I'm working on this problem for an algebra qualifying exam, and I haven't been able to make much progress.
For each positive integer $n$, let $R_n = \mathbb{Z}[2^{1/2}, \ldots, n^{1/n}]$ be a subring of the field of the real numbers. Prove that $R = \bigcup_{n \geq 1} R_n$ is not Noetherian.
My idea is to consider the ascending chain of ideals $(2^{1/2}) \subset (2^{1/2}, 3^{1/3}) \subset \cdots \subset R$ and show that the chain never stabilizes. Intuitively, I should show that adding an additional n-th root introduces new elements into the ideal each time so that the inclusion is always proper. However, I don't really know where to start with showing this.
Suppose $R$ is noetherian.
For each nonnegative integer $k$, if $x_k,n_k,y_k,e_k$ are given by $$ \left\{ \begin{align*} x_k&=2^{\bigl(2^{-k}\bigr)}\\[4pt] n_k&=2^{\bigl(2^k\bigr)}\\[4pt] y_k&=(n_k)^{\frac{1}{n_k}}\\[4pt] e_k&=2^{2^k-2k}\\[4pt] \end{align*} \right. $$ then identically we have $$ x_k=(y_k)^{e_k} $$ hence since $y_k\in R$, we get $x_k\in R$.
Then, noting that for all $k$ we have $x_k=(x_{k+1})^2$, it follows that $$ (x_0)\subseteq(x_1)\subseteq(x_2)\subseteq(x_3)\subseteq\cdots $$ is an ascending chain of principal ideals, hence, since R is noetherian, we must have $$ (x_m)=(x_{m+1}) $$ for some $m$.
Then since $x_m=(x_{m+1})^2$, it follows that $x_{m+1}$ is a unit of $R$.
But then, noting that $$ \bigl(x_{m+1}\bigr)^{2^{m+1}}=x_0=2 $$ it follows that $2$ is a unit of $R$, contradiction, since ${\large{\frac{1}{2}}}$ is not an algebraic integer, whereas all elements of $R$ are algebraic integers.