Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

1 Answer

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 $(x-1,y)=(y).$

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.