Consider the ideal subring of $ \scriptstyle Z[x] $ of all polynomials with even constant terms:
$ \scriptstyle I\ =\ \textstyle\{\scriptstyle a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0\textstyle\mid\scriptstyle\ 2\mid a_0\textstyle\} $.
Suppose $ \scriptstyle I $ is principal, such that $ \scriptstyle I\ =\ \{f(x)g(x)\ \mid\ g(x)\in Z[x]\} $.
Case 1: $ \scriptstyle f(x)\ =\ 0 $
- If $ \scriptstyle f(x)\ =\ 0 $ then $ \scriptstyle I\ =\ \{0\} $, which is a contradiction.
Case 2: $ \scriptstyle f(x)\ \neq\ 0,\ \ deg(f(x))\ =\ 0 $
- In this case, $ \scriptstyle f(x)\ =\ k\in \mathbb{Z}\setminus{\{0\}} $, so all coefficients of $ \scriptstyle f(x)g(x) $ are divisible by $ \scriptstyle k $. We know that $ \scriptstyle x\in I $, so $ \scriptstyle k\mid1 $. Therefore, $ \scriptstyle k\ =\ 1 $ or $ \scriptstyle k\ =\ -1 $. But, $ \scriptstyle 1,-1\notin I $, and again we reach a contradiction.
Case 3: $ \scriptstyle deg(f(x))\ \geq\ 1 $
- If $ \scriptstyle deg(f(x))\ \geq\ 1 $, then either $ \scriptstyle f(x)g(x)\ =\ 0 $ (signifying $ \scriptstyle g(x)\ =\ 0 $), or $ \scriptstyle deg(f(x)g(x))\ \geq\ deg(f(x))\ \geq\ 1 $. But, $ \scriptstyle 2\in I $, and $ \scriptstyle2 $ is neither $ \scriptstyle0 $ nor does it have a degree of $ \scriptstyle1 $ or higher. Here, too, there is a contradiction.
In all cases we reach a contradiction, so $ \scriptstyle I $ must not be principal. Therefore, $ \scriptstyle Z[x] $ cannot be a principal ideal domain. $ \scriptstyle\Box $
- --Nick Rupley 05:05, 2 April 2009 (UTC)
