Let $A$ be a commutative ring.
Then the following assertions are equivalent.
- $A$ is a field;
- $A[x]$ is a Euclidean domain;
- $A[x]$ is a principal ideal domain;
- $A[x]$ is a unique factorization domain of dimension 1;
- $A[x]$ 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?