# Kernel of direct sum of module homomorphisms

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?

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.