Show $\mathbb{R}[x]/\langle x^{3}-1\rangle \cong \mathbb{R} \times \mathbb{C}$

Question

I have to show $\frac{\mathbb{R}[x]}{x^{3}-1} \cong \mathbb{R} \times \mathbb{C}$ as rings. We know that $x^{3}-1 = (x-1)(x^{2}+x+1)$ and so by the Chinese remainder theorem, $\frac{\mathbb{R}[x]}{x^{3}-1} \cong \frac{\mathbb{R}[x]}{x-1} \times \frac{\mathbb{R}[x]}{x^{2}+x+1}$. Then we can give the map $\phi : (f(x),g(x))\mapsto (f(1),g(\frac{-1+\sqrt{3}i}{2}))$, which is well defined because $1,\frac{-1+\sqrt{3}i}{2}$ are roots and $\phi$ is an isomorphism. I was wondering if $\frac{\mathbb{R}[x]}{x^{3}-1} \cong \frac{\mathbb{R}[x]}{x-1} \times \frac{\mathbb{R}[x]}{x^{2}+x+1}$ is correct and if this is sufficient to claim $\frac{\mathbb{R}[x]}{x^{3}-1} \cong \mathbb{R} \times \mathbb{C}$.

Answer 1

Yes, the argument using the Chinese Remainder Theorem is correct. Then $\frac{\mathbb{R}[x]}{x-1} \cong \mathbb{R}$ is straight-forward (the isomorphism sends $x$ to $1$). Further $\frac{\mathbb{R}[x]}{x^2+x+1} \cong \mathbb{C}$ because $x^2+x+1$ is irreducible over the reals since it has no zeros. Thus $\frac{\mathbb{R}[x]}{x^2+x+1}$ is a proper field extension of the reals and therefore must be the complex numbers.

Answer 2

Your approach looks fine to me.

Because evaluating a polynomial at a point is always a homomorphism of rings, an alternative approach might be to consider the mapping $$F:\Bbb{R}[x]\to\Bbb{R}\times\Bbb{C}, f(x)\mapsto (f(1),f((-1+i\sqrt3)/2)).$$ It is a homomorphism of rings. It is relatively easy to show that the kernel is exactly the ideal generated by $x^3-1$. The codomain as well as the quotient $\Bbb{R}[x]/\langle x^3-1\rangle$ are both 3-dimensional vector spaces over the reals, so $F$ must also be surjective. Of course, CRT gives surjectivity just as well.