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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!

user26857
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Numbersandsoon
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  • 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 $k(t)$. – Jyrki Lahtonen Jul 13 '14 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)? – Jyrki Lahtonen Jul 13 '14 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 $X^2$. – Jyrki Lahtonen Jul 13 '14 at 12:44
  • Related (and helpful): http://math.stackexchange.com/questions/500315/working-out-the-normalization-of-mathbb-cx-y-x2-y3 – user26857 Jul 13 '14 at 13:26
  • Particular case of http://math.stackexchange.com/questions/678419/normalization-of-a-quotient-ring-of-polynomial-rings-reid-exercise-4-6 – user26857 Jul 13 '14 at 13:58
  • @JyrkiLahtonen I'm having trouble proving that $\text{Frac}A=k[t]$ :) I looked at the equation you mentioned, and we end up with $(Y/X)^2=1+X$. How does this lead me closer to proving $\text{Frac}A=k[t]$? Also, in order for $k[t]$ to be normal it must be integrally closed in its field of fractions, but isn't the field of fractions of $k[t]$ equal to $k[t]$? What I'm asking is that why is $k[t]$ normal since it is a UFD – Numbersandsoon Jul 14 '14 at 14:16
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    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 $k(t)\subseteq Frac(A)$. – Jyrki Lahtonen Jul 14 '14 at 15:33
  • @JyrkiLahtonen Ahhh, of course! I get it, thanks! :) – Numbersandsoon Jul 14 '14 at 17:25
  • More related: http://math.stackexchange.com/questions/1057771/normalisation-of-kx-y-y2-x2x-1 – user26857 Jul 02 '15 at 12:55

1 Answers1

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Hints.

  • 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 $K[T^2-1,T(T^2-1)]$.
user26857
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    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 :) – Numbersandsoon Jul 14 '14 at 14:23
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    @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? – user26857 Jul 14 '14 at 16:28
  • Why is $K[T]$ integrally closed? – hlcrypto123 Feb 04 '20 at 01:47
  • @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. – Elías Guisado Villalgordo Jul 24 '23 at 09:41
  • @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. – Elías Guisado Villalgordo Jul 24 '23 at 09:52