This is the followed-up from this question Show $E \cong E'$ as R-modules given $ h : N \to N'$ isomorphism, $\mathscr E : 0 \to N \to E \to M \to 0$ and its pushout which was about (a) and might or not be useful for this question
My question is about (b)
Let $R$ be a ring, and consider an extension $\mathscr E : 0 \to N \to E \to M \to 0$ of $R$-modules.
If $h : N \to N'$ is a homomorphism, we can form the pushout
$h_*(\mathscr E ) : 0 \to N' \to E' \to M \to 0 $.
On the other hand, $h$ induces a homomorphism $h_* : Ext^1_R(M, N) \to Ext^1_R(M, N')$.
Under this homomorphism, the class $[\mathscr E ] \in Ext^1_R(M, N)$ is mapped to the class $[h_*(\mathscr E)] \in Ext^1_R(M, N')$.
a)In the above situation, if $ h : N \to N'$ is an isomorphism, one can show that $E \cong E'$ as R-modules.
b) Take $R = k[t]$ where $k$ is a field. For $\lambda \in k$, let $M_{\lambda} = k[t]/(t -\lambda )$ and $M'_\lambda = k[t]/(t − \lambda )^2$. If we describe $R$-modules as pairs $(V, \phi )$ then $M_\lambda$ corresponds with $V = k$ with $\phi = \lambda \cdot id_k$, and $M'_\lambda$ corresponds with $V = k^2$ with $\phi = \begin{pmatrix} \lambda& 1 \\ 0 & \lambda\end{pmatrix}$
By what we have seen in Example 3.3.11 (see below) , Ext$^1_{k[t]}(M_\lambda , M_\lambda ) \cong M_\lambda$ .
Let $E$ be a $k[t]$-module that can be obtained as an extension of $M_\lambda$ by itself.
Show that either $E \cong M_\lambda \oplus M_\lambda$ or $E \cong M'_\lambda $.
I am overwhelmed with all the information given in (b), how should I start ?
Just in case:
source: https://docdro.id/IarlBl1
To be perfectly honest, I'm not sure what's going on with the matrices in $(b)$. It is true that $M_\lambda$ is isomorphic to $k$ with the module action given by $f(t)\cdot x:=f(\lambda)x$, and it is also true that $M'_\lambda$ is isomorphic to $k^2$ with the module action given by $f(t)\cdot(x,y):=(f(\lambda)x,f(\lambda)y+f'(\lambda)x)$ where we chose the basis $\{1,t-\lambda\}$ and consider we can always perform a division: $f(t)=f(\lambda)+f'(\lambda)(t-\lambda)+(t-\lambda)^2r(t)$. How that is captured by the notation $\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$, I couldn't tell you. Observe the module structure restricts to the standard $k$-vector space structures on $k,k^2$.
The question is essentially suggesting the only nonsplit extension classes $\xi:0\to M_\lambda\to E\to M_\lambda\to0$ satisfy $E\cong M'_\lambda$. To verify that, I think the best strategy is to create a candidate extension $0\to M_\lambda\to M'_\lambda\to M_\lambda\to0$ and then verify that, by tweaking the maps in the short exact sequence, we can obtain any element of the extension module $\mathsf{Ext}^1_{k[t]}(M_\lambda,M_\lambda)\cong M_\lambda$. Our life is made easier by the observation $M_\lambda$ is basically $1$-dimensional; if we find just one nonsplit extension using $M'_\lambda$ then by scaling we should be able to obtain all extension classes.
Well, of course $(t-\lambda)$ projects to an ideal of $M'_\lambda$ and we obtain a quotient $M'_\lambda/\overline{(t-\lambda)}\cong M_\lambda$. The kernel $(t-\lambda)M'_\lambda$ is easily isomorphic to $M_\lambda$; specifically we take the map $M_\lambda\hookrightarrow M'_\lambda$ which sends $x\mapsto(t-\lambda)x\sim(0,x)$. It's easy to see by my description above that this is a module morphism. We can also check that this doesn't split; a section $F:M'_\lambda\to M_\lambda$ would be determined by $a:=F(1,0),b:=F(0,1)$ and we require $f(\lambda)a+f'(\lambda)b=F(f(t)\cdot(1,0))=f(\lambda)a$ for all $f$, obviously forcing $b=0$; but then $x=F(0,x)=xb=0$ for all $x\in M_\lambda$ is forced so $F$ cannot be a section.
The last thing to do is show that we can modify the maps in our canonical $\Xi:0\to M_\lambda\to M'_\lambda\to M_\lambda\to0$, making a new extension $\xi$, to ensure the connecting homomorphisms are appropriately scaled; $\partial_{\xi}=z\cdot\partial_{\Xi}$, $z\in k^{\times}$ arbitrary. One can directly compute with the horseshoe lemma construction that adjusting the map $M_\lambda\to M'_\lambda$ so that it maps $x\mapsto(0,-z^{-1}x)$ will cause the connecting homomorphism to be scaled by $z$. I strongly recommend you check this for yourself (and feel free to disagree with me in case I've made a mistake).