There are so many equivalent definitions of projective module, that it’s hard to know which is the “right” definition. What about this: an $R$ module $M$ is projective if the functor ${\rm Hom}_R(M,-)$ is an exact functor. This one makes the most sense to me, of course there are others: $M$ is projective if and only if it is the direct summand of a free module, if and only if every short exact sequence $0\to A\to B\to M\to 0$ splits, etc.

Actually there is one equivalent condition for $M$ to be projective that doesn’t get too much attention, but can be quite useful: the dual basis condition. What is a dual basis? A dual basis for $M$ is a set of functions $\{ f_i : i\in I\}$ where $f_i\in {\rm Hom}_R(M,R)$ and a set of elements $\{ m_i : i\in I\}$ where $m_i\in M$ such that:

- For every $m\in M$, the set $\{ i\in I : f_i(m)\not=0\}$ is finite, and
- For every $m\in M$, we have $m = \sum_i f_i(m)m_i$

The name is a bit misleading, because a dual basis is *not* a basis necessarily! It is more like the next best thing to a basis. For although $\{m_i : i\in I\}$ certainly generates $M$ as an $R$-module, we certainly don’t require that every element can be written as a unique $R$-linear combination of the $m_i$ elements.

**Theorem.**An $R$-module $M$ is projective if and only if it has a dual basis.

This theorem is another way of looking at the fact that a dual basis isn’t really a basis in all cases, although it can be if $M$ were a free module. How do you prove this theorem? Well, it basically proves itself once you know how to construct the $f_i$ and $m_i$ elements. So, I won’t go over the details of the proof because it’s actually more instructive if you do it yourself. But I will tell you how to construct these elements for a projective $M$: take a surjection

$$\oplus_{i\in I}R\to M.$$ Because $M$ is projective, this surjection has a splitting $M\to \oplus_i R$. Define $f_i$ as the compostion

$$M\to \oplus_i R\to R$$ where the map $\oplus R\to R$ is the $i$th projection. Define the element $m_i$ as the image of $1\in R$ under the map

$$R\to \oplus_i R\to M$$ where $R\to\oplus_i R$ is the $i$th inclusion map.

With these constructions, it should be easy to go through the details required to prove the above theorem. As an exercise, you should use the theorem to prove the following:

**Theorem.**Suppose $R\to S$ is a ring homomorphism such that $S$ is projective as an $R$-module. If $M$ is a projective $S$ module, then $M$ is a projective $R$-module.

You can prove this using the existence of a dual basis for $M$ as an $S$ module and the existence of a dual basis for $S$ as an $R$-module. Actually, I find it easier to prove this statement using the fact that a module is projective if and only if it is the direct summand of a free module!

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