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

Is $\operatorname{Spec} R$ a Noetherian topological space?

Here is what I know.

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

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

3. $R$ is not Noetherian ring.

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

Is $\operatorname{Spec}R$ a Noetherian topological space?

Here is what I know.

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

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

3. $R$ is not a Noetherian ring.

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

Is $\operatorname{Spec} R$ a Noetherian topological space?

Here is what I know.

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

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 a strictly increasing chain of radical ideals.

3. $R$ is not Noetherian ring.

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

Is $\operatorname{Spec} R$ a Noetherian topological space?

Here is what I know.

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

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

3. $R$ is not Noetherian ring.

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.

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

Is $\operatorname{Spec} R$ a Noetherian topological space?

Here is what I know.

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

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 a strictly increasing chain of radical ideals.

3. $R$ is not Noetherian ring.