I've been trying to prove the following result:
Let $(R,\mathfrak{m})$ and $(S,\mathfrak{n})$ be (Noetherian) local rings, and let $\varphi:R\to S$ be a finite local homomorphism.
Then $\varphi(\mathfrak{m})S$ is $\mathfrak{n}$-primary.
by following the suggestion given in this comment on a similar question. However I seem to have ended up proving that it is equal to $\mathfrak{n}$ itself: Since $\varphi$ is finite, we have that $S=\varphi(R)s_1+\cdots+\varphi(R)s_n$ for some $s_i\in S$. Then we can define a surjection $\psi:\oplus_n R/\mathfrak{m}\twoheadrightarrow S/\varphi(\mathfrak{m})S$ given via $r_i+\mathfrak{m}\mapsto\varphi(r_i)s_i+\varphi(\mathfrak{m})S$, where $r_i$ is the $i$th coordinate of $r\in\oplus_nR/\mathfrak{m}$. Then $S/\varphi(\mathfrak{m})S\cong(\oplus_nR/\mathfrak{m})/\ker(\psi)$. But ideals of $\oplus_n R/\mathfrak{m}$ are of the form $\oplus_m R/\mathfrak{m}$ for some $m\leq n$, replacing the other copies with $0$, so $S/\varphi(\mathfrak{m})S\cong\oplus_{n-m}(R/\mathfrak{m})$. This will have $n-m$ maximal ideals, given by replacing one copy of $R/\mathfrak{m}$ with $0$. But $S/\varphi(\mathfrak{m})S$ is a local ring with maximal ideal $\mathfrak{n}/\varphi(\mathfrak{m})S$. We can't have $S/\varphi(\mathfrak{m})S=0$ since then $S/\varphi(\mathfrak{m})S=S$, but $\varphi$ is local so $\varphi(\mathfrak{m})S\subseteq\mathfrak{n}$, which cannot contain $1$. Then we must have $n-m=1$, so $S/\varphi(\mathfrak{m})S\cong R/\mathfrak{m}$. This is a field, so $\varphi(\mathfrak{m})S$ is maximal and therefore equal to $\mathfrak{n}$. I may be missing something obvious, but I can't see where I have made a mistake here. I would much appreciate any help in finding it, and seeing how to conclude the proof correctly.
Update: As pointed out by Angina in the comments, my error is that the function $\psi$ I define is only a module, and not a ring, homomorphism.
I have left my incorrect attempt up if anybody wants to click to see it, but I would much appreciate if anyone has a reference for, or proof of, the result itself.
Note: I originally forgot to specify that $R$ and $S$ are Noetherian, and this seems to be a necessary condition to guarantee that $\text{Ass}(S/\varphi(\mathfrak{m})S)\neq\varnothing$. Apologies for this, in the definitions of the notes I am following they require that local rings also be Noetherian, and so I neglected to state this explicitly.
However, assuming this, I believe that the following gives a proof of the result, following a suggestion in this comment on another question.
Since $\varphi$ is finite, we have that $S=\varphi(R)s_1+\cdots+\varphi(R)s_n$ for some $s_i\in S$. Furthermore, $S$ is an $R$-module under the action $r\cdot s\mapsto\varphi(r)s$.
Let $T=S/\varphi(\mathfrak{m})S$, and note that this is also a local ring with maximal ideal $\mathfrak{n}/\varphi(\mathfrak{m})S$.
We cannot have $T=0$ since we would have $S=\varphi(\mathfrak{m})S$, but $\varphi$ is local so $\varphi(\mathfrak{m})S\subseteq\mathfrak{n}$ which cannot contain $1$ since it is maximal and therefore proper.
We define a surjective $R$-module homomorphism $\psi:\oplus_n R/\mathfrak{m}\twoheadrightarrow T$ given via $r_i+\mathfrak{m}\mapsto\varphi(r_i)s_i+\varphi(\mathfrak{m})S$, where $r_i$ is the $i$th coordinate of $r\in\oplus_nR/\mathfrak{m}$.
Then $T\cong(\oplus_nR/\mathfrak{m})/\ker(\psi)$. Since $T$ is killed under this action by $\mathfrak{m}$, it has a compatible $R/\mathfrak{m}$-module structure, and so this is also an isomorphism of $R/\mathfrak{m}$-modules.
Now, $R/\mathfrak{m}$-submodules of $\oplus_n R/\mathfrak{m}$ are of the form $\oplus_m R/\mathfrak{m}$ for some $m\leq n$, replacing the other copies with $0$, so $T\cong\oplus_{n-m}(R/\mathfrak{m})$.
Let $I$ be an ideal of $T$. Then it is certainly closed under addition, and for any $r+\mathfrak{m}\in R/\mathfrak{m}$, $x+\varphi(\mathfrak{m})S\in I$, we have
$$(r+\mathfrak{m})\cdot(x+\varphi(\mathfrak{m})S)=\varphi(r)x+\varphi(\mathfrak{m})S\in I$$
Then $I$ is also an $R/\mathfrak{m}$-submodule of $T$. Since $T$ has only finitely many $R/\mathfrak{m}$-submodules, we see then that $T$ has only finitely many ideals also.
Let $\mathfrak{p}\in\text{Spec}(T)$, and suppose that $a\notin\mathfrak{p}$. The chain
$$aT\supseteq a^2T\supseteq a^3T\supseteq\cdots$$
must stabilise since $T$ has only finitely many ideals, and so $a^lT=a^{l+1}T$ for some $l\geq1$. Then there exists some $t\in T$ such that $a^l=a^{l+1}t$, so $a^l(1-at)=0\in\mathfrak{p}$. We know that $a\notin\mathfrak{p}$, and so $a^l\notin\mathfrak{p}$ since $\mathfrak{p}$ is prime. Then we must have $1-at\in\mathfrak{p}$, again since $\mathfrak{p}$ is prime, so $T=\mathfrak{p}+aT$. Then $\mathfrak{p}$ must be maximal, and so all prime ideals of $T$ are maximal. Then since $T$ is local, its only prime ideal is its maximal ideal.
This means that $\mathfrak{n}$ is the only prime ideal of $S$ which contains $\varphi(\mathfrak{m})S$. We know that $\text{Ass}(S/\varphi(\mathfrak{m})S)\neq\varnothing$ since $S$ is Noetherian, and clearly $\varphi(\mathfrak{m})S(x+\varphi(\mathfrak{m})S)=0$ for any $x+\varphi(\mathfrak{m})S\in S/\varphi(\mathfrak{m})S$, so any element of $\text{Ass}(S/\varphi(\mathfrak{m})S)$ must contain $\varphi(\mathfrak{m})S$.
Then we must have $\text{Ass}(S/\varphi(\mathfrak{m})S)=\{\mathfrak{n}\}$, and so $\varphi(\mathfrak{m})S$ is $\mathfrak{n}$-primary.