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2 sketched alternative argument

Partial answer: If $M\otimes\kappa(\mathfrak p)$ is infinite-dimensional for some prime $\mathfrak p$ in $A$, then the functor is not representable.

Proof: We may assume that $A=k$ itself is a field and $M=k^{(I)}$ for some infinite set $I$. Suppose there is a representing scheme $X$ (so $X(R)=R^{(I)}$), and let $x$ denote the rational point corresponding to $0\in M$. For a filtered inductive system of rings $R_i$, we may calculate $\varinjlim R_i$ as a inductive limit of abelian groups, so it commutes with the direct sum $R_i^{(I)}$, therefore $X(\varinjlim R_i)=\varinjlim X(R_i)$. Now EGA IV, 8.14.2 tells us that $X$ is locally of finite type over $k$, in particular $\mathcal O_{X,x}$ has finite dimension.

(It is tempting to conclude by comparing the finite-dimensional Zariski tangent space with the liftings of $x\in X(k)$ to an $k[\epsilon]/(\epsilon^2)$-valued point, and these liftings are given by $(k\cdot\epsilon)^{(I)}\subset(k[\epsilon]/(\epsilon^2))^{(I)}$. However, I do not see how to show that the two $k$-vector space structures agree.)

There is an obvious monomorphism $X\to\mathbb A^I=\mathrm{Spec}(k[T_i\mid i\in I])$, so $X$ is separated. Choose an embedding $\mathbb N\to I$. For each $n$, we get a monomorphism $i_n\colon\mathbb A^n\to X$ which is a section to the projection $X\to\mathbb A^n$, so $i_n$ is a closed embedding, whose image contains $x$. This implies that $\mathcal O_{X,x}$ has quotients of arbitrarily large dimension, contradiction.

(EDIT) A different way to conclude is by comparing the finite-dimensional Zariski tangent space with the liftings of $x\in X(k)$ to an $k[\epsilon]/(\epsilon^2)$-valued point, and these liftings are given by $(k\cdot\epsilon)^{(I)}\subset(k[\epsilon]/(\epsilon^2))^{(I)}$. Use the monomorphism to $\mathbb A^I$ to deduce that the two vector space structures agree.

1

Partial answer: If $M\otimes\kappa(\mathfrak p)$ is infinite-dimensional for some prime $\mathfrak p$ in $A$, then the functor is not representable.

Proof: We may assume that $A=k$ itself is a field and $M=k^{(I)}$ for some infinite set $I$. Suppose there is a representing scheme $X$ (so $X(R)=R^{(I)}$), and let $x$ denote the rational point corresponding to $0\in M$. For a filtered inductive system of rings $R_i$, we may calculate $\varinjlim R_i$ as a inductive limit of abelian groups, so it commutes with the direct sum $R_i^{(I)}$, therefore $X(\varinjlim R_i)=\varinjlim X(R_i)$. Now EGA IV, 8.14.2 tells us that $X$ is locally of finite type over $k$, in particular $\mathcal O_{X,x}$ has finite dimension.

(It is tempting to conclude by comparing the finite-dimensional Zariski tangent space with the liftings of $x\in X(k)$ to an $k[\epsilon]/(\epsilon^2)$-valued point, and these liftings are given by $(k\cdot\epsilon)^{(I)}\subset(k[\epsilon]/(\epsilon^2))^{(I)}$. However, I do not see how to show that the two $k$-vector space structures agree.)

There is an obvious monomorphism $X\to\mathbb A^I=\mathrm{Spec}(k[T_i\mid i\in I])$, so $X$ is separated. Choose an embedding $\mathbb N\to I$. For each $n$, we get a monomorphism $i_n\colon\mathbb A^n\to X$ which is a section to the projection $X\to\mathbb A^n$, so $i_n$ is a closed embedding, whose image contains $x$. This implies that $\mathcal O_{X,x}$ has quotients of arbitrarily large dimension, contradiction.