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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.

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    $\begingroup$ Can you think of an inverse of the polynomial $x$?\ $\endgroup$
    – user61527
    Apr 28, 2014 at 2:28
  • $\begingroup$ That has a solution for every x in F, but that is not really the same x as the indeterminate of the polynomial ring. $\endgroup$
    – rschwieb
    Apr 28, 2014 at 2:38

3 Answers 3

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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.)

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  • $\begingroup$ Very neat proof ! :) $\endgroup$
    – DaveWasHere
    Dec 10, 2018 at 18:29
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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).

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

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