Let $\mathbb{F}$ be some finite field, and let $R := M_n(\mathbb{F})$ be the set of $n$-by-$n$ matrices over $\mathbb{F}$. Then $R$ is finite. Does there exist some pair $(\varphi, S)$ such that $S$ is a ring with fewer elements than $R$ (but more than 1 element) and $\varphi : R \rightarrow S$ is a ring homomorphism?
The motivation for this question is practical rather than mathematical – the element $\varphi(A)$ would function like a MAC (message authentication code) for $A$. Thus I would want it to have a smaller representation than $A$ (in terms of bits) but not so small that one could just "guess" it. A rng homomorphism would also be sufficient, i.e. the requirement $\varphi(1_R) = 1_S$ is not important, what matters is that $\varphi$ is homomorphic wrt both addition and multiplication.
But I assume the answer is negative either way, because if $R$ were a field, then $\varphi$ would have to be injective, and $R$ seems like it's "mostly" a field, having more invertible than non-invertible elements. However, I haven't managed to turn the idea into a proof. The trace is an additive map, the determinant is a multiplicative map, but netiher is both, and in fact $\mathbb{F}$ seems like it cannot be the target set at all, having no nilpotent elements except 0.
Best Answer
The answer to both questions is no.
First, there can be no (unit-preserving) ring homomorphism from $R = M_{n}(\Bbb F)$ to a smaller ring $S$. We note that for any ring-hom $\varphi:R \to S$, $\ker\varphi$ must form a two-sided ideal in $R$. However, $R = M_{n}(\Bbb F)$ is a simple ring: its only two-sided ideals are $R$ and $\{0\}$. Thus, the image of $\varphi$ must be isomorphic either to $R/\{0\} \cong R$, or to $R/R \cong \{0\}$.
This also rules out the possibility of a rng-homomorphism. Note that we can always restrict the codomain so that $\varphi:R \to S$ is surjective. We note that for any $s \in S$, there exists an $r \in R$ with $\varphi(r) = s$ and we have $$ \varphi(1_R) \, s = \varphi(1_R) \varphi(r) = \varphi(r) = s\\ s\,\varphi(1_R) = \varphi(r)\varphi(1_R) = \varphi(r) = s. $$ That is, $S$ must have identity element $\varphi(1)$, which is to say that any onto rng-homomorphism from $R$ to $S$ is necessarily identity-preserving.