Intersection of ideals $I=(2x)$ and $J=(2x^2)$ of $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ is not finitely generated. Consider the subring $\mathbb{Z}[2x,2x^2,2x^3,\dots]\subset \mathbb{Z}[x]$. Then show that the intersection of ideals $I=(2x)$ and $J=(2x^2)$ of $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ i.e., $I\cap J\subseteq \mathbb{Z}[2x,2x^2,2x^3,\dots]$ is not finitely generated. Also $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ is not a unique factorization domain.
I bumped on this question few days ago while trying to learn about Euclidean domain, PID and UFDs. I was able to do the unique factorization part. But could not solve the previous part, any hint would be highly appreciated.
Thank you
 A: $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ is not a UFD because $4x^2 = (2x)^2 = 2\cdot 2x^2$ where none of the two latter can be factorized further.
Now for the ideals: The monomials contained in $I$ are either of the form $2ax$ or of the form $4ax^n$ for $n \geq 2$ (and any monomial on this form is in $I$). Similarily for $J$ we have $2ax^2$ and $4ax^n$ for $n \geq 3$.
That means that the monomials in $I \cap J$ are all the monomials of the form $4ax^n$ for $n \geq 2$. Say it can be finitely generated, and let $S$ be a fiunite set of generators. Then there is one element of S with largest degree, say $n$. The monomial $4x^{n+1}$ cannot be generated from elements of $S$, and therefore $S$ cannot be a complete set of generators. Therefore any set of generators of $I \cap J$ must be infinite.
A: Put $A = \mathbb{Z}[2X, 2X^2, \dots]$. Suppose $I\cap J = (4x^n:\, n \geq 2)$ has a finite generating set $f_1, \dots, f_n$. Put $g = 4x^m$ with $m > \deg f_i$ for all $i$, and write $g = p_1 f_1 + \cdots + p_n f_n$ for some $p\in A$. For any $f = a_0 + a_1 X + \dots + \alpha_k X^k\in A$, let $\alpha = (\alpha_1, \dots, \alpha_k)\subset \mathbb{Z}$. Then every $\alpha(p_i)\subset (2)$ and every $\alpha(f_i)\subset (4)$. Since $g$ has no terms of degree less than $m > \deg f_i$, it follows that $\alpha(g)\subset \alpha(p_1 f_1) + \cdots + \alpha(p_n f_n)\subset(8)$. But clearly $\alpha(g) = (4)$.
