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I have to do the problem 2.50, I assume the problem 2.30 done.

So I did the following:

b) I know that (t)=m that is equal to ${t,t^2,...}$ is the generator so I have that $R/m^n={u(1,t,t^2,...)/u(t^n,...)}$ where u is the unity, so $R/m^n$ has n elements so its dimension is n.

a) $m^n/m^{n+1}={(t^n,t^{n+1},...)/(t^{n+1},t^{n+2}...)}$ so $m^n/m^{n+1}=t^n$ that's why their dim is 1

c) Since $R/(z)=R/m^n=n$ by part b) so if the ord(z)=n then $dimR/(z)=ord(z)$

I have this but I'm not sure if that is right, and I think I miss something because the dimension is with k, I appreciate your help.

Kc2
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1 Answers1

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It is easier to prove first $a)$ and then $b)$.

For $a)$ you can write down an explicit isomorphism, namely multiplication by the uniformizer $t$ (raised to the right power).

\begin{align} R &\to m^{n}/m^{n+1} \\ x &\mapsto xt^{n} \end{align}

This map is surjective and the kernel is $m$.

For $b)$ use the obvious filtration of $R/m^{n}$ with $k$-vector subspaces $$ 0\subseteq m^{n-1}/m^{n}\subseteq \ldots \subseteq m/m^{n} \subseteq R/m^{n} $$ Since the quotients have dimension $1$ over $k$ by part $a)$, the dimension of $R/m^{n}$ over $k$ is $n$ (this is standard linear algebra).

And then use $b)$ to conclude that $$ \operatorname{ord}(z)=\operatorname{dim}_{k}(R/(z)) $$ which is (I would) say the goal of the exercise, since the title "algebraic curves" (whose local rings are in fact DVR's when they are smooth) suggests that $\operatorname{ord}(z)$ represents the vanishing/pole order of some divisor on the curve defined locally by some rational function.

Pedro
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  • Why is $R/m$ 1 dimensional? – Not Euler Mar 12 '20 at 14:18
  • The $k$-vector space structure on $R/m$ is induced by the ring isomorphism written in exercise 2.30, so $k$ and $R/m$ are also isomorphic as $k$-vector spaces – Pedro Mar 12 '20 at 15:02
  • Oh I see. Then do you know why it is true if local ring of an irreducible curve at a point happens to be a DVR? – Not Euler Mar 12 '20 at 16:10
  • @HritRoy it has to be a regular curve. Intuitively, a single parameter is enough to parametrize the curve around each point (this doesn't happen on a node, say). Formally, an irreducible variety is regular at a point if the corresponsing maximal ideal can be generated by as few elements as its dimension. But a local noetherian domain which is not a field and in which the maximal ideal is principal is a DVR (this is one of the many equivalent definitions you can find on Wikipedia) – Pedro Mar 12 '20 at 16:21