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I'm reading the notes on Elliptic Curves from this MIT course, more specifically this part where the local ring of a curve $C$ at point $P$ is defined, as the set of rational functions $f$ on $C$ such that $f(P) \neq \infty$.

It is stated (see §23.3) that a local ring is a principal ideal domain, without a proof.

I can see that it is a domain, but I don't see why it should be principal (e.g. every ideal from this local ring is generated by a single element).

As this is something important for my understanding of those notes, I would like to have a proof of that fact. I would especially appreciate an answer that does not involve too many abstractions besides what is included in those notes.

An answer showing how to exhibit a generator of an ideal of a local ring on a particular example would also be appreciated.

Weier
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    For more general curves, it's essential to assume the curve is non-singular. If you are allowing curves to have singular points (for example the "elliptic curve" $y^2 = x^3 - x^2$) then it's no longer true that local rings at those singularities are PIDs. – Daniel Schepler Jan 30 '23 at 20:32
  • If any example will do, consider the curve $y=0$. – Mariano Suárez-Álvarez Jan 30 '23 at 20:34
  • The local rings $\mathcal{O}_P$ are DVRs, so local PIDs, see wikipedia. It is mentioned that many results have a detailed proof in Silverman's book on elliptic curves. Did you have a look there? – Dietrich Burde Jan 30 '23 at 20:34
  • Anyway, for nonsingular elliptic curves of the form $y^2 = x^3 + ax + b$, $y$ is a generator at points $(x_0, 0)$, and otherwise $x$ is a generator at points $(x_0, y_0), y_0 \ne 0$. At least for points within the affine plane - for the "point at infinity" on the projective elliptic curve $y^2 z = x^3 + a x z^2 + b z^3$, you'll have to fill in that case yourself. – Daniel Schepler Jan 30 '23 at 20:43
  • @DietrichBurde how do I prove that they are DVR then? Reading the notes, I was under the impression that we had first to prove they were PID. (Do not hesitate to answer the question if you can elaborate on it)

    Also, I managed to have a look at Silverman's book and it doesn't prove any of it.

     – Weier Jan 30 '23 at 20:47
  • I think this is answered in https://math.stackexchange.com/questions/3871349/uniformizer-is-a-generator-for-the-maximal-ideal-from-silvermans-aec/3871886 – djao Jan 30 '23 at 20:59
  • Potential duplicate of https://math.stackexchange.com/questions/236545/local-ring-at-a-non-singular-point-of-a-plane-algebraic-curve – KReiser Jan 30 '23 at 21:47
  • @MarianoSuárez-Álvarez on the curve $y = 0$ I would say that the curve is isomorphic to the base projective field $K$. The local ring is the ring of fractions with positive exponent of $X-a$ (where $P = a \in K$).

    If I consider an ideal of the local ring, I can consider a nonzero element $M$ with minimal exponent of $X-a$. Then any other element of the ideal is of the form $M \times Q$ where $Q$ is a fraction with nonnegative exponent of $X-a$ (hence in the local ring), which shows that $M$ generates the ideal.

    Is that correct?

     – Weier Jan 30 '23 at 22:25
  • Also, this is an "easy" case since we deal with univariate fractions and it doesn't show how this can be generalised to other curves. – Weier Jan 30 '23 at 22:26
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    @weler, the curve is isomorphic to the line, not to base field! But yes. Try a conic next. Do a couple of examples and try to see what element generates the maximal ideal. (In order to prove that you have a PID you have to do this for all ideals, of course) – Mariano Suárez-Álvarez Jan 30 '23 at 22:36
  • @MarianoSuárez-Álvarez If I consider the parabola $Y = X^2$ I end up with an univariate function field and it's similar to the previous case. If I consider the circle $X^2 + Y^2 = 1$ I end up with a function field composed of fractions of the form $\frac{YA(X) + B(X)}{YC(X) + D(X)}$. If I consider for example the point $(1, 0)$ the condition on the local ring is $D(1) \neq 0$. Then I'm stuck because I can't express things in terms of factor as in the univariate case. – Weier Jan 31 '23 at 12:48
  • In that case, $(X - 1, Y)$ generates the ideal $I$ of the local ring composed of functions which evaluate to $0$ on $P = (1, 0)$.

    Also, in the function field we have $X - 1 = -Y^2/(X+1)$ and this is of the form $QY$ where $Q$ is in the local ring if $1 + 1 \neq 0$ (which we'll assume).

    Hence, $Y$ generates $I$ which is then principal.

    @MarianoSuárez-Álvarez Does this make sense?

     – Weier Feb 01 '23 at 13:32
  • You might find my answer here helpful: Normality of localizations in polynomial rings? – Viktor Vaughn Feb 01 '23 at 16:19

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