For a field $K$, is there a way to prove that $K[x]$ is a PID without mentioning Euclidean domain?

I know that if $K$ is a field then $K[x]$ is a Euclidean domain and every Euclidean domain is a PID. In this way I can prove that $K[x]$ is a PID.

But is there a method to show $K[x]$ is a PID directly from the definition?

I mean a usual procedure is to design the concept of Euclidean norm and shows that $K[x]$ is Euclidean domain, taking advantage of that concept.

But the concept of PID and ideal does not really look related with Euclidean (division) algorithm structure on the surface. So there might be a method to show some structure is a PID without mentioning Euclidean algorithm structure.

But maybe it's impossible? For the Euclidean algorithm concept is so basic?

Solution 1:

You need to be able to establish that any pair of polynomials $p, q \in K[x]$ have a greatest common divisor $d$. Then, the ideal generated by $p$ and $q$ will in fact be generated by $d$.

How to do this? It seems that you can't get away from consideration of degree of the polynomials. As you well know, the degree function $$ \deg: K[x] \setminus \{0\} \to \mathbb{Z}_{\ge 0} $$ is a norm. It is in fact a Euclidean norm, but you don't need quite that much strength in order to establish that $K[x]$ is a PID.

You need it to be a Dedekind-Hasse norm. (Of course, any Euclidean norm can be used to construct a D-H norm; hence, any Euclidean domain is a PID).

As far as I can tell, the answer to your question is: yes, you can get away with not proving that $K[x]$ is a Euclidean domain, but it requires a bit of looking the other way. :-)

Incidentally, It may be worthwhile to show that the ring $\mathbb{Z}[(1 + \sqrt{-19})/2]$ is a PID using a D-H norm, since it is not Euclidean.