Is $F[x]/(x-\alpha)\approx F$?

Question

Let $F$ be an algebraically closed field.

I am supposed to find all the prime ideals of $F[x]$. It is a PID so all prime ideals are maximal, and all maximal ideals are generated by irreducible polynomials. The only irreducible polynomials in an algebraically closed field are of degree 1 and 0. Thus $(x-\alpha)$ is irreducible for all $\alpha \in F$. Now is $F[x]/(x-\alpha)\approx F$?. It would seem that is the case by the map $f(x) \to f(\alpha)$. Is this correct? Then this would mean the only quotient rings are $F$ and $0$.

Answer 1

Yes, that is correct. Let $\operatorname{ev}_\alpha\colon F[X]\longrightarrow F$ be the map defined by $\operatorname{ev}_\alpha\bigl(P(x)\bigr)=P(\alpha)$. Then:

1. $\operatorname{ev}_\alpha$ is a ring homomorphism;
2. $\ker\operatorname{ev}_\alpha=\langle X-\alpha\rangle$;
3. $\operatorname{ev}_\alpha\bigl(F[X]\bigr)=F$.

So, $\operatorname{ev}_\alpha$ induces a field isomorphism from $F[X]/\langle X-\alpha\rangle$ onto $F$.