I've found issues with questions in my algebra textbook before, and I believe that I've found another erroneous question.
L. E. Sigler Algebra Chapter 2 "Rings: Basic Theory" Section 7 "Morphisms and Quotient Rings" Exercise 4:
Let $m$ be different from $0$ and $1$ and belong to $\mathbb{N}$. Show that $f: \mathbb{Z} \to \mathbb{Z}_n$ such that $f(x) = mx$ is a morphism if and only if $n = m(m-1)$.
As a first thing, I think that $f(x) = \overline{mx}$ or $f(x) = mx + n\mathbb{Z}$ is what the textbook meant. So, with that adjustment, I think that the forward direction of the implication is false.
A counterexample to the contrapositive of the forward implication can be found with $m = 9$ and $n = 12 \neq 9(9-1)$. Let $x,y \in \mathbb{Z}$. Then $$ f(x + y) = 9(x + y) + 12\mathbb{Z} = 9x + 9y + 12\mathbb{Z} = (9x + 12\mathbb{Z}) + (9y + 12\mathbb{Z}) = f(x) + f(y) $$
$$ f(xy) = 9xy + 12\mathbb{Z} = (9 + 12\mathbb{Z}) (xy + 12\mathbb{Z}) = (9 + 12 \cdot 6 + 12\mathbb{Z}) (xy + 12\mathbb{Z}) = (81 + 12\mathbb{Z}) (xy + 12\mathbb{Z}) = (9^2 + 12\mathbb{Z}) (xy + 12\mathbb{Z}) = 9^2xy + 12\mathbb{Z} = (9x + 12\mathbb{Z}) (9y + 12\mathbb{Z}) = f(x)f(y). $$
So $f: \mathbb{Z} \to \mathbb{Z}_{12}$ such that $f(x) = \overline{9x}$ is a morphism of rings, counter to what the textbook claims.
My questions are: Did I do anything wrong? If not, how can this question be edited into something which works?
Indeed, if $1 \mapsto m$ (I neglect overlines), then $ab \mapsto abm$ on the one hand, and on the other this has to equal $(am)(bm) = abm^2.$ So you would need $m^2 = m$ in the domain. This is what you try to show in the proof: $9^2$ is really equal to $9$ modulo $12$: their difference is $72,$ which is divisible by $12$ indeed.
What does $m^2 \equiv m \text{ mod} (n)$ mean? This means that $m^2 - m = m(m-1)$ is divisible by $n$, so it's equal to $kn$ for some $k$ rather than just $n$.