),p(x)operatorname{mod} (x^2+1)big)$$
and
$$Psibig(t,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)t-x^2b(x) Big)operatorname{mod} x(x^2+1).$$