Showing isomorphism of Quotient Ring to direct product of Complex numbers

Question

So I need to prove $$\mathbb{C}[x]/(x^3+1)$$ is isomorphic to $\mathbb{C} \times \mathbb{C} \times \mathbb{C}$, where $\mathbb{C}$ is the field of complex numbers. Based on an example in the book I feel like the answer might have to do with representing the elements of the field using the division algorithm of $\mathbb{C}$, but I'm honestly not sure and don't really now how to start here. Quotient groups tend to be kind of unintuitive for me.

Answer 1

I don't know if you still need the answer, but here it goes: First notice that $x = -1$ is a root of $x^3 + 1$ and then you can divide $x^3 + 1$ by $x+1$ using your favorite method to obtain:

$$x^3 + 1 = (x+1)\left(x-\frac{1+i\sqrt{3}}{2}\right)\left(x - \frac{1-i\sqrt{3}}{2}\right)$$

Now notice that you wrote your polynomial as product of irreducible/prime elements, which means that the ideals $\langle x+1 \rangle$, $\left\langle x-\frac{1+i\sqrt{3}}{2} \right\rangle$ and $\left\langle x - \frac{1-i\sqrt{3}}{2} \right\rangle$ are coprimes and you can use the Chinese remainder theorem to obtain:

\begin{align} \frac{\mathbb{C}[x]}{\langle x^3+1 \rangle} & = \frac{\mathbb{C}[x]}{\left\langle (x+1)\left(x-\frac{1+i\sqrt{3}}{2}\right)\left(x - \frac{1-i\sqrt{3}}{2}\right)\right\rangle} \\ & = \frac{\mathbb{C}[x]}{\left\langle x+1\right\rangle \left\langle x-\frac{1+i\sqrt{3}}{2} \right\rangle \left\langle x - \frac{1-i\sqrt{3}}{2} \right\rangle} \\ & \overset{CRT}{\cong} \frac{\mathbb{C}[x]}{\langle x+1 \rangle} \times \frac{\mathbb{C}[x]}{\left\langle x-\frac{1+i\sqrt{3}}{2} \right\rangle} \times \frac{\mathbb{C}[x]}{\left\langle x-\frac{1-i\sqrt{3}}{2} \right\rangle} \\ & \overset{ev}{\cong} \mathbb{C} \times \mathbb{C} \times \mathbb{C} \end{align}

Here $ev_{\alpha} : \mathbb{C}[x] \to \mathbb{C}$ is the evaluation map, ie, $ev_{\alpha}(p(x)) = p(\alpha)$. Can you see that $ev_{\alpha}$ induces an isomorphism between $\mathbb{C}[x]/\langle x-\alpha \rangle$ and $\mathbb{C}$?