If $F[x]$ is a principal domain does $F$ have to be necessarily a field? [duplicate]

Solution 1:

Yes it does. For if not the ideal $(x)$ would not be maximal, and every prime ideal in PID is maximal. This goes through the characterization of a maximal ideal as $M\subseteq R$ is maximal iff $R/M$ is a field.

In our case it's easy to see that:

$$F[x]/(x)\cong F$$

So the result is pretty immediate.

If you'd like the proof of "PID implies all (non-zero) prime ideals are maximal." It's pretty straightforward:

Proof: Let $\mathfrak{p}=(p)$ be a non-zero prime ideal of a commutative ring, $R$ with $1$. then if $\mathfrak{p}\subseteq M\subseteq R$, then if $M=(x)$ we have that $x|p$ hence is either a unit or $p$ itself by definition of a prime.

Addendum (where you went wrong): You do not necessarily have a division algorithm in an arbitrary polynomial ring, $R[x]$, since $R$ not a field means that the coefficients are not all units (i.e. the ring is not Euclidean). Take $R=\Bbb Z$, then you do not have the desired factorization with only integer polynomials for something like $3x$ and $2x$, since the difference cannot be made to have lower degree.

This generalizes to any non-field since you can find some non-unit somewhere to pull the same thing on.