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I'm trying to solve exercise 4.7 in Reid's UCA:

"Find the normalisation of $A=k[X,Y]/(Y^3 - X^5)$."

I can easily show $A$ is not normal: let $x$ and $y$ denote the images of $X$ and $Y$ in $A$. Thus $y^3 = x^5$. So there exists $t=y/x\in\operatorname{Frac} (A)$ which is a root of the monic polynomial $T^3 - x^2 \in A[T]$. But since $t\notin A$, $A$ is not normal.

However, I don't know how to compute the normalisation of $A$; I have not seen dimension theory developed, so this answer to a similar question doesn't really help me. Similarly I cannot try and apply something like this answer to a related question, because in this case the map $k[X,Y]\rightarrow k[t]$ sending $X\mapsto t^3, Y\mapsto t^5$ is injective so the image isn't isomorphic to $A$

UCA never really gives a good explanation on how to actually compute normalisations, so I'd be grateful if anyone could walk me through it.

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Alex Saad
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1 Answers1

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In fact, the kernel of that map is the ideal $(Y^3-X^5)$, and therefore we have $$k[X,Y]/(Y^3-X^5)\simeq k[t^3,t^5].$$ The ring extension $k[t^3,t^5]\subset k[t]$ is integral and both rings have the same field of fractions, so the integral closure of $k[t^3,t^5]$ is $k[t]$. (If you want to come back to your ring, then the integral closure is $k[x,y,\sqrt{y/x}]$, where $x,y$ are the residue classes of $X,Y$.)

user26857
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  • @user26857 can you give a proof of the fact that kernel is precisely the ideal $(Y^3-X^5)$. It is clear that $(Y^3-X^5) \subset \text{ Kernel }$ , but I don't think that the other inclusion is that much obvious! – Brozovic Nov 04 '19 at 18:32
  • @Brozovic 1) https://math.stackexchange.com/questions/2270989/usual-k-algebra-isomorphisms-kx-y-y3-x5-cong-kt3-t5. 2) https://math.stackexchange.com/questions/1904175/finding-generators-for-the-kernel-of-kx-y-to-kt3-t5 – user26857 Nov 04 '19 at 19:08
  • @user26857 Is it true for general $(X^i - Y^j) $ with $i,j$ distinct primes? – Brozovic Nov 04 '19 at 19:11
  • @BrozovicB It's true even for $i,j$ coprime. – user26857 Nov 04 '19 at 19:16
  • Can you provide a link for proof of the fact for $i,j$ coprime ? – Brozovic Nov 04 '19 at 19:28
  • @Brozovic The proof goes similarly to the case $i=5$ and $j=3$. – user26857 Nov 04 '19 at 20:49