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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?

Karl Kroningfeld
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le4m
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  • So you want to prove that every ideal $I$ has a generator without just defining it to be the monic polynomial of lowest degree in $I$? Isn't this kind of meaningless? What other structure on polynomials do you want to allow to use? What definition of the generator do you expect? If you accept Hilbert's Basis Theorem (which is, of course, totally overkill here), it suffices to treat finitely generated ideals, which reduces to ideals $(f,g)$. Then you have to prove that $f,g$ have a ggT $h$, and that $h$ is a linear combination of $f,g$. (cont'd) – Martin Brandenburg Oct 18 '13 at 20:04
  • But this is exactly the Euclidean algorithm. It stops because in each step the degree decreases, and the degree is a natural number. What other reason do you want to use? Remark that this is not entirely formal. If you work with polynomials whose exponents have values in $\mathbb{Q}$ or some crazy stuff, the algorithm might never end, and you won't get a PID. – Martin Brandenburg Oct 18 '13 at 20:09
  • Well, given a nonzero ideal $I$ you could prove that $I$ has a unique nonzero monic element of minimal degree (if $I$ is nonzero, then dividing a nonzero element by its leading coefficient gives at least one monic element; and if there are two monic elements of that minimal degree, the difference has smaller degree, and you can divide by the leading coefficient to get a contradiction). Now clearly $\langle f \rangle\subseteq I$... – Daniel Schepler Jan 23 '24 at 00:57
  • For the other direction, if $g \in I$, then $g + \langle f \rangle$ has an element of minimal degree. If that element is nonzero, then that element minus a multiple of $f$ is in $I$ and has smaller degree, giving a contradiction. – Daniel Schepler Jan 23 '24 at 00:59

1 Answers1

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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.

Sammy Black
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