For every $n$ we define $R_n=\mathbb{Z}[2^{1/2}, \dots, n^{1/n}]$, viewed as a subring of the real numbers. How can we prove that $R=\bigcup_n R_n$ is not Noetherian?
I think we need to find a sequence of ideals which does not stabilizes, but I can't find such a sequence.
Consider the ideal $I_n=\langle2^{\frac12}, (2^2)^{\frac1{2^2}}, \cdots,(2^n)^{\frac1{2^n}}\rangle$ in $R$.
Let $v$ be an extension of the $2$-adic valuation of $\Bbb Z$ to $R$. ($v$ can be any extension of the $2$-adic valuation of $\Bbb Z$ to $\bar{\Bbb Q}$ then restricted to $R$.)
Since the sequence $\frac12, \frac2{2^2}, \frac 3{2^3},\cdots,$ decreases to $0$ and $\min\{v(a)\mid a\in I_n\}=\frac n{2^n}$, the sequence $I_1, I_2, I_3, \cdots,$ does not stabilize.