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I am trying to construct a ring that contains this chain of principal ideals: $$(2)\subsetneq (2^{1/2})\subsetneq (2^{1/3})\subsetneq \cdots$$ How can I show that it gives a ring?

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Exhibit it as a subring of $\mathbb{R}$. – Qiaochu Yuan Mar 3 '12 at 19:25
I think OP wants $(2) \subset (2^{1/2}) \subset \dots$ to be a strictly increasing chain of ideals, which does not hold for $\mathbb R$. – sdcvvc Mar 3 '12 at 19:28
@Dylan: I assume the intent is to get a sequence of strict inclusions (for which $\mathbb{R}$ doesn't work). – Qiaochu Yuan Mar 3 '12 at 19:28
That was my guess as well, but there's no harm in spelling it out :) I'll edit. – Dylan Moreland Mar 3 '12 at 19:31
@sdcvvc: All that means is that we're not going to make use of the lattice of ideals of $\mathbb{R}$. Subrings of $\mathbb{R}$ have plenty of ideals: $\mathbb{Z}$, for example. :) – Hurkyl Mar 3 '12 at 19:52
up vote 3 down vote accepted

Try $\mathbb{Z}[2^{\frac12},2^{\frac13},...]$.

This is a ring, and you have strict inclusions

$(0) \subsetneq (2) \subsetneq (\sqrt{2}) \subsetneq (2^{\frac 13}) \subsetneq ...$

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