Let $R = \mathbb{Z}/2\mathbb{Z}[X,X^{1/2},X^{1/4},\cdots,X^{1/{2^n}},\cdots]$

Is Spec R is Noetherian a topological Space?

Here are what I know.

1.R is integral over $\mathbb{Z}/2\mathbb{Z}[X]$ and hence 1 dimentinal.

2.If we replace $\mathbb{Z}/2\mathbb{Z}$ with $\mathbb{C}$, the corresponding Spectrum is Noetherian since $(X-1)\subset (X^{1/2}-1)\subset (X^{1/4}-1)\subset \cdots (X^{1/{2^n}}-1)\subset \cdots$ is strictly increasing chain of radical ideals.

3.R is not Noetherian ring.