Suppose that we have a field $p^k$, and we want to express values in this field modulo smaller fields $(q_1)^k$, $(q_2)^k$,... I believe that there is a way to do this using the Chinese remainder theorem, if we can find a polynomial that is irreducible modulo $p^k$, and completely reducible modulo the smaller fields.
My question is, can we do this, and how small can we make the largest prime $q^k$, if we know what $p$ and $k$ are?
Although I am not sure what you mean by "shrinking" a field, it seems like you mean some sort of (quotient?) map from a field to a "smaller" field.
In fact, this isn't really possible for any sensible notion of "being smaller". Recall that in set theory, we can define a notion of cardinality in the following way: For $M, N$ sets, $|M| \leqslant |N|$ means that there exists an injective map $M \rightarrow N$. Then, $|M| < |N|$ means that there does not exist an injective map $N \rightarrow M$.
Now, let $F, G$ be fields and $\varphi : F \rightarrow G$ be a field homomorphism. $\varphi$ is necessarily injective! To see this, consider $\ker \varphi$. If there is a non-zero element $a \in \ker \varphi$, then $1 = \varphi(1) = \varphi(aa^{-1}) = \varphi(a)\varphi(a^{-1}) = 0$. Thus, the kernel needs to be trivial, and $\varphi$ is injective. (Alternatively, you can see this by considering that $\ker \varphi$ needs to be an ideal of $F$ as a ring, and fields only have $F, (0)$ as ideals).
Putting this result together with our notion of "being smaller", this means that if there exists a mapping $F \rightarrow G$, we always have $|F| \leqslant |G|$.