# If $F$ is a field show that $F[x]$ is not a field. [duplicate]

I know that $ax=1$ has a solution in $F$ so that every element must be a unit but then I'm not sure how to proceed.

• Can you think of an inverse of the polynomial $x$?\ – user61527 Apr 28 '14 at 2:28
• That has a solution for every x in F, but that is not really the same x as the indeterminate of the polynomial ring. – rschwieb Apr 28 '14 at 2:38

Hint $$\rm\rm\,\ x \; f(x) = 1 \,$$ in $$\rm\ F[x]\, \Rightarrow\, 0 = 1 \,$$ in $$\rm F \,$$ by evaluating at $$\rm\ x = 0.\$$
Can you find a polynomial in $p \in F[x]$ such that $p(x)\cdot x = 1$? Why not?
It's obvious that $F[x]/(x)$ is isomorphic to $F$, and hence $(x)$ is a non trivial proper ideal of $F[x]$, and hence $F[x]$ can't be a field.