# 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.

• Is that supposed to be a quotient? I mean, is all that stuff $\;(x-1)(x^2+4)\Bbb Q[x]\;$ , as an ideal in the polynomial ring, supposed to be in the denominator part of the quotient ring? It may be wise to use parentheses here. – Timbuc Dec 16 '14 at 13:32
• I would consider looking at the Chinese Remainder theorem... – Simon Rose Dec 16 '14 at 13:34

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 $P$.
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.
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 $(1,0),(0,1),(0,i)$.
• Sure but (assuming it's not obvious) we would have to show that $(x-1)$ and $(x^2+4)$ are coprime first, right? Without it we couldn't say that the first one has basis $1,x,x^2$, right? – Mitkel Dec 17 '14 at 20:50
• No, it is not necessary. That is exactly the beauty of this method: you don't need to check anything. In general, if $p$ is a monic polynomial of degree $n$ in $k[x]$, then $k[x]/p$ has dimension $n$ as a vector space. This proof means that almost all the information to construct the isomorphism is encoded in the existence of the ring homomorphism $\phi$. Once this is done, all the algebra is encoded in $\phi$, and you only have to check that the dimensions are right. The fact that $(x-1)$ and $(x^2+4)$ are coprime can be seen as a consequence. – Giulio Bresciani Dec 18 '14 at 1:32
• To be clearer: try to do the stupid case $\mathbb{Q}[x]/(x-1)^2\to \mathbb{Q}\times \mathbb{Q}$ exactly in the same way, defining $\phi(w)=(w(1),w(1))$. The fact that $x-1$ is not coprime with $x-1$ is already encoded in the fact that $\phi$ is not surjective! – Giulio Bresciani Dec 18 '14 at 1:47