Extensions of $M$ by $M$ with same derivation

Question

If $M$ is a $A$-module where $A$ is a complete noetherian local ring, for any extension of $A$-modules $0 \rightarrow M \rightarrow X \rightarrow M \rightarrow 0$, we get an object $X$ with a derivation $d:X \rightarrow M \rightarrow X$ ($d$ is the composition of the maps in the exact sequence in the opposite order, so $d\circ d=0$ which is the definition of a differential). If two extensions give isomorphic $(X,d)$, must they be equivalent?

Answer 1

No. For instance, let $A=\mathbb{Z}/(4)$ and $M=\mathbb{Z}/(2)\oplus\mathbb{Z}/(2)$. Let $X=\mathbb{Z}/(2)\oplus\mathbb{Z}/(2)\oplus\mathbb{Z}/(4)$ be considered as an extension of $M$ by itself in two different ways: one where the $\mathbb{Z}/(4)$ is obtained from the first summand in each copy of $M$, and one where the $\mathbb{Z}/(4)$ is obtained from the second suumand in each copy of $M$. These two extension induce the same differential $X\to X$ (namely multiplication by $2$). However, they are not isomorphic as extensions of $M$ by $M$. (For instance, in the first extension the image of the second summand of $M$ under the inclusion $M\to X$ is a direct summand of $X$, but in the second extension it is not.)

More generally, if you start with an extension $0\to M\to X\to M\to 0$ with differential $d$ and apply an automorphism of $M$ to get a new extension $0\to M\to X'\to M\to 0$ with differential $d'$, then obviously $(X,d)\cong (X',d')$, but the extensions may not be isomorphic. In fact, every example arises in this way: given an isomorphism between $(X,d)$ and $(X',d')$, we get an isomorphism between the short exact sequences $0\to M\to X \to M\to 0$ and $0\to M\to X'\to M\to 0$ since the inclusion $M\to X$ is just a kernel of $d$ and the quotient $X\to M$ is just a cokernel of $d$, and similarly for $d'$. Moreover, the two automorphisms $M\to M$ induced on the kernel and cokernel are equal, since it must commute with the isomorphisms $\operatorname{coker}(d)\to \ker(d)$ and $\operatorname{coker}(d')\to \ker(d')$ induced by $d$ and $d'$. This means exactly $0\to M\to X'\to M\to 0$ is isomorphic to the extension $0\to M\to X\to M\to 0$ obtained by modifying our first extension by this automorphism of $M$.