TITLE: Is $m$ a projective $A$-module?
QUESTION [5 upvotes]: $A$ is a Noetherian local ring and $m$ be its maximal ideal. Then is $m$ a projective $A$-module? 

I got this problem while solving another problem. Can anyone please help me to figure it out?

REPLY [4 votes]: Over a local commutative ring, projective modules coincide with  free modules, so the question is whether $\mathfrak m$ is free.
If it is free it must be of rank one, because two elements of $a,b\in A$ are necessarily $A$-linearly dependent: $a\cdot b-b\cdot a=0$  (duh!)
Freeness of dimension one means that for some $m\in \mathfrak m$ the $A$-linear map $A\to \mathfrak m:a\mapsto am$ is bijective. Hence we get:  
Theorem:
The maximal ideal  of a commutative local ring is projective iff it can be generated by an element which is not a zero-divisor .  
Important  example:
If $(A,\mathfrak m)$ is a noetherian local domain, then  $\mathfrak m$ is projective iff $A$ is a discrete valuation ring.