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

Question

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.

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 02:38

score 5 · Answer 1 · answered Apr 28 '14 at 02:43

5

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.)

answered Apr 28 '14 at 02:43

voldemort

13,182

Very neat proof ! :) – DaveWasHere Dec 10 '18 at 18:29

Bill Dubuque · Answer 2 · 2019-08-27T19:04:15.997

5

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).

edited Aug 27 '19 at 19:04

answered Apr 28 '14 at 02:46

Bill Dubuque

272,048

score 4 · Answer 3 · answered Apr 28 '14 at 02:32

4

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

answered Apr 28 '14 at 02:32

Ben Grossmann

225,327

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

3 Answers3