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?

Thank you for your help.

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    $\begingroup$ How do you propose to divide $\,x\,$ by $\,2\,$ in $\,\Bbb Z[x],\, $ in (fruitless) attempt to prove $\,(x,2)\,$ is principal? $\endgroup$ – Bill Dubuque Jul 21 '14 at 0:29
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    $\begingroup$ I see where I went wrong. Thank you :) $\endgroup$ – MathMan Jul 21 '14 at 0:39
  • $\begingroup$ 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". $\endgroup$ – 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.

  • $\begingroup$ Thanks. Can you please tell where I could have gone wrong in the proof above. $\endgroup$ – MathMan Jul 21 '14 at 0:26
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    $\begingroup$ Further, PIDs are precisely the UFDs of dimension $\le 1,\,$ i.e. UFDs where every nonzero prime ideal is maximal. $\endgroup$ – Bill Dubuque Jul 21 '14 at 0:26
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    $\begingroup$ 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 $x$. $\endgroup$ – Andreas Blass Jul 21 '14 at 0:29
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    $\begingroup$ @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. $\endgroup$ – Adam Hughes Jul 21 '14 at 0:37
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    $\begingroup$ Non-zero prime ideals in PIDs are maximal =] $\endgroup$ – Ragib Zaman Jul 21 '14 at 1:20

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