On math.se Bill Dubuque gave a proof that given a ring $R$, the
polynomial ring $R[x]$ cannot then be a field. This is one of the nicest
proofs I've seen and I wanted to go further into how it works.
So all credit to Bill as the author, this is
just my analysis of his words and finishing off his proof from the hint.
So without further ado, the hint he gave is this:
$$xf(x) = 1 \in R[x] \Rightarrow 0 = 1 \in R$$ by evaluating $x=0$.
I blogged about this proof on a previous blogs and made some mistakes, so,
time for correction.
Firstly, an element $u$ of a ring $R$ is a unit iff there exists some
$v \in R$ such that $uv = vu = 1$.
Now let $x, f(x)$ be elements of $R[x]$, where $f$ is some polynomial
function. Then, and this is the beauty of this particular proof,
$f(x)$ is simultaneously an element of the ring $R$ by evaluation
of $x=r$ and of $R[x]$ by its expression of the unknown $x$ as a polynomial.
Bringing the two ideas together, if $x$ some element in $R[x]$ and it
is a unit, then the formula for units holds, so $xf(x) = 1$. As we have
observed, $f(x)$ can be an element in $R$ by evaluation at
$x=r$ and so if $x$ is
an unit in $R[x]$ then all $x=r$ are units in $R$ by evaluation of $f(r)$.
Now we have everything we need to finish the proof. Suppose $R[x]$ is a
field. Then since every element is a unit, for $x \in R[x]$ there is some
$f(x)\in R[x]$ such that $xf(x)=1$, that is $x$ is a unit.
We can then use the previous idea that if $x$ is a unit then $x=r$ is a
unit by evaluating $xf(x)=1$ in $R$. So let $x=0$. Then $0\cdot f(0)=1
\Rightarrow 0=1 \in R$. This is clearly not true and so we have a
contradiction. Thus $R[x]$ is not a field. $\square$
If Bill happens to be reading this, yes, I upvoted your answer, thank
you very much.