Determining whether a ring is a principal ideal ring or not

Question

I have been attempting to attack the following problem off and on for a few weeks now, without much success:

Is the ring $R=\mathbb{Z}_{4}[x]$ of polynomials with coefficients in $\mathbb{Z}_{4}$ a principal ideal ring?

I keep going back and forth on this one. In the moments where I think the answer is "no", I think of the nilradical or the ideal $I=\left\langle 2,x\right\rangle$ as potential counterexamples. When I think the answer is "yes", I keep running in circles trying to prove it. Any hints would be appreciated.

Answer 1

Let's assume that $R=\mathbb Z_4[X]$ is a PIR. Then the ideal $(2,X)$ of $R$ is principal generated by a polynomial $f=2a+a_1X+\cdots+a_nX^n$. Since $2\in(f)$ there is $g=b_0+b_1X+\cdots+b_mX^m$ such that $2=fg$. It follows: $2=2ab_0$, $2ab_1+a_1b_0=0$, and so on. Now note that the ideal $2\mathbb Z_4$ is prime (maximal), and then, from $a_1b_0\in2\mathbb Z_4$ it follows $a_1\in2\mathbb Z_4$ or $b_0\in2\mathbb Z_4$, but the last isn't possible since then from $2=2ab_0$ we get $2=0$, a contradiction. Proceeding this way we get $a_i\in2\mathbb Z_4$ for all $i\ge1$, so $2f=0$. Since $X\in (f)$ we get $2X=0$, a contradiction.

As a general conclusion one can record this result

Let $R$ be a commutative ring with unity. Then $R[X]$ is a PIR if and only if $R$ is a finite product of fields.

which follows easily from the structure theorem of PIRs: every PIR is a finite product of PIDs and of artinian local PIRs (see Theorem 33, page 245, in Zariski and Samuel, Commutative Algebra, volume 1).