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I'm currently trying to understand Silvermans example for the valuation on curves discussed in the answer to this post: Definition and example of "order of a function at a point of a curve"

Silverman has already shown that $M_P / M_P^2$ is generated by (y), but why does $M_P = (y)$ and then $ord_P(y) = 1$ follow from this fact?

This probably boils down to a deeper problem: Why are the elements if the function field with order one exactly the generators of $M_P$. I can see why $M_P = (t) \implies ord_P(t) = 1$ but I'm not really sure about the other implication. And how do the elements / generators of $M_P$ and $M_P^2$ relate to each other?

Zedssad
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If $(R, \mathfrak{m}, k)$ is a noetherian local ring and $M$ is a finite $R$-module, a special case of Nakayama's lemma states that if $\mathfrak{m}M = M$ then $M = 0$.

In particular if $[m_1], \dots, [m_n]$ is a $k$-basis for $M/\mathfrak{m}M$, we can choose representatives $m_1, \dots, m_n \in M$. Consider the submodule $N \subset M$ generated by these representatives. Then, $\mathfrak{m}(M/N) = M/N$, since $M = N + \mathfrak{m}M$, so by Nakayama's lemma $M = N$ and $M$ is generated by $m_1, \dots, m_n$.

Now, in your situation, $M_p/M_p^2$ is generated by $[y]$, so $y$ lifts to a generator of $M_p$, and since $M_p \neq M_p^2$, we must have $y \notin M_p^2$. Hence $\operatorname{ord}_p(y) = 1$. Similarly if $\operatorname{ord}_p(t) = 1$ for some rational function $t$ then $t \in M_p \setminus M_p^2$, so $[t] \neq 0$ in $M_p/M_p^2$. As long as $\dim_k M_p/M_p^2 = 1$ (ie. the curve is nonsingular at $p$) it follows that $t$ lifts to a uniformizing parameter.

Daniel
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    Could you please explain why $\dim_K M_P / M_P^2 = 1$ implies that t lifts to a unformizer? – Zedssad Mar 27 '24 at 09:58
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    @Zedssad Sure, sorry for the delay. My definition of a uniformizer is a generator of $M_p$. If $[y]$ generates $M_p/M_P^2$ the argument above implies $y$ generates $M_P$. Do you have a different definition? – Daniel Mar 28 '24 at 17:01
  • Sorry, should have been more precise: With $\dim_K M_P / M_P^2 = 1$ we have that $M_P / M_P^2 = ([t])$. Trying to complete your argument, we have that $M_P = (t) + M_P^2$ and now $M / (t) = M_P^2$ which implies by Nakayama that $M / (t) = 0$ (great) but also $M_P^2 = 0$ (not so great). Where have I went wrong? – Zedssad Mar 28 '24 at 18:04
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    The equality $M_P/(t) = M_P^2$ is not right. If you take the quotient of the equation $M_P = (t) + M_P^2$ by $(t)$, we get that $M_P \mod (t) \equiv M_P^2 \mod (t)$, so that $M_P/(t) = 0$ by Nakayama's lemma, so that $M_p \mod (t) = [0]$. But then $M_p^2 \mod (t) = [0]$ as well, since $t$ clearly divides $t^2$. – Daniel Mar 28 '24 at 21:58
  • I can understand it very clearly now, thanks! – Zedssad Mar 29 '24 at 08:24