2 questions about the ring $\mathbb Q[X]/(X^{3}-1)$ I am unable to solve this particular question in Ring Theory. This was asked in a masters exam for which I am preparing .
Let $A =\mathbb Q[X]/(X^{3}-1)$ .
(a) Prove that $A$ is direct product of two integral domains.
(b) Is the ring $A$ isomorphic to $\mathbb Q[X]/(X^{3}+1)$?
I can know by $X^{3}-1$ that now elements would be $ax^2+bx+c$, $a,b,c$ belonging to $\mathbb{Q}$. But I have no clue direct products of which integral domain will make this ring.
Also for 2nd I am having problems in defining a map as $X^3$ will act as -1 in 2nd ring. I don't think map like $\phi( ax^2+bx+c )=px^2 +qx+r$ would work as this map is not $1-1$.
So, can anyone please tell how should I approach both of these problems.
 A: HINT:
(a) Use the Chinese Remainder Theorem, which says that for a ring $A$ and ideals $\mathfrak a,\mathfrak b$ of $A$ such that $\mathfrak a+\mathfrak b=(1)$, $A/\mathfrak{ab}\cong A/\mathfrak a\times A/\mathfrak b$. Furthermore, a quotient ring $\mathbb Q[X]/(f(X))$ is an integral domain iff $(f(X))$ is a prime ideal iff $f(X)$ is irreducible (since $\mathbb Q[X]$ is a PID).
(b) I claim $\mathbb Q[X]/(X^3+1)\to\mathbb Q[X]/(X^3-1):X\mapsto-X$ is an isomorphism. Check all the axioms hold.
A: (a) As Kenta S stated, since $1=(x^2-x+1)+x(x-1)$ and $(x^2-x+1)(x-1)=x^3-1$, we have $\langle x^2-x+1\rangle+\langle  x-1\rangle=\mathbb Q[x]$ and so $\mathbb Q[x]/\langle x^3-1\rangle\cong \mathbb Q[x]/\langle x^2-x+1\rangle\times \mathbb Q[x]/\langle x-1\rangle$ by the Chinese Remainder Theorem. Clearly, $x^2-x+1$ and $x-1$ are irreducible. Hence, $\mathbb Q[x]/\langle x^2-x+1\rangle$ and $\mathbb Q[x]/\langle x-1\rangle$ are domains.
(b) Clearly, $\mathbb Q[x]/\langle x-1\rangle\cong \mathbb Q\cong\mathbb Q[x]/\langle x+1\rangle$. Also, $\mathbb Q[x]/\langle x^2-x+1\rangle\cong\mathbb Q[x]/\langle x^2+x+1\rangle$ by $x\to -x$. Hence, $\mathbb Q[x]/\langle x^3-1\rangle\cong \mathbb Q[x]/\langle x^2-x+1\rangle\times \mathbb Q[x]/\langle x-1\rangle\cong \mathbb Q[x]/\langle x^2+x+1\rangle\times \mathbb Q[x]/\langle x+1\rangle\cong\mathbb Q[x]/\langle x^3+1\rangle$.
