This is exercise 3.4.4 from "An Invitation to Algebraic Geometry": Show that the affine varieties V(x) and V($x-y^4-z^4$) is isomorphic in $A^3$ but their projective closures are not in $P^3$.
What I did:
They're isomorphic in $A^3$ since they're both isomorphic to $A^2$, and an isomorphism between them is $[0;y;z] \mapsto [y^4+z^4;y;z]$. Then I tried to show that the projective varieties V(x) and V($w^3x-y^4-z^4$) is not isomorphic in $P^3$. The first one is now a $P^2$, the second one is also topologically a $P^2$ so I'm trying to find somewhere the second one is not smooth but failed. Could anybody tell me a point where the second one is not smooth?

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    $\begingroup$ $(1:0:0:0)$ doesn't seem to be in the smooth locus (where the first coordinate is $x$) $\endgroup$ – zcn Mar 13 '14 at 1:23
  • $\begingroup$ You are absolutely right, @user115654. $\endgroup$ – Georges Elencwajg Mar 13 '14 at 8:32
  • $\begingroup$ @user115654: the affine variety in the x=1 hyperplane is $w^3=y^4+z^4$. Its real part is smooth at (0,0,0) since all partial derivatives vanish at the point, it has w=0 as its tangent plane. Could you explain more details? Thank you very much. $\endgroup$ – Xipan Xiao Mar 14 '14 at 14:09
  • $\begingroup$ @XipanXiao: What is your working definition of a smooth point: one where the partial derivatives all vanish? Also, it's not necessary to pass to an affine patch: the Jacobian criterion still applies to projective varieties (see e.g. Hartshorne exercise I.5.8) $\endgroup$ – zcn Mar 14 '14 at 17:24
  • $\begingroup$ @user115654: being smooth means there is a well-defined tangent plane. If the partial derivatives all vanish we get a tangent plane $w=0$. I'll check the Jacobian criterion. Thanks. $\endgroup$ – Xipan Xiao Mar 14 '14 at 17:59

For a projective hypersurface $V(f) \subset \mathbb P^n$ defined by a homogenous polynomial $$f(x_0,...,x_n)$$ there is a simple criterion to characterize a singular point of $V(f)$: The point $x = (x_0:\ ...\ :x_n) \in \mathbb P^n$ is a singular point $x \in V(f)$ iff all partial derivatives vanish, i.e.

$$\frac {\partial f}{\partial x_i}(x) = 0, i = 0,...,n.$$

Note. Euler's Lemma (see Hartshorne Ex. I,5.8) implies that $x \in V(f)$.


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