For example, does the "minimal set" $\{1,\sqrt 2,\sqrt 3,\sqrt 6 \}$ form a basis over $\mathbb{Q}$?
Does any linearly independent set form a basis over a field
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abstract-algebra
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0...a basis for what? When you say "basis" you generally mean "basis for the vector space $X$". – 2012-03-28
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4Basis 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 $\mathbb R$. – 2012-03-28
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0Sorry about that I'll go back and take that out. What I'm really asking is if the minimal set listed above is also a basis in $\mathbb{Q}$ – 2012-03-28
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2A basis of *what*?! – 2012-03-28
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2It is a basis **of the $\mathbb Q$-subspace of $\mathbb R$ generated by it**, i. e. of $\mathbb Q + \mathbb Q \sqrt 2 + \mathbb Q \sqrt 3 + \mathbb Q \sqrt 6$. – 2012-03-28
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0Well I know that the minimal set is a basis for $\mathbb\{\mathbb{Q(\sqrt(3),\sqrt(2)}\}$ over $\mathbb{Q}$. Now I'm just wondering if that is also a basis for $\mathbb{Q}$. I understand that people commonly use vectors to span a particular vector space like $(1,-1)$ and $(-1,2)$ which forms a basis for $\mathbb{R^2}$. – 2012-03-28
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1It cannot be a basis for $\mathbb{Q}$ because $\sqrt{2}$ is not in $\mathbb{Q}$. – 2012-03-28
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0Also, there is really no need to write $\mathbb{Q}(\sqrt{2},\sqrt{3})$ within braces. – 2012-03-28
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0Thank you for noting that Rankeya – 2012-03-28
1 Answers
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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).