Finding all homomorphisms and proving a solution exists Had some troubles solving this type of questions.
Let $L = \Bbb{Z}[x] / \langle x^2+x+1 \rangle$. 
Find all homomorphisms from $L$ to $\mathbb C$.
Show that the equation $y^2+3=0$ has a solution in $L$.
Since $x^2+x+1$ is of degree 2, it has two roots of the complex field, and therefore there are two homomorphisms. But how do I describe them?
Thanks for any assistance! 
 A: It seems you got it: a homomorphism $f$ from $L$ to $\Bbb C$ is uniquely determined by $\alpha:=f([x])$ where $x$ is the coset of $x\in\Bbb Z[x]$. Moreover, $\alpha$ has to be a root of the given equation, as $[x^2+x+1]=[0]$ has to be mapped to $0$.
Prove that for a general polynomial $p\in\Bbb Z[x]$, we must have $f(p)=p(\alpha)$.
Now, calculate the complex roots of $x^2+x+1=0$, and express one root by the other one, and $\sqrt{-3}$ by the roots, using ring operations.
A: Your $L=\{ax+b \, | \, a,b \in \mathbb{Z} \text{ and } x^2+x+1 \equiv 0\}$. So if you know where $1$ and $x$ maps to then you know everything about the homomorphism. Say $x \mapsto z$, then since $f(x^2+x+1)=f(0)=0$, therefore $z^2+z+1=0$. Now solve this in $\mathbb{C}$ to get the possible values where $x$ can map to. 
The solutions to the equation $z^2+z+1=0$ are $\omega$ and $\omega^2$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ (cube root of unity). Thus one homomorphism will be where $x \mapsto \omega$ and $1 \mapsto 1$, thus $f(ax+b)=a\omega+b$. The other will be given by $f(ax+b)=a\omega^2+b$. 
