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I have worked through most details in user26857's outline:

Normalisation of $k[x,y]/(y^2-x^2(x-1))$,

for dealing with similar questions. In reference to the comment:

"Since $\ker\varphi$ is a prime ideal of height one, all we have to do now is to show that $(Y^2−X^2(X−1),XT−Y,T^2−X+1)$ is also a prime ideal.",

I would like to know if I can replace the part involving any reference to Krull dimension and height of the kernel by more "elementary" means? For example, can I deduce that the kernel is principal without invoking these things for now?

I have seen that there are alternatives to such an approach, but right now I would like to stick with this. My reference is Reid.

user26857
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undefined
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1 Answers1

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First notice that $$(Y^2−X^2(X−1),XT−Y,T^2−X+1)=(XT−Y,T^2−X+1).$$ This follows from $Y^2−X^2(X−1)=-(XT-Y)(XT+Y)+X^2(T^2-X+1)$.

Now suppose that $f\in\ker\varphi$, that is, $f(x,y,y/x)=0$. Write $f(X,Y,T)=(XT-Y)g(X,Y,T)+r(X,T)$. Then write $r(X,T)=(T^2-X+1)s(X,T)+a(X)+b(X)T$. Since $f(x,y,y/x)=0$ we get $a(x)+b(x)y/x=0$ which is equivalent to $xa(x)+b(x)y=0$. This means that $Xa(X)+b(X)Y\in(Y^2−X^2(X−1)$ which implies $a=b=0$. We have shown that $f\in(XT−Y,T^2−X+1)$.

user26857
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  • Yes - thank you. And I add to the discussion that you already gave the first part in response to the follow up question in the previous post: "Use repeatedly the following isomorphism...". – undefined Apr 27 '22 at 21:04