$\langle x\rangle$ is a maximal ideal of $R[x]$, where $R$ is a field 
Show that $\langle x\rangle$ is a maximal ideal of $R[x]$, where $R$ is a field.

This is one of my assignment questions. It is supposed to be hard but what I did below is so easy but I couldn't find anything wrong. Can anybody help me check it out?
Since $R$ is a field and $f(x) = x$ is irreducible in $R[x]$, so $R[x]/\langle x\rangle$ is a field, and $\langle x\rangle$ is a ideal of $R[x]$, so $\langle x\rangle$ is maximal.
 A: You are correct that if you can prove that $R[x]/\langle x\rangle$ is a field, then you can conclude that $\langle x\rangle$ is maximal.
However, it is not true in general that if $a$ is irreducible in a domain $D$, then $D/\langle a\rangle$ is a field. For example, $2$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$, but $\langle 2\rangle$ is not a maximal ideal of $\mathbb{Z}[\sqrt{-5}]$. So you need more than just "$f(x) = x$ is irreducible in $R[x]$" to conclude "$R[x]/\langle x\rangle$ is a field."
(For another, simpler example, consider the case of $\langle x \rangle$ in $R[x,y]$. You can still say that "since $R$ is a field, $x$ is irreducible", but it is no longer true that $R[x,y]/\langle x\rangle$ is a field; in fact, $\langle x\rangle$ is not maximal in this ring, since it is properly contained in the proper ideal $\langle x,y\rangle$.)
But perhaps you can show that the homomorphism $R[x]\to R$ given by "evaluation at $0$" is onto, and has kernel $\langle x\rangle$? That will show that $R[x]/\langle x\rangle \cong R$ is a field.
A: If you know that $R[x]$ is a PID when $R$ is a field, then any ideal above $\langle x\rangle$ must be generated by some monic polynomial $f$ which divides $x$. Hence $f=x$ or $f=1$ and so the ideal is either $\langle x\rangle$ or $R[x]$, and $\langle x\rangle$ is maximal.
A: Hint $\rm\ f\not\in\! (x)\ \Rightarrow\ (x,f) = (x,\: f\ mod\ x) = (x,f(0)) = (1)\:$ by $\rm\:f(0)\neq 0\ \Rightarrow\ f(0)$ unit   
Or: $\rm\ mod\ x\!:\:\ x\equiv 0\ \Rightarrow\ f(x)\equiv f(0) =$ unit or $0,\:$ so $\rm\:R[x]\ mod\ x\:$ is a field.
