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Let $X$ be the affine curve $\operatorname{Spec}k[t,s]/(s^2 -t^2(1+t))$.

How to verify that the normalization $X'$ of $X$ is $\operatorname{Spec}k[u]$ with $u:= s/t$ and $u^2=1+t$?

We know that since the normalization $n: X' \to X$ is an affine map, $X'$ is an affine scheme $\operatorname{Spec}A$ with $A$ integral closure of $k[t,s]/(s^2 -t^2(1+t))$ in $\operatorname{Frac}\bigl(k[t,s]/(s^2 -t^2(1+t))\bigr)$.

Is there any standard strategy to calculate integral closure of rings $R:= k[x,y]/(P(x,y))$ in $\operatorname{Frac}(R)$?

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  • There is no standard way to find out the normalization of $R:= k[x,y]/(P(x,y))$. (I guess that $P$ is an irreducible polynomial.) – user26857 Sep 04 '19 at 18:32
  • Can you not use the integral closure $k[t]$ instead? The big field of fractions is a quadratic extension of $k(t)$, making it easy to show that $k[u]$ is the integral closure of $k[t]$. – Jyrki Lahtonen Sep 04 '19 at 18:51
  • How much more to this problem is there than showing that $u$ is integral over the original ring, and that this ring and $k[u]$ have the same fraction field? – Lubin Sep 04 '19 at 22:38
  • @Lubin: going this stategy the main obstacle seems for me to verify that $Frac(k[t,s]/(s^2 -t^2(1+t))) = k(u)=Frac(k[u])$. is there any argument telling immediately that it's "obvious" or do all ways lead through finding an explicite embedding $k[t,s]/(s^2 -t^2(1+t)) \to k[u]$? –  Sep 05 '19 at 09:44

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Prove that the normalisation of $A=k[X,Y]/(Y^2-X^2-X^3)$ is $k[t]$ where $t=Y/X$ (Reid, Exercise 4.5)

Characterizing the field of fractions of $\mathbb Q[x,y]/(x^2+y^2-1)$.

Working out the normalization of $\mathbb C[X,Y]/(X^2-Y^3)$

$\mathbb{Q}[x,y]/(x^2+y^2)\cong \mathbb{Q}[y,yi]$?

As in the above links, I think it seems easier when we find an isomorphic subring in a polynomial ring. In your case, we first define a ring homomorphism, $\phi : k[t,s]\rightarrow k[u]$ given by $\phi(t)=u^2-1,\phi(s)=u(u^2-1)$. Then $\ker\phi=(s^2-t^2(1+t))$ and $k[t,s]/(s^2-t^2(1+t))\cong k[u^2-1,u(u^2-1)]\hookrightarrow k[u]$ and $\mathrm{Frac}(k[u^2-1,u(u^2-1)]=\mathrm{Frac}(k[u])=k(u)$. So we first may need to parametrize variables very well. Proving the equality of the kernel is not really obvious to me and the method used in the following link and comments on that would be helpful The kernel of $\mathbb{Q}[x,y]\rightarrow \mathbb{Q}(t)$ is $(x^2+y^2-1)$.

Another strategy is

The field of fractions of $\mathbb{R}[x,y]/(x^2+y^2-1)$ is $\mathbb{R}(x)[y]/(x^2+y^2-1)$

Showing $F[x,y]/(ax^2+by^2-1)$ is a Dedekind domain

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    https://math.stackexchange.com/questions/2212770/normalization-of-a-plane-curve-showing-equality-of-ideals may be relevant – user347489 Sep 05 '19 at 20:41