Is $\mathbb{Q}[x]/(x-1)(x^2+4)\mathbb{Q}[x]$ isomorphic to $\mathbb{Q}\times\mathbb{Q}[i]$? 
Problem: Check if $\mathbb{Q}[x]/(x-1)(x^2+4)\mathbb{Q}[x]$ is isomorphic to $\mathbb{Q}\times\mathbb{Q}[i]$. If not, find what is it isomorphic to.

My guess: they're isomorphic.
My attempt:
I define $\varphi: \mathbb{Q}[x] \to \mathbb{Q} \times \mathbb{Q}[i]$, $\varphi(w) = (w(1),w(2i))$. It is easy to show that $\varphi$ is a surjective homomorphism but i struggle with showing that $\ker \varphi = (x-1)(x^2+4)\mathbb{Q}[x]$.
I know it should be easy but find it hard to proceede. I'd be thankful for some hints. 
 A: It's clear that $\ker \varphi$ contains $(x-1)(x^2+4)$, so it contains the ideal generated by $(x-1)(x^2+4)$. Conversely let $P \in \ker \varphi$. Then since $\mathbb{Q}[x]$ is a Euclidean domain, there exist unique polynomials $Q,R \in \mathbb{Q}[x]$ such that
$$P = (x-1) (x^2+4) Q + R,$$
with $\deg R < 3$. Now $P(1) = 0 = R(1)$ and $P(\pm 2i) = 0 = R(\pm 2i)$, hence $R$ has three distinct roots in $\mathbb{C}$ (here I used the fact that if $z \in \mathbb{C}$ is a root of some real polynomial, then so is $\bar z$). Being of degree less than three it can only be the zero polynomial. It follows that $(x-1) (x^2+4)$ divides $P$.
A: Since Najib Idrissi has given a direct method, I'll add an answer based on the Chinese remainder theorem. In fact, this is an immediate application of the theorem. We just need to show that the ideals $(x - 1)$ and $(x^2 + 4)$ are coprime. By the polynomial division algorithm,
$$
x^2 + 4 = (x + 1)(x - 1) + 5.
$$
Since $5$ is a unit in $\mathbb Q$, we are done.
A: They both have dimension 3 as $\mathbb{Q}$ vector spaces, so a surjective homomorphism is also injective. In fact, the first has basis $1,x,x^2$, while the second $(1,0),(0,1),(0,i)$.
