Let $R$ be a Discrete Valuation Ring with unique maximal ideal $M$. Show that $M^n/M^{n+1}$ is a vector space.

Question

This is a question from Dummit and Foote Chapter 16:

"Suppose $R$ is a Discrete Valuation Ring with unique maximal ideal $M$ and quotient $F=R/M$. For any $n\geq 0$, show that $M^n/M^{n+1}$ is a vector space over $F$ and that $\dim_F(M^n/M^{n+1})=1$."

I think this question should be straight-forward, but I am having a hard time picturing what $F=R/M$ looks like. In general, I am also wondering if I have to prove all 8 vector axioms, or if there is another way to approach this question.

Answer 1

With the right level of abstraction, it takes only one axiom to define vector space:

A vector space is an abelian group on which a field acts.

Let's check: $M^n$ is an ideal, hence in particular an abelian group. $M^{n+1}$ is a subgroup thereof, hence $M^n/M^{n+1}$ is an abelian group. $F$ is a field because $M$ is maximal. $R$ acts on $M^n$ and $M^{n+1}$ (or any ideal) by left multiplication, hence also on $M^n/M^{n+1}$. Left multiplication by an element of $M$ takes $M^n$ to $M^{n+1}$, i.e., acts as $0$ on $M^n/M^{n+1}$. Thus we obtain an action of $F$, as desired.

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The dimension claim is clear for $n=0$. Can you see how to do induction?

Answer 2

Pick some element $a$ of order $n$, then the map $$m_x:R/\mathfrak{m}\to \mathfrak{m}^{n}/\mathfrak{m}^{n+1}\qquad x\bmod \mathfrak{m}\mapsto ax\bmod \mathfrak{m}^n$$ is an $R$-isomorphism (thus an $R/\mathfrak{m}$-isomorphism).