Are the rings $\mathbb{Q}[X]/(X^3-1)$ and $\mathbb{Q}[X]/(X^3+1)$ isomorphic?

Question

Let $A=\mathbb{Q}[X]/(X^3-1)$

(a) Prove that $A$ is a direct product of two integral domains.
(b) Is the ring $A$ isomorphic to $\mathbb{Q}[X]/(X^3+1)$? Justify your answer.

I was guessing the first part will be isomorphic to direct product $\mathbb{Q}[x]/\langle x-1\rangle$ and $\mathbb{Q}[x]/\langle x^2 + x +1\rangle$. But I cannot prove they are equal. I think here equality means in terms of isomorphism of rings. I will be happy if anyone helps me in solving this.

Answer 1

To show the isomorphism between $\mathbb{Q}[X]/(X^3-1)$ and $\mathbb{Q}/(X-1)\oplus \mathbb{Q}/(X^2+X+1)$ apply the Chinese remainder theorem, or use the fact that there exist polynomials $A$ and $B$ such that $A(X-1)+B(X^2+X+1)=1$ and consider the canonical map $\mathbb{Q}[X]\rightarrow \mathbb{Q}[X]/(X-1)\oplus \mathbb{Q}[X]/(X^2+X+1)$ and show that its kernel is the ideal generated by $X^3-1$.

Answer 2

Since $\gcd(X-1,X^{2}+X+1)=1$, so $(X-1)$ and $(X^{2}+X+1)$ are coprime, so Chinese Remainder Theorem applies, just as what @Tsemo Aristide has noted.

Answer 3

The answer to (b) is affirmative. Here we give an explicit isomorphism.

Let $A=\mathbb Q[X]/(X^3+1)$ and $B=\mathbb Q[X]/(X^3-1)$. Define a map $T\colon \mathbb Q[X]\to \mathbb Q[X]$ by $$T(P)(X):=P(-X).$$ Clearly, $T$ is a ring automorphism. The restriction of $T$ to the ideal $(X^3+1)$ is a ring isomorphism of $(X^3+1)$ onto $(X^3-1)$, so the map $\tilde{T}\colon A\to B$ given by $$\tilde{T}( P+ (X^3+1) ) := T(P)+(X^3-1)$$ is well-defined an it is a ring isomorphism.