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?
Constructing a ring with a chain of ideals $(2) \subsetneq (2^{1/2}) \subsetneq (2^{1/3}) \subsetneq \cdots$
2
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abstract-algebra
ring-theory
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1Exhibit it as a subring of $\mathbb{R}$. – 2012-03-03
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0I think OP wants $(2) \subset (2^{1/2}) \subset \dots$ to be a strictly increasing chain of ideals, which does not hold for $\mathbb R$. – 2012-03-03
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0@Dylan: I assume the intent is to get a sequence of strict inclusions (for which $\mathbb{R}$ doesn't work). – 2012-03-03
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0That was my guess as well, but there's no harm in spelling it out :) I'll edit. – 2012-03-03
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0@DylanMoreland: Sorry for the ambiguity! – 2012-03-03
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0@sdcvvc: I know that this is a daft question, but why does that not hold for R? – 2012-03-03
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0$\mathbb R$ is a field, so all the ideals $(2), (2^{1/2}), \dots$ are the same. – 2012-03-03
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0@Genevieve $\mathbf R$, being a field, has only two ideals: $(0)$ and $(1)$. – 2012-03-03
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1@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. :) – 2012-03-03
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0That said, it might make more sense just to give a canned example, such as the ring of algebraic integers. – 2012-03-03
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0Yeah, I was responding to a now deleted comment. – 2012-03-03
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0It might be slightly trickier to prove that this ring does what you want. Do you see how to do it? – 2012-03-04
1 Answers
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Try $\mathbb{Z}[2^{\frac12},2^{\frac13},...]$.
This is a ring, and you have strict inclusions
$(0) \subsetneq (2) \subsetneq (\sqrt{2}) \subsetneq (2^{\frac 13}) \subsetneq ...$