A Projective Module that is not Free

A Projective Module that is Not Free

We work over a fixed commutative unital ring R. An R-module is projective if given any surjection f : M \to N and map g : P \to N there exists a (not necessarily unique) lift h : M \to P such that g = fh.

Proposition. All free modules are projective.

Pf. We need the axiom of choice. Lift the image of each generator f(e_i) to some m_i if the fiber over f(e_i). Q.E.D.

The converse is not true in general.

Counterexample. R \times \{0\} is projective but not free (as a R-module).

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s