# 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)

This is a problem about finding the normalisation of a quotient polynomial ring. So I have to find the integral closure of the ring in its field of fractions. The problem statement is as follows:

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

Can I do this by showing that the field of fractions $\text{Frac} A$ of $A$ is equal to $k[t]$, and subsequently showing that the field of fractions is normal? (This could be done by showing that $k[t]$ is a UFD?)

I am lost at calculating/determining $\text{Frac}A$, or similarly proving that $k[t]=\text{Frac}A$. Also, how do I show that it is normal?

I hope you can help!

• You have, indeed, that the field of fractions of $A$ is $k(t)$. Also in the mapping $i:A\to k(t)$ we actually have that $i(A)\subset k[t]$. Therefore you should show that 1) $k[t]$ is integral over $A$, 2) $k[t]$ is integrally closed in $k(t)$. Commented Jul 13, 2014 at 12:31
• But I don't understand the idea of showing that $Frac(A)$ is a UFD. $Frac(A)$ is a field, so there is no useful divisibility/factorization concept there. Did you mean to use that $k[t]$ is a UFD (which it is by virtue of being a PIDC)? Commented Jul 13, 2014 at 12:35
• To make sure: If you had problems to get started check what do you get when you divide the equation $$Y^2=X^2+X^3$$ by $X^2$. Commented Jul 13, 2014 at 12:44
• Related (and helpful): math.stackexchange.com/questions/500315/… Commented Jul 13, 2014 at 13:26
• The field of fractions of $k[t]$ is $k(t)$, not $k[t]$ itself. The equation $(Y/X)=1+X$ show that $X=t^2-1$. Consequently $Y=(Y/X)X=t^3-t$. This means that $A\subset k[t]$. Also $Y/X=t$ obviously belongs to $Frac(A)$, so consequently all of $k(t)\subseteq Frac(A)$. Commented Jul 14, 2014 at 15:33

• Consider $\varphi:K[X,Y]\to K[T]$ given by $\varphi(X)=T^2-1$, $\varphi(Y)=T(T^2-1)$. Prove that $\ker\varphi=(Y^2-X^2-X^3)$.
• We also have $A\simeq\operatorname{Im}\varphi=K[T^2-1,T(T^2-1)]\subset K[T]$, and $T$ is integral over $K[T^2-1,T(T^2-1)]$.
• Sorry I haven't had time to look at your answer! Very nice, I just need to understand how this shows that $k[t]$ is the normalisation of $A$. So $K[T^2-1,T(T^2-1)] \subset K[T]$, and since $T$ is integral over $K[T^2-1,T(T^2-1)]$, $K[T]$ is the normalisation of $A$? This needs to be the integral closure of $A$ in its field of fractions. Don't we need to show that $K[T]$ is the field of fractions of $A$? Or is that trivial :) Commented Jul 14, 2014 at 14:23
• @BoSchmidt If $A\subset B$ is an integral extension of integral domains, $B$ is integrally closed, and $\operatorname{Frac}A=\operatorname{Frac}B$, then the integral closure of $A$ is $B$, right? Commented Jul 14, 2014 at 16:28
• Why is $K[T]$ integrally closed? Commented Feb 4, 2020 at 1:47
• @hlcrypto123 every DFU is integrally closed (aka normal). To prove this, one generalizes the proof done for $\mathbb{Z}$ in p. 59 of Atiyah-MacDonald. Commented Jul 24, 2023 at 9:41
• @Numbersandsoon Yes, we need to show that $K(T)$ is the field of fractions of $A\cong K[T^2-1,T(T^2-1)]\subset K[T]$. This is how you do it: we have an induced $K$-algebra map between fields of fractions $$\label{1}\tag{1} Q(K[T^2-1,T(T^2-1)])\to Q(K[T])=K(T),$$ which is injective. On the other hand, the formal fraction $\frac{T(T^2-1)}{T^2-1}$ in the domain of \eqref{1} is sent to $T$ in the target. Hence, \eqref{1} is onto; thus, an isomorphism. Commented Jul 24, 2023 at 9:52