I'm reading this PDF: Discrete Valuation Rings and Function Fields of Curves. I'm trying to understand in this theorem why $\mathfrak m/\mathfrak m^2$ is a $R/\mathfrak m$-module (see number 4 below).
I don't think $\mathfrak m/\mathfrak m^2$ is an ideal of $R/\mathfrak m$, it's not even a subset.
Thanks.
On
You're right, $\mathfrak{m}/\mathfrak{m}^2$ is not a subset of $R/\mathfrak{m}$. However, for any ring $A$, the definition of "$A$-module" makes no requirement that it be a subset of $A$, so your observation is irrelevant to the question.
Informally, an $A$-module $N$ is just an abelian group equipped with a notion of "multiplication by elements of $A$". That is, we should be able to add elements of $N$ together (e.g., $n_1+n_2$) and multiply an element of $N$ by an element of $A$ (e.g., $an$), and these operations should be compatible (e.g., $a(n_1+n_2)=an_1+an_2$).
We can add elements of $\mathfrak{m}/\mathfrak{m}^2$ together as usual; it is an abelian group quotiented by a subgroup. The addition operation on such cosets is $$(m_1+\mathfrak{m}^2)+(m_2+\mathfrak{m}^2)=(m_1+m_2)+\mathfrak{m^2}$$ Now you can check that this is a well-defined operation of multiplying an element of $\mathfrak{m}/\mathfrak{m}^2$ by an element of $R/\mathfrak{m}$ (i.e., it doesn't depend on the coset representatives $r$ and $m$ that are chosen): $$(r+\mathfrak{m})\cdot (m+\mathfrak{m}^2)=rm+\mathfrak{m}^2$$ This demonstrates that $\mathfrak{m}/\mathfrak{m}^2$ is an $R/\mathfrak{m}$-module.
Lemma. Let $I$ be an ideal of $R$ and let $M$ be an $R$-module such that $I$ annihilates $M$. Then $M$ is an $R/I$ module (See Here, for proof).
Note hat $m$ annihilates $m/m^2$.
All ideals are modules, but it is not true that all modules are ideals; The fact is that if a subset of $R$ is a module it will be an ideal.