Originally Posted by

**ThePerfectHacker** (I think we should mention a commutative ring with unity first).

We need to show that,

$\displaystyle S=\{2 n+xm|n,m\in \mathbb{Z}[x]\}$ is not equal to some $\displaystyle <r>, r\in \mathbb{Z}[x]$

Let us assume that it is,

$\displaystyle 2\in S$ thus, $\displaystyle 2=ar$ since 2 is prime and $\displaystyle a,r$ are polynomials we conclude they must be constant and that $\displaystyle a=1,r=2$, $\displaystyle a=2,r=1$.

Thus, if it is an ideal it must be either,

$\displaystyle S=<1>,S=<2>$

But, $\displaystyle <1>=\mathbb{Z}[x]$

And $\displaystyle S$ contains all polynomials with the constant term being even, thus, $\displaystyle S\not = <1>$

Thus, if it is an ideal it must be,

$\displaystyle S=<2>$

But, $\displaystyle <2>$ contains all the polynomials with even coefficients. While $\displaystyle S$ does contains some that are not, for example, $\displaystyle 2(1)+x(x)=1x^2+2$. Thus, $\displaystyle <2>\not = S$

An impossibility.

(What we actually shown is that,

$\displaystyle <2>\subset S\subset <1>$)

Now, what ails me is that you say,

"conclude Z[x] is not a principal ideal ring"

That is not true, any commutative ring with unity always is a principal ideal ring. What you probabaly wanted to say is that, "conclude that (2,x) is not a principal ideal ring in Z[x]".

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That is impossible, the kernel of any ring homomorphism is an ideal!

$\displaystyle (2,x)$ is not an ideal in $\displaystyle R$.