# Rings such that $A[x]$ is a principal ideal domain

Let $A$ be a commutative ring.

Then the following assertions are equivalent.

1. $A$ is a field;
2. $A[x]$ is a Euclidean domain;
3. $A[x]$ is a principal ideal domain;
4. $A[x]$ is a unique factorization domain of dimension 1;
5. $A[x]$ is an integral domain of dimension $1$.

The implication 1->2 is well-known. 2->3->4->5 are also standard.

The only difficulty might be 5->1. I thought of the following incomplete argument.

Firstly, one can show that $\dim A[x] = \dim A +1$. Therefore, $\dim A = 0$. Since $A[x]$ is an integral domain, it follows that $A$ is a zero-dimensional integral domain. Thus, we conclude that $A$ is a field.

I have two problems with this argument.

Firstly, I can't seem to prove in an easy way that $\dim A[x] = \dim A +1$. Does somebody have an easy argument?

Secondly, I don't like the dimension argument. Is there an easier argument not relying on Krull dimensions?

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If $A$ is not a field, we can find a non-zero maximal ideal $I$ of $A$. Then the ideal $J$ generated by $I$ and $x$ in $A[x]$ is not all of $A[x]$, and is prime since the quotient $A[x]/J$ is isomorphic to the field $A/I$. The ideal $(x) \subset A[x]$ is also prime since $A[x]/(x) \cong A$, which is a domain. So, $0\subset (x) \subset J$ is a chain of prime ideals of length 3 in $A[x]$, whence $\dim A[x] \geq 2$. –  user15464 Apr 1 '12 at 13:26

The ideal generated by $x$ is prime. Since the dimension of the ring is one and $0 \subset (x)$ is a chain of prime ideals of length one and since each prime ideal is contained in some maximal ideal, it follows that $(x)$ is maximal (since $\dim A[x] = 1$). But $A = A[x]/(x)$, so it is a field.