Is $m$ a projective $A$-module? 
$A$ 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?
 A: It depends on the ring. For instance, if $R=\mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime ideal $p$, then the maximal ideal is principal, so isomorphic as a module to the ring itself, hence projective.
In case the ring is $\mathbb{Z}/4\mathbb{Z}$, the maximal ideal is not projective.

Added from comment.
If your aim is to discuss $\def\Tor{\operatorname{Tor}}\Tor^A_1(A/m,M)$, then projectivity of $m$ is not involved. Given any $A$-module $M$, from the exact sequence
$$
0\to m\to A\to A/m\to0
$$
we get the following final fragment of the long exact sequence obtained by applying $-\otimes_AM$:
$$
\Tor^A_1(A,M)\to\Tor^A_1(A/m)\to m\otimes_AM\to A\otimes_AM\to
A/m\otimes_AM\to0
$$
Since $A$ is projective (hence flat), $\Tor^A_1(A,M)=0$, so we can state that

$\Tor^A_1(A/m,M)=0$ if and only if $m\otimes_AM\to A\otimes_AM$ is injective.

Projectivity of the maximal ideal $m$ is not involved here and neither is the fact that $A$ is a local ring. This holds for any ring and any (right) ideal $m$ of $A$.
Of course, if you know from other results that $m\otimes_AM\to A\otimes_AM$ is injective in your specific case, then you have as a consequence that $\Tor^A_1(A/m,M)=0$.
A: 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.  
