17
$\begingroup$

Projective modules were introduced in 1956 by Cartan and Eilenberg in their book Homological Algebra. Does anyone know why they chose the word "projective"? Does it have something to do with the notion of projection?

$\endgroup$

1 Answer 1

24
$\begingroup$

The term "projection" has a few possible meanings in linear algebra, and they are equivalent to the property of being a projective module.

  1. A linear operator $f \colon {\mathbf R}^n \rightarrow {\mathbf R}^n$ is geometrically a projection to a subspace in some coordinate system if and only if $f^2 = f$.

  2. For two vector spaces $V$ and $W$, the function $V \oplus W \rightarrow V$ where $(v,w) \mapsto v$ is called a projection (out of the direct sum).

For a commutative ring $R$ and $R$-module $P$, the following properties that abstract the above two conditions are both equivalent to $P$ being a projective module.

  1. There is a free $R$-module $F$ and an $R$-linear map $f \colon F \rightarrow F$ such that $f^2 = f$ and $f(F) \cong P$ (so $P$ is isomorphic to the image of a "projection").

  2. Any way of making $P$ look like a quotient module essentially amounts to making it the image of the natural projection out of a direct sum module to one of its direct summands: if $f \colon M \twoheadrightarrow P$ is any surjective $R$-linear map then there is an $R$-module isomorphism $h \colon M \cong P \oplus K$ for some $R$-module $K$ such that $h(m) = (f(m),*)$ for all $m \in M$ (thus $f$ looks like the natural projection map $P \oplus K \rightarrow P$).

$\endgroup$
1
  • $\begingroup$ Makes sense. Thanks for the explanation. $\endgroup$
    – PatrickR
    Jan 23, 2013 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.