Can someone give me a hint why in the quotient ring $\mathbb{F}_p[x]/{\langle f\rangle}$, where $f$ is an irreducible polynomial of degree $k$ and $p$ is a prime (and $\langle f\rangle$ the ideal generated by $f$), every element is invertible ?
[Math] Invertible elements in polynomial quotient rings
abstract-algebrapolynomials
Best Answer
As pointed out above, 0 is not invertible. However, for any field $K$, $K[x]$ is a PID. Since $f$ is irreducible and $\mathbb{F}_p[x]$ is a PID, $f$ is a prime element of $\mathbb{F}_p$ (if this isn't clear, then sit down and hammer out a simple proof).
The only thing that we need now is the following proposition.
Proposition: In a PID, every nonzero prime ideal is maximal.
Sketch of proof: Suppose that $D$ is a PID and that we have $0\subsetneq P\subseteq M$ for a maximal ideal $M$ of $D$ and a (nonzero) prime ideal $P$ of $D$. Since $D$ is a PID, we have $P=(p)$ and $M=(q)$ for some nonzero $p,q\in M$. Also, since both $M$ and $P$ are prime ideals, it follows that $p$ and $q$ are prime (and hence irreducible) elements of $D$ (if this isn't immediately clear, then sit down and write out a short proof as to why).
However, $(p)\subseteq (q)$ implies that $q \vert p$, implying that $q=up$ for some $u\in U(D)$. Therefore $M=P$ and every nonzero prime ideal of a PID is maximal. $\blacksquare$
So, going back to $f\in \mathbb{F}_p[x]$, we have that $f$ is a prime polynomial in $\mathbb{F}_p[x]$, hence $(f)$ is a maximal ideal. Therefore $\mathbb{F}_p[x]/(f)$ is a field.