To be brief, here's my question: Suppose we have a ring $R$ (say commutative with unit) with an $R$-linear map $M\xrightarrow{f}N$ where $M,N$ are finitely generated projective (FGP) $R$-modules. Suppose that the categorical kernel of $f$ in the category of FGP $R$-modules exists: denote it by $P$. Letting $\ker{f}=\{m\in M: f(m)=0\}$ be the usual module-theoretic kernel, why must we have that $P\cong \ker{f}$ (or is this not true)?
For a bit more background: I am ultimately trying to find examples of (non-obvious) categories which fail to have kernels. After some reading, it looks like FGP modules over appropriate rings form a common example. From what I've read, the usual way of showing this is to produce a morphism for which the module-theoretic kernel fails to be projective or finitely generated (e.g. see mdp's great answer here).
There's one subtle piece of this method that I don't understand, which is why the categorical kernel of such a morphism must be (isomorphic to) the module-theoretic kernel.
Here's what I know:
- For any non-trivial commutative ring $R$, the category of $R$-modules, $R\textsf{-Mod}$, certainly has kernels, and the category of FGP $R$-modules is a full subcategory of $R\textsf{-Mod}$
- For any category ${\mathscr{C}}$ and full subcategory $\mathscr{D}$ of $\mathscr{C}$, a morphism $X\xrightarrow{f}Y$ of $\mathscr{D}$ can have a kernel in $\mathscr{D}$ but no kernel in $\mathscr{C}$. However, if $f$ has a kernel in $\mathscr{C}$ and that kernel is a morphism in $\mathscr{D}$, it is also the kernel in $\mathscr{D}$.
- If $\mathscr{C,D}$ are as before, there are examples where a morphism $X\xrightarrow{f}Y$ of $\mathscr{D}$ has a kernel in both $\mathscr{C}$ and $\mathscr{D}$, but these kernels are not isomorphic.
- If you denote the module theoretic kernel by $\text{ker}_R(f)$, and the categorical kernel of $f$ by $\text{ker}_{FGP}(f)$, there is a monomorphism $\text{ker}_{FGP}(f)\hookrightarrow\text{ker}_R(f)$ which commutes with the kernel morphisms, so you may replace $\text{ker}_{FGP}(f)$ by a submodule of $\text{ker}_R(f)$ if necessary.
With all this said, I think the proof (if it's true) has much more to do with the algebra at play, but I don't see how.
Any and every hint/bit of help is greatly appreciated, thank you!