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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?

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Dear user, if indeed $M^{r} \cong M^{s}$ implied $r=s$, how would you then define the rank of $M$ if $M$ is not free? (If $M$ is free then all is clear since you reduce to $M=R^n$) – Georges Elencwajg Apr 1 '12 at 9:08
Dear user: all right, I'll post an answer in the commutative case – Georges Elencwajg Apr 1 '12 at 9:28
Dear user: Take $M=0$. – Pierre-Yves Gaillard Apr 1 '12 at 10:59
up vote 2 down vote accepted

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 $r = s$.

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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 $r=s$.

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 $\phi(\sum q_g \cdot g)=\sum q_g$.

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Dear Georges, Two minor points: you are implicitly assuming that $M$ is f.g., and that $M$ is non-zero. Best wishes, – Matt E Apr 1 '12 at 11:37
Dear @Matt, you are absolutely right and have outbourbakied me! I have now edited my answer. Thanks for your interest. – Georges Elencwajg Apr 1 '12 at 12:29

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