Why can't the Polynomial Ring be a Field? I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set. 
The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?
 A: Consider $\mathbb{C}[x]$ the ring of polynomials with coefficients from $\mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicative inverse.
A: The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.
A: 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.
A: 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.
A: 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.
A: Let $R$ be a field. If $f$ is a polynomial whose degree is at least 1 in $R[x]$, then $f$ cannot have an inverse.
For if $f(x)g(x) = 1$, where $f$ and $g$ are polynomials whose degree is at least 1 in $R[x]$, then the leading coefficient of $g$ would have to be $0$, which is
impossible.
