Prove or disprove that the ideal $(2+4\mathbb{Z},x)$ is a principal ideal in $(\mathbb{Z}/4\mathbb{Z})[x]$ Prove or disprove that the ideal $(2+4\mathbb{Z},x)$ is a principal ideal in $(\mathbb{Z}/4\mathbb{Z})[x]$.
I know $\mathbb{Z}/4\mathbb{Z}$ is not a field. have something help?
 A: Suppose $(2+4\mathbb Z, x)$ is principal, generated by $p(x) \in (\mathbb Z/4\mathbb Z)[x]$. In order for $2$ to be generated by $p(x)$, we must have a nonzero constant term in $p$, so we can write $p(x) = a+xq(x)$, where $a \neq 0+4\mathbb Z$. So we must have $2+4\mathbb Z = (a+xq(x))(b+xf(x))$ and $x = (a+xq(x))(c+xg(x))$ (no assumptions on $b$ or $c$ yet), which requires $ab=2+4\mathbb Z$ and $ac = 0+4\mathbb Z$.
We first consider the case $c=0$. Then we have $(a+xq(x))g(x) = 1$, and therefore $p(x)$ is invertible. To eliminate this case, verify that $(2+4\mathbb Z, x) \neq (\mathbb Z/4\mathbb Z)[x]$, e.g. by verifying that the quotient ring $(\mathbb Z/4\mathbb Z)[x]/(2+4\mathbb Z, x)$ is isomorphic to $\mathbb Z/2\mathbb Z$.
So in fact, we must have $a=c=2+4\mathbb Z$, because that's the only other way to solve $ac=0+4\mathbb Z$. But now it's easy to verify that the equation $x = (2+4\mathbb Z+xq(x))(2+4\mathbb Z+xg(x))$ cannot be solved.
A: If it were principal it would have to be generated by a constant in $\mathbb{Z}/4\mathbb{Z}$, and since $2$ is in this ideal then it would have to be generated by 2. But $x$ is not in the ideal generated by 2.
