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Every vector space is a free module, [1] but, if the ring of the coefficients is not a division ring (not a field in the commutative case), then there exist non-free modules. Given any set S and ring R, there is a free R-module with basis S, which is called the free module on S or module of formal R-linear combinations of the elements of S.
On July 2, 2006, Answers.com released a trivia game known as blufr. [citation needed] In November 2006, Answers.com acquired the question and answer site FAQ Farm. [6] Following the acquisition, the product was renamed WikiAnswers. [7] In the fall of 2009, Answers.com launched a revamped version of their website that fully integrated ...
Contributions owned by the author. Quora granted "worldwide, non-exclusive, royalty-free license" to content use, distribution, or modification. [9] Yes, except to view single answers Free Reddit: 2005 — All topics: Depends on subreddit: No to browse, yes to contribute Sharecare: 2009 — Health and wellness: English: No Spring.me (formerly ...
Asked if he talked to his now-former Yankees teammates during the free-agent process and what might have been said during the competition for his services, Soto said there was no correspondence.
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Hence when n = 1, R is an R-module, where the scalar multiplication is just ring multiplication. The case n = 0 yields the trivial R-module {0} consisting only of its identity element. Modules of this type are called free and if R has invariant basis number (e.g. any commutative ring or field) the number n is then the rank of the free module.
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