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Show that the normalization of $A = k[x_1,x_2] / (x_2^2 - x_1^3)$ is isomorphic to $k[x]$ and describe (for $k$ algebraically closed) the induced map $Spec(k[x]) \to Spec(A)$

I know that $A$ is a non closed integral domain, because (explicit calculations) the integral element $\frac{\bar{x_2}}{\bar{x_1}} \in Frac(A)$ does not belong to $A$. Furthermore, I am almost sure that the isomorphism should be searched putting $x = \frac{\bar{x_2}}{\bar{x_1}}$ - but despite these facts, I don't manage to write a clear and rigorous proof.

I know that there this related post (Working out the normalization of $\mathbb C[X,Y]/(X^2-Y^3)$), but I don't understand clearly how to completely solve my problem.

Thank you in advance for any kind of suggestion!

Cheers

2 Answers2

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Note that $k[x]$ is normal, and hence if $A := k[X,Y]/(X^2-Y^3)$ were isomorphic to $k[x]$, then $A$ would be normal too. Thus, what you wanted to ask was to show that the NORMALIZATION of $A$ is isomorphic to $k[x]$.

Anyway, the link you provided basically gives you the answer. The point is that in $A$, $Y = X^2/Y^2$, so $\sqrt{Y} = X/Y$. Further, if you adjoin $\sqrt{Y}$ to $A$, then $A$ becomes a $k$-algebra generated by $X,Y$, and $\sqrt{Y} = X/Y$, but of course any such algebra is also generated by $\sqrt{Y}$, since as soon as you have $\sqrt{Y}$, you also have $(\sqrt{Y})^2 = Y$, and you also have $\sqrt{Y}*Y = (X/Y)*Y = X$.

oxeimon
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  • Dear @oxeimon, thank you for your answer, but I actually already knew what you say. My problem is not about the intuition beyond the problem, but in formalize in a clear and rigorous way this idea. So the point is: I have $\bar{A}$ and $k[t]$. How can I write explicitly an isomorphism? –  May 24 '15 at 21:30
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    @user233650 Well if you accept the fact that $\overline{A} = k[X,Y,\sqrt{Y}]/(X^2-Y^3) = k[\sqrt{Y}]$, then the isomorphism is obvious: Just send $\sqrt{Y}$ to $t$. – oxeimon May 26 '15 at 04:34
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This isomorphism can also be visualized:

The zero locus of the normalization $\tilde{A}:=\mathbb{C}[x_1,x_2,x_3]\,/\,(x_3^2-x_1,x_1x_3-x_2)$ is an affine curve, and its real part (on $\mathbb{R}^3$) is the intersection of the surfaces $x_1=x_3^2$ (a "parabolic cylinder") and $x_2=x_1x_3$ (a "hyperbolic paraboloid").

If we project this affine curve onto the $x_1$-$x_2$-plane, we get the curve defined by $x_1^2-x_2^3=0$.

Open the links for pictures: image1 image2 image3

Ooooscar
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