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Corollary 3.10. Let $A$ be a local ring with maximal ideal $\mathfrak{m}$ and residue field $k:=A/\mathfrak{m}$, and let $M$ be a finitely generated module over $A$. The action of $A$ on $M/\mathfrak{m}M$ factors through $k$, and elements $a_1,...,a_n$ of $M$ generate it as an $A$-module if and only if the elements $$a_1+\mathfrak{m}M,...,a_n+\mathfrak{m}M$$ generate $M/\mathfrak{m}M$ as $k$-vector space.

In the corollary 3.10 of this book, 'The action of $A$ on $M/\mathfrak{m}M$ factors through $k$' so they view $M/\mathfrak{m}M$ as a vector space over $k$. Then how does the operation defined? Is it $(a+\mathfrak{m})(x+\mathfrak{m}M) = ax+\mathfrak{m}M$? And in general, if we have such form (some given action of a ring of given module factors through some field obtained by quotient), then can we always view such module as a vector space over that field? And is the operation defined as I wrote above?

one potato two potato
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    Please avoid images: they are not searchable, most screen readers cannot parse them, and they are not accessible. – Arturo Magidin Feb 12 '21 at 04:13
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    @ArturoMagidin should I type in LaTeX and reference the book? – one potato two potato Feb 12 '21 at 04:14
  • You should avoid images (or make them accessible... which turns into typing out what they contain). In any case, when you are refering to a book it is good practice to actually say what book you are refering to, and where. – Arturo Magidin Feb 12 '21 at 04:25
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    Yes, that’s the action; you should verify that it is indeed well-defined. A module over a field is exactly a vector space. – Arturo Magidin Feb 12 '21 at 04:25

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The action of $A$ on $M/\mathfrak mM$ is the ring homomorphism $\mu : A \to \operatorname{End}_{\textsf{Ab}}(M/\mathfrak mM)$ such that for each $a \in A$, $\mu(a) \in \operatorname{End}_{\textsf{Ab}}(M/\mathfrak mM)$ is left-multiplication by $a$.

Note that if $a \in \mathfrak{m}$, then $am \in \mathfrak mM$ for every $m \in M$, and so $$ (\forall m \in M) \quad \mu(a)(m+\mathfrak mM) = a(m+\mathfrak mM) = am+\mathfrak mM = \mathfrak mM, $$ that is, $\mu(a)$ is the trivial map in $\operatorname{End}_{\textsf{Ab}}(M/\mathfrak mM)$, which means that $a \in \ker \mu$.

Thus $\mathfrak m \subseteq \ker \mu$, and then the ring homomorphism $\tilde\mu : A/\mathfrak m \to \operatorname{End}_{\textsf{Ab}}(M/\mathfrak mM)$ given by $\tilde\mu(a+\mathfrak m) = \mu(a)$ is well-defined. Moreover, it is precisely the action that you proposed.

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