Sum of commuting nilpotent elements

Detailed Solution

Since $a$ and $b$ commute, the binomial theorem applies: $(a+b)^{m+n-1}=\sum_{k=0}^{m+n-1}\binom{m+n-1}{k}a^k b^{m+n-1-k}$. For every term, either $k\ge m$ (making $a^k=0$) or $k\le m-1$, which forces the exponent of $b$ to satisfy $m+n-1-k\ge n$ (making $b^{m+n-1-k}=0$). So every term vanishes, giving $(a+b)^{m+n-1}=0$ — proving (A). This argument relies essentially on $ab=ba$ to justify the binomial expansion; without commutativity, sums of nilpotents need not be nilpotent in general noncommutative rings, so (B) is false. Since $a,b$ commute, $(ab)^{\max(m,n)}=a^{\max(m,n)}b^{\max(m,n)}=0$, so $ab$ is always nilpotent here, ruling out (C). There is no general reason $a-b$ must be a unit (e.g. taking $a=b$ gives $a-b=0$, never a unit in a nontrivial ring), ruling out (D). Thus the correct option is $\boxed{\mathrm{A}}$.