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We study free modules in a Modern Algebra course or by reading a book on Algebra. In any case a free module looks like a vector space, for we consider the generating set and basis... My questions are

  • Who first presented the definition of a free module?

  • What problem led people to care about free modules (or just think about the similarity of vector space over a field ?)

Thank for reading.

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Here's one upper bound: free modules do appear in the 1956 text (which I believe was actually written and informally distributed several years earlier) of Cartan and Eilenberg (this is the text in which projective modules were first introduced). –  Pete L. Clark Dec 2 '11 at 15:45
    
Thank Pete L.Clark for noticing that. I also know that injective modules were first introduce in 1940, but projective modules appeared after quite long time, 16 years. It looks like at that time mathematicians did not pay much attention to homological algebra. –  N.Z.K Dec 2 '11 at 16:30
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@Nguyen: you're welcome. Note though that the chronology of C&E is a little strange: the concept of projective modules appears (critically!) in Serre's FAC paper, which was published in 1955 and submitted in October, 1954. (One must also add the amount of time it takes for someone -- even for Serre -- to do and write up 80 substantial pages of mathematics. Let's add, say, another six months for this.) Projective modules are duly attributed to C&E, and their text is cited explicitly, even though it had not yet been published! So "early 1950's" seems a safe upper bound for free modules. –  Pete L. Clark Dec 2 '11 at 17:55
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A next step (for people who don't know the answer already, as I don't!) might be to look at the "classic" algebra books from the 1900's to the 1940's to see which of them mention free modules. For instance, van der Waerden's 1930 text apparently covers modules over a PID, so surely must deal in some ways with the concept of a free module. It would be interesting to see whether he uses the name. –  Pete L. Clark Dec 2 '11 at 18:07
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In the 1948 edition of Bourbaki's Algebra II, Chapter III tensor products of modules appear. In order to construct it, they introduce free modules on a set. The first version of that I was able to find in the Bourbaki archives, was document R100_nbr_011, entitled Appendices I et II where free modules (modules libres) appear on page 98. Unfortunately, I couldn't find when exactly that document was written. –  t.b. Dec 8 '11 at 6:01

1 Answer 1

up vote 3 down vote accepted

B.L. van der Waerden "Modern Algebra, Volume II" (that volume was published in 1931) clearly identifies the notion of a free module (of finite rank), without using the term. I cite (English edition, p.98), where module means right K-module over a not necessarily commutative ring K:

The module $\mathfrak M$ is said to be finite (over K) if its elements may be represented linearly in the form $$ u_1\lambda_1+\cdots+u_n\lambda_n $$ by means of a finite number of elements $u_1,\ldots,u_n$. In this case we write $$ \mathfrak M = (u_1\mathrm K,\ldots,u_n\mathrm K)\hbox{ or } \mathfrak M = (u_1,\ldots,u_n). $$ If we further assume that the $u_i$ are linearly independent, that is $\sum u_i\alpha_i=0$ implies $\alpha_i=0$, then $\mathfrak M$ is called a ($n$-termed) module of linear forms or an $n$-dimensional vector space (cf. Section 14).

So clearly the notion is known by then, and the notion of vector space is even considered to encompass this rather general case. Given the quest for generality (he doesn't initially even assume that the identity of K acts as the identity on the module, though he quickly shows a reduction to that case!) I don't see why finite rank is assumed here from the start.

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Hm, an unfortunate adoption of a "dimension" which isn't well-defined for arbitrary rings... The presentation is very easy to read considering its age, though! –  rschwieb Nov 12 '13 at 17:04

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