# 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, 2014 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. Apr 28, 2014 at 2:38

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.

(Note that there are other trivial ways of doing this problem, as mentioned in the comments, and an answer above, but I thought of doing this problem in a bit different way, just for fun.)

• Very neat proof ! :) Dec 10, 2018 at 18:29

Hint $$\rm\rm\,\ x \; f(x) = 1 \,$$ in $$\rm\ F[x]\, \Rightarrow\, 0 = 1 \,$$ in $$\rm F \,$$ by evaluating at $$\rm\ x = 0.\$$

If you know the universal (mapping) property of the polynomial ring then you may find it instructive to interpret the above from that viewpoint (see here).

Can you find a polynomial in $p \in F[x]$ such that $p(x)\cdot x = 1$? Why not?