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For example, does the "minimal set" $\{1,\sqrt 2,\sqrt 3,\sqrt 6 \}$ form a basis over $\mathbb{Q}$?

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...a basis for what? When you say "basis" you generally mean "basis for the vector space $X$". –  Alex Becker Mar 28 '12 at 5:53
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Basis of what? Your set isn't $\mathbb R$-linearly independent as $\sqrt 2 - \sqrt 2 \cdot 1 = 0$ is a non-trivial $\mathbb R$-linear combination. But it forms a basis of a 4-dimensional $\mathbb Q$-subspace of $\mathbb R$. –  martini Mar 28 '12 at 5:53
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A basis of what?! –  Mariano Suárez-Alvarez Mar 28 '12 at 6:05
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It is a basis of the $\mathbb Q$-subspace of $\mathbb R$ generated by it, i. e. of $\mathbb Q + \mathbb Q \sqrt 2 + \mathbb Q \sqrt 3 + \mathbb Q \sqrt 6$. –  martini Mar 28 '12 at 6:10
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It cannot be a basis for $\mathbb{Q}$ because $\sqrt{2}$ is not in $\mathbb{Q}$. –  Rankeya Mar 28 '12 at 6:16
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up vote 3 down vote accepted

If $S$ is a set of vectors linearly independent over a field $F$, then $S$ is a basis over $F$ for the span of $S$ (which is an $F$-vector space).

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