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Let $M, N, P, Q$ be $R$-modules and $\varphi:M\to P$, $\psi:M\to Q$, $\phi:N\to P$ and $\lambda:N\to Q$ module homomorphisms. Consider

$$ M\oplus N \xrightarrow{f:=\left(\begin{matrix} \varphi & \phi \\ \psi & \lambda \\ \end{matrix}\right)} P\oplus Q $$ that is for $(a,b)\in M\oplus N$ we have

$$f(a, b) = \left(\begin{matrix} \varphi & \phi \\ \psi & \lambda \\ \end{matrix}\right)\left(\begin{matrix}a \\ b \end{matrix}\right) = \left(\begin{matrix}\varphi(a)+\phi(b) \\ \psi(a) +\lambda(b) \end{matrix}\right)$$

(Please excuse the slight abuse of notation here, just trying to make the matrix notation look neat) I'm trying to describe $\ker (f)$ in terms of the kernels of $\varphi$, $\psi$, $\phi$ and $\lambda$. I know that the kernel of

$$\left(\begin{matrix}\varphi & 0 \\0 & \lambda \\\end{matrix}\right)$$

is $\ker(\varphi)\oplus\ker(\lambda)$ since this is just the generic direct sum of maps. But is it possible to have a similar representation with this more general case? I can see that $$(\ker(\varphi)\cap\ker(\psi))\oplus(\ker(\phi)\cap\ker(\lambda))\subsetneq \ker(f)$$ but past that I'm unsure. Anyone have any ideas?

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There's not much else you can say in general, as the following example illustrates: Let $R = k$ be a field, and let $M = N = P = Q = k$, so that $\varphi, \psi, \phi, \lambda$ are just scalars and the matrix is literally a $2\times 2$ matrix with entries in $k$. If we know that all of the kernels of $\varphi, \psi, \phi, \lambda$ are zero, that means the scalars are nonzero—but what does a matrix having all nonzero entries tell us about its kernel? Really not much. We know the kernel of $f$ isn't all of $k^2$, and more generally we can say that $\ker(f)$ doesn't contain the standard basis vectors $(1, 0)$ or $(0, 1)$, but I can't see what more there is to say in general. The kernel could be trivial, or it could be any one-dimensional subspace other than the coordinate factors.

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  • $\begingroup$ Ah that's a pain, I was really hoping to be able to extract some more useful information out this scenario. I did suspect that it was maybe a dead end though. Thanks for your answer! $\endgroup$
    – SeraPhim
    Nov 10 '20 at 17:26

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