Show the ring $(\mathbb Z/4)[x]$ is not a principal ideal ring Show the ring $(\mathbb Z/4)[x]$ is not a principal ideal ring.
What I know:
The ideal $(2,x)$ is an ideal of $(\mathbb Z/4)[x]$, and if it were principal, then its image in $(\mathbb Z/(4,x^2))[x]$ is such that $(2,x)=(2+x)$ in this image, then where do I go from here? $(\mathbb Z/4)$? If so, not sure how. 
Or maybe there is a better way than the above.
 A: Good idea to look at $(2,x)$.  
I would recommend approaching it by writing out the equations that $(2,x)$ principal would imply, i.e. we would have polynomials $f,g,h,u,v$ such that $$2f + xg = h$$ $$hu = 2$$ $$hv = x$$
The important thing to note is that the ideal $(2)$ is prime in $(\mathbb{Z}/4)[x]$ (quotienting at (2) yields $(\mathbb{Z}/2)[x]$, clearly a domain). So either $2 \mid h$ or $2 \mid u$.  
If you write down all the information that these three equations give you about the constant term of $h$, you will arrive at the needed contradiction.  
A: *

*$(2,x)$ is a proper ideal, it does not contain $1$. if $1=2a(x)+xb(x)$, $2=2(2a(x)+xb(x))=2xb(x)$. If $2b(x)=0$, $2xb(x)=0$, contradiction. If $2b(x)\neq 0$, the degree of $2xb(x)$ is superior to $1$, contradiction.


Suppose that $(2,x)=(a(x))$ in $(\mathbb{Z}/4)[x]$. The quotient $(\mathbb{Z}/4)[x]/(2)=\mathbb{Z}/2[x]$. Let $p$ be the quotient map, $p(2,x)$ is the ideal of $\ mathbb{Z}[x]$ generated by $x$, this implies that we can suppose that $p(a(x))=x$, we deduce that $a(x)=x+2b(x)$.
Write $2=(x+2b(x))u(x)$, in $\mathbb{Z}/4[x]$, by mutiplying the equation by $2$,we deduce that $(x+2b(x))2u(x)=2u(x)=0$, we deduce that $u(x)=2v(x)$ and $2=(x+2b(x))2v(x))=2xv(x)=x(2v(x))$ contradiction.
