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.
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.