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Is there a subfield $K$ such that $\mathbb{Q} \subset K \subset \mathbb{R}$ (proper subset) as follows:

$\mathbb{R}$ is a vector space over $K$ and has no finite generating set and $K$ is a vector space over $\mathbb{Q}$ and has no finite generating set.

Sum and multiplication are both defined as usual in $\mathbb{R}$.

My guess is that there is one: $K$ should not contain continuum as it is field so in this case it would be equal to $\mathbb{R}$, so $K$ is countable, and $\mathbb{R}$ has no finite generating set over countable sets. On the other hand, $K$ can contain countable but infinite number of elements from $\mathbb{R} - \mathbb{Q}$ such that $\forall a \in \mathbb{R} - \mathbb{Q} \,\,\,\,a \in K \Rightarrow \exists b \in K, q \in \mathbb{Q}:a\neq q\cdot b$, i.e. contains infinite number of irrational numbers so that pairwise their division is not rational. In this case number of elements in $K$ basis is countable, but infinite (each irrational number is in basis, if its product with some rational number isn't in it). $K$ seems to be field and not equal to $\mathbb{R}$, but I'm not sure.

Thanks in advance!

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1  
Does like... $\overline{\mathbb{Q}}$ work? –  Dylan Wilson Nov 29 '12 at 2:29
    
If $\,\overline{\Bbb Q}\,$ means "the algebraic closure of the rationals" then it is not contained in the reals. –  DonAntonio Nov 29 '12 at 2:40
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Hehe whoops- amend that to $\overline{\mathbb{Q}} \cap \mathbb{R}$ which is what Ittay has as well. :) –  Dylan Wilson Nov 29 '12 at 2:44

1 Answer 1

up vote 5 down vote accepted

Take $K$ to be the field of real algebraic numbers over $\mathbb {Q}$. It is countable so $\mathbb {R}$ is not finitely generated over $K$. Further, using Eisenstein's Criterion, it is simple to show that $[K:\mathbb {Q}]=\infty$ and so $K$ is not finitely generated over $\mathbb{Q}$.

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But $\,K\rlap{\,\;/}\subset\Bbb R\,$... –  DonAntonio Nov 29 '12 at 2:40
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I just edited the correction. Thanks DonAntonio! –  Ittay Weiss Nov 29 '12 at 2:43
    
+1 Very nice indeed, @Ittay –  DonAntonio Nov 29 '12 at 2:45
    
It seems that you could also take something like $\mathbb Q(\pi)$. –  JSchlather Nov 29 '12 at 3:50
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@starovoitovs That's not a basis for $\mathbb{Q}(\pi)$ as a vector space over $\mathbb{Q}$. Remember that any finite extension is algebraic, nevertheless $\pi$ is transcendental. The extension $\mathbb{Q}(\pi)/ \mathbb{Q}$ is infinite. –  Adrián Barquero Nov 29 '12 at 4:43

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