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

If $$F$$ is a field, then $$F[x]$$ is a principal ideal domain.

Conversely, if $$F[x]$$ is a principal domain does $$F$$ have to be necessarily a field?

My Thoughts: Suppose instead of $$F$$, we take the set of polynomials $$R[x]$$ over a commutative ring $$R$$ with unity.

Then, suppose $$I$$ is an ideal of $$R[x]$$. Let $$g(x) \in I$$ such that $$g(x)$$ is the polynomial of the lowest degree in $$I$$.

Then: $$\langle g(x) \rangle \subseteq I .......... (1)$$

Let $$f(x) \in I$$. Then $$f(x) = p(x)g(x) + r(x)~~|~~p(x),r(x) \in R[x], \deg r(x) < \deg g(x)$$

Since, $$I$$ is an ideal $$\implies f(x) - p(x)g(x) = r(x) \in I$$

But, $$g(x)$$ is of the lowest degree in $$I \implies r(x) = 0 \implies f(x) \in \langle g(x) \rangle \implies I \subseteq \langle g(x) \rangle ......(2)$$

Then from $$(1),(2) : I = \langle g(x) \rangle$$

Does the Presence of zero divisors in $$R[x]$$ really make a difference? The only advantage I see is that if there are no zero divisors in $$R[x]$$ then $$I=\{0\} \implies I= \langle 0 \rangle$$.

But, every ideal contains the zero element. Why is there the condition of a field specifically given in textbooks for $$F[x]$$ to be a principal ideal domain?

• How do you propose to divide $\,x\,$ by $\,2\,$ in $\,\Bbb Z[x],\,$ in (fruitless) attempt to prove $\,(x,2)\,$ is principal? – Bill Dubuque Jul 21 '14 at 0:29
• I see where I went wrong. Thank you :) – MathMan Jul 21 '14 at 0:39
• I believe it can be improved to from "field" to "division ring", if we change the conclusion "PID" to "all ideals are principal, but not asking integral domain". – Santropedro Jul 15 '17 at 17:09

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.

• Thanks. Can you please tell where I could have gone wrong in the proof above. – MathMan Jul 21 '14 at 0:26
• Further, PIDs are precisely the UFDs of dimension $\le 1,\,$ i.e. UFDs where every nonzero prime ideal is maximal. – Bill Dubuque Jul 21 '14 at 0:26
• To see where your proof goes wrong, look at it in the special case of the following easy counterexample: $R=\mathbb Z$ and $I$ is the ideal generated by $2$ and $x$. – Andreas Blass Jul 21 '14 at 0:29
• @VHP glad to help. I think I once wondered this myself early in my undergraduate career and was happy to learn the answer as well. – Adam Hughes Jul 21 '14 at 0:37
• Non-zero prime ideals in PIDs are maximal =] – Ragib Zaman Jul 21 '14 at 1:20