Counting ring homomorphisms from $ \mathbb{Q}[X]/(X^{3}-1) \rightarrow \mathbb{C} $

Question

I've come across a question on finding all possible ring homomorphisms from $ \mathbb{Q}[X] / (X^{3} - 1) \rightarrow \mathbb{C}$. My initial approach was showing that if we have a ring homomorphism $\psi :\mathbb{Q}[X] / (X^{3} - 1) \rightarrow \mathbb{C} $, then since homomorphisms preserve the additive identity we must have $ \psi ((X^{3} - 1)) = 0$ (the zero quotient element is mapped to zero).

I then aim to construct a corresponding homomorphism $ \phi $ where $ker \phi = (X^{3}-1)$, $ \phi: \mathbb{Q}[X] \rightarrow \mathbb{C}$. Because a homomorphism from $\phi: \mathbb{Q}[X] \rightarrow \mathbb{C}$ would be completely determined by $1$ and $X$, it' s then easy to show that we have three choices for $ \phi(X) $ and hence three homomorphisms. However, my question is how do I show rigourously that there is indeed a bijection between the two different sets of homomorphisms $ \{ \phi \mid \phi: \mathbb{Q}[X] \rightarrow \mathbb{C}, ker \phi = (X^{3} - 1) \}$ and $\{ \psi \mid \psi :\mathbb{Q}[X] / (X^{3} - 1) \rightarrow \mathbb{C} \} $? Thank you very much.

Answer 1

There isn't such a bijection! And, in fact, there are no ring homomorphisms $\mathbb{Q}[X] \to \mathbb{C}$ with kernel $(X^3 - 1)$.

What you do have, however, is a bijection between:

• The set of homomorphisms $\varphi : \mathbb{Q}[X] \to \mathbb{C}$ with $(X^3 - 1) \subseteq \ker(\varphi)$
• The set of homomorphisms $\varphi : \mathbb{Q}[X]/(X^3 - 1) \to \mathbb{C}$

Or, more generally, for arbitrary rings $R,S$ and ideal $I$ of $R$, you have a bijection between

• The set of homomorphisms $\varphi : R \to S$ with $I \subseteq \ker(\varphi)$
• The set of homomorphisms $\varphi : R/I \to S$

Answer 2

The ring homomorphisms $\phi$ from $\Bbb Q[X]/(X^3-1)$ and $\Bbb C$ are determined by the image of $X$, for which as you say, there are three possibilities, namely $\phi_0(X)=1$, $\phi_1(X)=\exp(2\pi i/3)$ and $\phi_2(X)=\exp(4\pi i/3)$. But the kernel of $\phi_0$ is $(X-1)$ (or more strictly $(X-1)\Bbb Q[X]/(X^3-1)$) and the kernels of both $\phi_1$ and $\phi_2$ are $(X^2+X+1)$.