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The algebraic elements of $\mathbb{R}$ are those elements which are roots of nonzero polynomials with coefficients in $\mathbb{Q}$. In fact, by multiplying through by denominators, we can even take the polynomials to have coefficients in $\mathbb{Z}$. My question is, in a sense, about the converse of this situtation in a more general setting.

Let $\mathbb{L}$ be a field extension of a field $\mathbb{K}$, and let $\mathbb{S}$ be a subring of $\mathbb{K}$ such that every algebraic element of $\mathbb{L}$ is the root of some nonzero polynomial with coefficients in $\mathbb{S}$. Is $\mathbb{K}$ the field of quotients of $\mathbb{S}$?

Note, as $\mathbb{S}$ is a subring of the field $\mathbb{K}$, it is automatically an integral domain so we can construct the field of quotients of $\mathbb{S}$.


As Mariano and pritam's examples show, the answer is no. However, in all of these examples, $\mathbb{K}$ is an algebraic extension of the field of quotients of $\mathbb{S}$. Is this always the case?

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Take $\mathbb K=\mathbb L=\mathbb Q(\sqrt 2)$ and $\mathbb S=\mathbb Z$. –  Mariano Suárez-Alvarez Sep 6 '12 at 6:39
    
If that is too annoying, take $\mathbb K=\mathbb Q(\sqrt2)$, $\mathbb L=\mathbb Q(\sqrt2,\pi)$ and $\mathbb S=\mathbb Z$ :-) –  Mariano Suárez-Alvarez Sep 6 '12 at 6:42
    
@Mariano: Thank you for your counterexamples. Maybe instead I should ask whether $\mathbb{K}$ must necessarily be an algebraic extension of the field of quotients of $\mathbb{S}$. –  Michael Albanese Sep 6 '12 at 7:59
    
@Michael Precisely what do you mean by an "algebraic element of $\Bbb L$", i.e. algebraic over what field? –  Bill Dubuque Sep 6 '12 at 14:22
    
@Bill: I mean algebraic over $\mathbb{K}$. –  Michael Albanese Sep 7 '12 at 5:19
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3 Answers

up vote 2 down vote accepted

Concerning the new question. $K$ is a field containing $S$, so it is an extension of the field $F$ of quotients of $S$. If it's not algebraic over $F$, then it has some element, call it $\pi$, transcendental over $F$. $L$ is an extension of $K$, so it contains $\pi$, and $\pi$ is algebraic over $K$, since it's actually in $K$, but it's not a root of a polynomial with coefficients in $S$, since it's transcendental over $F$. So, for your hypotheses to be met, $K$ must be algebraic over $F$.

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Excellent. Thank you. –  Michael Albanese Sep 6 '12 at 13:12
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No. Take $\mathbb{L}=\mathbb{C},\mathbb{S}=\mathbb{R},\mathbb{K}=\mathbb{C}$

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Thanks pritam. In your example, as with Mariano's examples, $\mathbb{K}$ is an algebraic extension of the field of quotients of $\mathbb{S}$. Is this necessarily the case? I will edit my post to include this question. –  Michael Albanese Sep 6 '12 at 11:52
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Yes to your new question. Your hypothesis asserts that every element of $L$ algebraic over $K$ is algebraic over the subfield $Q =$ fraction field of $S.$ In particular this is true for all elements of $K$.

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