Let $A$ be a local (not necessarily noetherian) ring with maximal ideal $\mathfrak{m}$ and residue field $k$. Let $M$ be a finitely generated $A$-module such that $\mathfrak{m}\otimes_A M\rightarrow A\otimes_A M = M$ is injective. Let $\varphi:A^n\rightarrow M$ be a surjection.
Is $i\otimes \text{id.}:\mathfrak{m}\otimes_A \ker\varphi\rightarrow A\otimes_A\ker\varphi$ (necessarily) injective?
Here $i:\mathfrak{m}\rightarrow A$ is the embedding.
In the case that (i) $A$ is noetherian and (ii) $n$ is the dimension of $M\otimes k$ as a $k$-vector space, where $k=A/\mathfrak{m}$, I have convinced myself that this is true and in fact $\ker \varphi = \mathfrak{m}\otimes \ker\varphi = 0$. Argument below. The argument does not compel me to think any such thing if I drop either assumption, but I have neither an example to show it is possible for it to fail to be injective, nor enough of an intuition about homological algebra (yet) to have a sense of whether to expect that there is some other argument that shows it must be injective. Can you show me a counterexample to injectivity or a proof of injectivity?
Thanks in advance!
Argument in case that $A$ is notherian and $\dim_k M\otimes k = n$:
There is a commutative diagram
$$\require{AMScd} \begin{CD} 0@>>> \mathfrak{m}\otimes \ker\varphi @>j'>> \mathfrak{m}\otimes A^n@>\varphi'>> \mathfrak{m}\otimes M@>>> 0\\ @. @VaVV @VbVV @VcVV @.\\ 0@>>> A\otimes \ker\varphi @>>j> A\otimes A^n @>>\varphi> A\otimes M@>>> 0 \end{CD} $$ where the bottom row is exact and the top row is too except possibly at $\mathfrak{m}\otimes \ker\varphi$. $b$ is an injection because $A^n$ is free and therefore flat over $A$. Since $j$ is also an injection and $b\circ j' = j\circ a$, $j'$ and $a$ must have the same kernel. Thus they both factor though the quotient of $\mathfrak{m}\otimes \ker\varphi$ by this kernel, call this quotient $K$, and we thus have a commutative diagram $$\require{AMScd} \begin{CD} 0@>>> K @>\bar j'>> \mathfrak{m}\otimes A^n@>\varphi'>> \mathfrak{m}\otimes M@>>> 0\\ @. @V\bar aVV @VbVV @VcVV @.\\ 0@>>> A\otimes \ker\varphi @>>j> A\otimes A^n @>>\varphi> A\otimes M@>>> 0 \end{CD} $$ where now both rows are exact and the images of $\bar a$ and $a$ are equal. Because both rows are exact, the snake lemma applies, and we have an exact sequence $0\rightarrow\ker \bar a\rightarrow\ker b\rightarrow \ker c\rightarrow \operatorname{coker} \bar a\rightarrow\operatorname{coker} b\rightarrow \operatorname{coker} c\rightarrow 0$. The cokernels are the tensors with $k$ (this is where I use that $\bar a$ has the same image as $a$), and the assumption about the injectivity of $\mathfrak{m}\otimes M\rightarrow A\otimes M$ is the statement that $\ker c$ is zero. Thus the sequence is
$$\dots\rightarrow 0\rightarrow k\otimes \ker\varphi \rightarrow k\otimes A^n\rightarrow k\otimes M\rightarrow 0$$
and because $k\otimes A^n\rightarrow k\otimes M$ is a surjection between vector spaces of the same dimension, we must have $k\otimes\ker\varphi = \ker\varphi/\mathfrak{m}\ker\varphi = 0$. Since $A$ is noetherian, $\ker\varphi$ is finitely generated, so Nakayama's lemma applies to prove $\ker\varphi = 0$. Therefore $\mathfrak{m}\otimes\ker\varphi = K = \ker\varphi = 0$ after all, and a map between zero modules is injective.
You can see why this argument made me want to ask this question: it constructs $K$ to get around the not-necessarily-exact-ness of the first row in the first diagram above, but then in the end, that row was exact after all. So I want to know if (dropping the conditions that $A$ is noetherian and $\dim_k M\otimes k = n$) it can ever be non-exact, given the key assumption that $c$ is injective. (The counterexample to exactness I came up with is also a counterexample to $c$ being injective.)
Addendum:
Jake Levinson points out below that the snake lemma does not require exactness of the left term in the top row. Therefore the construction of $K$ in the above proof is superfluous: one can apply the snake lemma to the first diagram.
For some reason I don't quite have my finger on, this makes me suspect that the answer to my question is no. I'm still hoping for an example.