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I am trying to figure out the normalisation of $k[x,y]/(y^2-x^2(x-1))$, for an algebraically closed field $k$.

I can show that it is not normal and I have the information that the normalisation is $k[t]$, where $t=\frac{y}{x}$. I can't figure out how to prove that this is the normalisation i.e. that the fraction fields of $k[x,y]/(y^2-x^2(x-1))$ and $k[t]$ are the same, that $k[t]$ is integrally closed and that $k[t]$ is an integral extension of $k[x,y]/(y^2-x^2(x-1))$.

Also, how does one get to the idea of $t=\frac{y}{x}$? Algebraically if possible, since this is from a commutative algebra class, not algebraic geometry.

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  • $\begingroup$ It sounds like you know that $t$ is integral over $R=k[x,y]/(y^2-x(x-1)$, in other words that $t^2=x-1$. Can you show that this implies that $R[t]=k[t]$? As $R[t]$ is contained in the field of fraction of $R$ it follows that the fields of fractions of $R$ and $k[t]$ agree. Can you show that $k[t]$ is integrally closed in its field of fractions? $\endgroup$ – Jyrki Lahtonen Dec 8 '14 at 18:00
  • $\begingroup$ Closely related: math.stackexchange.com/questions/865875/… $\endgroup$ – user26857 Dec 8 '14 at 19:17
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    $\begingroup$ Let's suppose you have an integral domain $R$ which is not integrally closed. Now adjoint $t$ an integral element over $R$ such that $R[t]$ is integrally closed (if this is iso with a polynomial ring over a field then you win), and let $K$ be the field of fractions of $R$. Now you have the following tower of rings: $R\subset R[t]\subset K$. Can you see now that $R[t]$ is the integral closure of $R$? $\endgroup$ – user26857 Dec 8 '14 at 19:27
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    $\begingroup$ @user26857 that's because I am trying to solve a more general problem for a homework assignment (the ideal considered is $(y^2 - f(x))$, where $f(x)$ is of odd degree), and one of the subtasks was to determine maximal ideals for which the localisation is a PID. Since in this case the localisations are of dimension 1 they are normal iff they are PIDs, and the original ring is normal iff the localisations are all normal. I can prove that whenever there is a multiple root in the polynomial $f(x)$ one of these localisations is not normal. But finding the normalisation is the last part of the task. $\endgroup$ – baltazar Dec 8 '14 at 19:53
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    $\begingroup$ I see. Probably you mean this: math.stackexchange.com/questions/678419/… $\endgroup$ – user26857 Dec 8 '14 at 20:04
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Let $R=k[X,Y]/(Y^2-X^2(X-1))$. It is easily shown that $R$ is an integral domain. Moreover, $R$ is not normal since the Jacobian matrix $(-3x^2+2x \;\; 2y)$ has rank zero modulo the maximal ideal $\mathfrak m=(x,y)$, while $2-\dim R_{\mathfrak m}=1$.

Now notice that $t=y/x$ satisfies $t^2=x-1$, so $t$ is integral over $R$. (Now we can provide another reason for $R$ is not integrally closed: $t\notin R$, otherwise $t=f(x,y)$ which is equivalent to $y=xf(x,y)$. By lifting the classes we get $Y-Xf(X,Y)\in(Y^2-X^2(X-1))$ which leads quickly to a contradiction by sending $X$ to $0$.)

Let's show that $R[t]$ is integrally closed. We have $R[t]=k[x,y,y/x]$ and this ring is isomorphic to $$k[X,Y,T]/(Y^2-X^2(X-1),XT-Y,T^2-X+1).$$ In order to show this define a surjective ring homomorphism $\varphi:k[X,Y,T]\to k[x,y,y/x]$ by $X\mapsto x$, $Y\mapsto y$, $T\mapsto y/x$, and note that $(Y^2-X^2(X-1),XT-Y,T^2-X+1)\subseteq\ker\varphi$. 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. But $$k[X,Y,T]/(Y^2-X^2(X-1),XT-Y,T^2-X+1)\simeq k[T],$$ and thus we get two things: $(Y^2-X^2(X-1),XT-Y,T^2-X+1)=\ker\varphi$ and $R[t]\simeq k[T]$.

From the tower of rings $R\subset R[t]\subset K$ (where $K$ is the field of fractions of $R$) we deduce that the integral closure of $R$ is $R[t]$.

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    $\begingroup$ I have a quick followup. I don't understand why $K[X,Y,Z]/(Y^2-S^2(X-1),XT-Y,T^2-X+1) \cong k[T]$? I guess it should be obvious but I am missing it. $\endgroup$ – baltazar Dec 8 '14 at 23:39
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    $\begingroup$ @baltazar Use repeatedly the following isomorphism: $R[Y]/(Y-a)\simeq R$ for some $a\in R$. (Start with $a=XT$ and $R=k[X,T]$.) $\endgroup$ – user26857 Dec 8 '14 at 23:45

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