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It is a well-known theorem that over a field $\mathbb{F}$, any $n+1$ vectors in $\mathbb{F}^n$ are linearly dependent.

Does this theorem hold over commutative rings as well? Meaning, if $R$ is a commutative ring, are any $n+1$ vectors in $R^n$ linearly dependent? I know that over division rings there is a counterexample, but I couldn't find an answer for commutative rings.

Guy
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  • Look here: http://math.stackexchange.com/questions/79726/cardinality-of-a-minimal-generating-set-is-the-cardinality-of-a-basis/79798#79798 And here: http://math.stackexchange.com/questions/375263/bases-and-linearly-independent-sets-in-free-r-modules – LeviathanTheEsper Sep 17 '15 at 16:21
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    And what would happen if $R=\prod_{i\in\mathbb{N}}\mathbb{Z}_2$ and $R^2=R$? – LeviathanTheEsper Sep 17 '15 at 16:42

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yes use the notion of basis. if the n vectors are linearly independent then they are the full set of a basis and the remaining one will surely be a spanned by the other n elements.

Shan Bynnud
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