Why can't the Polynomial Ring be a Field?

Hint $\rm\quad\rm x \, f(x) = 1 \,$ in $\,\rm R[x]\ \Rightarrow \ 0 = 1 \, $ in $\,\rm R, \, $ by evaluating at $\rm\ x = 0 $

Remark $\ $ This has a very instructive universal interpretation: if $\rm\, x\,$ is a unit in $\rm\, R[x]\,$ then so too is every $\rm\, R$-algebra element $\rm\, r,\,$ as follows by evaluating $\ \rm x \ f(x) = 1 \ $ at $\rm\ x = r\,.\,$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $\rm R$-algebra. $ $ A natural choice is the nonunit $\,\rm 0\in R,\,$ which yields the above proof.

Because by definition, the only polynomial that can have a negative degree is $0$, which is defined to have a degree of $-\infty$. Non-zero constants have degree $0$. You then have the degree equation: $\deg (fg) = \deg (f) + \deg (g)$ for any polynomials $f,g$. By inspection, any polynomial of degree $n \geq 1$ would need as an inverse a polynomial of degree $-n$, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the field of rational functions, and is precisely the set of ratios of polynomials. It is constructed the same way that the field of rational numbers is from the ring of integers.

For $F[x]$ to be a field, you need to show there is an inverse for each element that isn't 0. Now $x \in F[x]$, and clearly $x \ne 0$ (considered as a polynomial). But if you multiply $x$ by any non-zero polynomial, the result will always contain $x$ or higher powers, so it has no inverse.