Is $\mathbb{Z}[x]/(2x^2)$ a finitely generated $\mathbb{Z}$ - module, so I'm thinking not, since the scalars come from $\mathbb{Z}$ and so it must be generated by a finite number of elements not in $\mathbb{Z}$.
So this means that the elements in $\mathbb{Z}[x]/(2x^2)$ must have a maximum power of $x$ they can take and if we were looking at $\mathbb{Z}[x]/(x^2)$ this would be the case since it is generated by $\{1, x\}$, but since it is $2x^2$ and 2 does not have an inverse in $\mathbb{Z}$, then is it the case that $\mathbb{Z}[x]/(2x^2)$ consists of polynomials with coeffecients in $\mathbb{Z}$ such that the for powers of $x$ higher than 1, the coeffecient is odd.
So it would be: $\{ a_nx^n + a_{n-1}x^{n-1}..... a_1x + a _0 | a_k \equiv 0 \text{ (mod }2) \text{ for all } 2 \leq k \leq n \}$
I'm not sure if this is completely wrong, if it is can anyone point out where I've gone wrong please?
Taking up your idea: suppose the module is finitely generated, meaning
$$M:=\Bbb Z[x]/(2x^2)=\left\langle\,\overline f_1(x),\ldots,\overline f_n(x)\,\right\rangle_{\Bbb Z}\,,\,\;\text{with}\;\;\overline f_k(x)=f_k(x)+I\;,$$
$$\,I:=(2x^2)\;,\;\,f_k(x)\in\Bbb Z[x]\,,\;\,\deg f_k=m_k$$
For simplicity, suppose we chose the indexes in such a way that $\;m_1\le m_2\le\ldots\le m_n\;$
Thus, there exist $\;a_1,...,a_n\in\Bbb Z\;$ s.t. that
$$\;\overline{x^{2m_n+1}}=\sum_{k=1}^na_k\overline f_k(x)\iff x^{2m_n+1}-\sum_{k=1}^na_kf_k(x)\in I\implies$$
$$\;\implies x^{2m_n+1}-\sum_{k=1}^na_kf_k(x)=2h(x)x^{2}\;,\;\;h(x)\in\Bbb Z[x]$$
But this is impossible since the principal coefficient of $\;2x^2h(x)\;$ is even, whereas the principal coefficient of $\;x^{2m_k+1}-\sum\limits_{k=1}^na_kf_k(x)\;$ is $\;1\;$ since $\;\deg\left(\sum\limits_{k=1}^na_kf_k(x)\right)\le m_n<2m_n+1\;$