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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!

Charuvinda
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    You can search for an ideal $I$ containing $2$ such that the quotient $R/I$ is obviously not the zero ring. – reuns Jan 15 '22 at 05:00
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    $1/2$ is not integral. – Kenta S Jan 15 '22 at 05:02
  • Is it clear what essentially distinguishes between this problem and the problem of whether $1/2$ is in $\mathbb Z[2^{1/2},2^{1/3},\dots,2^{1/10}]$? – Alex Ortiz Jan 15 '22 at 05:49
  • @AlexOrtiz a solution to the problem you stated would probably be easily adapted to solve my problem. – Charuvinda Jan 15 '22 at 05:53
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    Every number in your ring is an algebraic integer, i.e. a root of a monic poly with integer coef's, but $1/2$ is not (by the Rational Root Test). Generally integral extensions can't change a nonunit to a unit (similarly for ideals, i.e. nonunit ideals survive, e.g. see the paper cited here for various characterizations of integral extensions) – Bill Dubuque Jan 15 '22 at 09:29

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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.

user26857
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Charuvinda
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