Rank of Noetherian modules Suppose $M$ is a Noetherian left-module, why is the rank of $M$ unique? that is if $M^{r} \cong M^{s}$ as left modules then why $r=s$? Is this true if $M$ is Artinian?
 A: The property you ask about is true if  the base ring $R$ is commutative.
Indeed,  if $\mathfrak m\subset R$ is maximal, consider the field $k=R/ \mathfrak m$.
If $M$ is a finitely generated non-zero $R$ module, then from $M^r\cong M^s$ , you deduce that $(M\otimes_R k)^r=M^r\otimes_R k\cong M^s\otimes_R k=(M\otimes_R k)^s$.
Since $M\otimes_R k$ is finite-dimensional, the isomorphism  $(M\otimes_R k)^r\cong (M\otimes_R k)^s$ forces $r=s$.    
Generalization
The above proof generalizes to the case where there exists  a ring morphism $\phi: R\to k$, where $k$ is a commutative field.
For example this is the case when $R=k[G]$ is the group ring over an arbitrary group $G$ and $\phi(\sum q_g \cdot g)=\sum q_g$.
A: For the general (not necessarily commutative case), the key concept seems to be that of Hopfian module, a concept that has appeared several times before on this forum (although I don't have time to find the links now).
Label the numbers $r$ and $s$ so that $r > s$.
Since $M$ is Noetherian, so is $M^s$; thus it is also Hopfian (see the above link). The composite $M^s \cong M^r \to M^s$, in which the first arrow is the presumed isomorphism, and the second arrow is projection onto the first $s$ places, is a surjective endomorphism of $M^s$, which is thus necessarily injective, since $M^s$ is Hopfian.  The only way this can happen is if $M = 0$, or else  if $r = s$.
Thus if $M^r \cong M^s$, then either $M = 0$ or $r = s$.  
