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I'm trying to practice with localization of rings. In particular, I want to determine the generator t of the principal maximal ideal of the localization of the coordinate ring of a curve at a point. Let's say I have a curve C given by $y^2=(x-1)(x-2)(x-3)$.

The coordinate ring of C is $k[C]=k[x,y]/( y^2-(x-1)(x-2)(x-3) )$

I want to localize $k[C]$ at say $p=(1,0)$. I'd like to determine $t$, so I'm thinking I should know how $k[C]_p$ looks like explicitly. This is where I'm stuck. I'd appreciate someone showing me how this is done.

The curve $C$ has a point at infinity $p=[1:0:0]$. Why is $p$ the point at infinity? I think I don't understand this concept with respect to curves.

For $p=[1:0:0]$, I want to determine the generator of the maximal ideal $\mathfrak{m}_p$ in the localization $k[C]_p$. So, I'll have to do this in the affine chart that contains $p$. So, the chart will be $U=\{\,[1,y,z]\, |\, y,z\in k\,\}\cong\{\,(y,z)\,|\,y,z\in k\,\}=\mathbb{A}^2$. In this chart, the curve is given by the equation: $y^2=z(1-z)(1-2z)(1-3z)$

Then what? I'm thinking to localize $k[C]$ at the point $(0,0)$ because it corresponds to $p$. So $(0,0)$ corresponds to the maximal ideal $(y,z)$ in $k[C]$, also the maximal ideal in $k[C]_p$ is $\mathfrak{m}_p=(y,z)$. Now, $(1-z)(1-2z)(1-3z)\notin (y,z)$ which means that $(1-z)(1-2z)(1-3z)$ is a unit $u$ in $k[C]_p$. We write $y^2=uz$ which means that $z=u^{-1}y^2$. This shows that $z\in(y)$, and we conclude that $(y,z)=(y)$.

I think this is correct, but I doubt it. I think my problem is with the concept of points at infinity on curves.

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Silverman‘s The Arithmetic of Elliptic Curves gives a few examples of this calculation in its first chapter. With the technique that Andrew gives below, they should be easy for you now! – Dylan Moreland Aug 31 '12 at 20:59
up vote 2 down vote accepted

The point $p$ corresponds to the maximal ideal $(x-1,y)\subseteq k[C],$ thus, the localization $k[C]_p$ is the ring obtained by inverting elements in $k[C]\setminus (x-1,y).$

The maximal ideal of $k[C]_p$ is obviously $\frak{m}_p$ $= (x-1,y).$ Note that $(x-2)(x-3)\notin\frak m_p,$ so in the localization we have $y^2=u(x-1)$ for $u$ a unit. This shows that $x-1=u^{-1}y^2,$ which implies that $x-1\in (y).$ Thus $(x-1,y)=(y).$

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Thanks. That was pretty simple. I'm realizing the importance of paying attention to details when reading math. – Marty Aug 31 '12 at 20:57

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