Computation of Normalization of $\frac{k[X,Y]}{(Y^2-f(X))}$

Question

How to compute the normalization of $R:=\frac{k[X,Y]}{(Y^2-f(X))}$ with $f(X)\in k[X]$ of odd order?

I proved that $R$ is normal iff $f(X)$ is square free. What are the methods and the ideas that bring me to the normalization of a ring?

How can i describe the map $\Omega_{R/k}\rightarrow \Omega_{\bar{R}/k}$?

Thank you :)

Probably a duplicate of https://math.stackexchange.com/questions/678419/normalization-of-a-quotient-ring-of-polynomial-rings-reid-exercise-4-6 — user26857, Jan 25 '18 at 16:21
Thank you. I've another question, how can i describe the map induced by $R\rightarrow \bar{R}$ in the differentials $\Omega_{R/k}\rightarrow \Omega_{\bar{R}/k}$ — Acuo95, Jan 25 '18 at 17:00

score 0 · Answer 1 · answered Jan 25 '18 at 16:33

0

In general if $f(x)=p(x)^2\cdot h(x)$ where $h(x)$ is the square-free part of $f(x)$ we have $\frac{y^2}{p(x)^2}=h(x)$ in the fraction field of $R$ hence you have to adjoin the inverse of $p(x)$ to the ring to get the normalization.

Hence the normalization is $k[x,y,z]/(y^2-f(x),z\cdot p(x)^2-1)$

answered Jan 25 '18 at 16:33

Levent

4,804

This looks like a story without any proof. I'm sure you can explain it better than this. – user26857 Jan 25 '18 at 16:39
Yeah, it's not complete I guess. But I just noticed your comment and in that answer everything is much more rigorous. – Levent Jan 25 '18 at 16:41
Thank you so much, i'll read the answer linked above – Acuo95 Jan 25 '18 at 16:59
I've understood that is normal, but why is it the normalization? – Acuo95 Jan 25 '18 at 17:30
The corresponding variety projects onto the curve $y^2=f(x)$ and it is normal. There are few things to check. – Levent Jan 25 '18 at 18:06

Computation of Normalization of $\frac{k[X,Y]}{(Y^2-f(X))}$

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