# Answer to Question #12541 in Algebra for Hym@n B@ss

Question #12541

Prove that K[X] over integral domain K is principal ideal ring iff K is a field

Expert's answer

Let we have K[X] - PID. Then for any a<>0 we consider ideal

I=<a,X>.

As it is PID we have b in K that I=<b>.

X belongs to

I then X=(dX)b. Then db=1. Hence b=d^(-1) => I=<b>=K[X].

So, 1=

u(X)X+v(X)a. => 1=v(0)a => a - invertible => K is

field.

Reverse implication is obvious.

I=<a,X>.

As it is PID we have b in K that I=<b>.

X belongs to

I then X=(dX)b. Then db=1. Hence b=d^(-1) => I=<b>=K[X].

So, 1=

u(X)X+v(X)a. => 1=v(0)a => a - invertible => K is

field.

Reverse implication is obvious.

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