Is $1/2$ in $\mathbb{Z}[2^{1/2},2^{1/3},2^{1/4},...]$? I was wondering if $\frac{1}{2}$ is in the ring $R=\mathbb{Z}[2^{1/2},2^{1/3},2^{1/4},...]$. I don't think it is, and I've been trying to prove by contradiction. So far I've shown that if this is true, then $R=\mathbb{Z}[...,2^{-1/3},2^{-1/2},2^{1/2},2^{1/3},...]$, which seems unusual. Do I need more techniques from field theory to make headway? Any hints are appreciated!

Going off of Kenta's and Bill Dubuque's comments, we note that $\mathbb{Z}[2^{1/2},2^{1/3},...]$ is an integral extension of $\mathbb{Z}$, since every root of $2$ is integral over $\mathbb{Z}$. If $1/2\in \mathbb{Z}[2^{1/2},2^{1/3},...]$, then $\mathbb{Z}[1/2]$ must be an extension of $\mathbb{Z}$ contained in $\mathbb{Z}[2^{1/2},2^{1/3},...]$, and thus an integral extension. However, $1/2$ is not integral over $\mathbb{Z}$, and we arrive at a contradiction.
