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Learn more about Stack Overflow the company endobj 9 0 obj Let C be an irreducible curve of bidegree (1, 2) on . stream I'm particularly interested in the case $X = P^N,$ a projective space over a field. Let C be a cuspidal cubic with a cusp at a point P on \({\mathbb {P}}^2\) and let T be the Zariski tangent line to C at the point P. Then is isomorphic to .

Alternately, by $E$ do you mean $\pi^{-1} I_{Y,X} \cdot O_Z$? Then the blow up of $X$ along $Y$ is the same as the blow up of the vertex and $\pi_*\mathscr I_{E\subseteq Z}$ is the ideal sheaf of the vertex, not of the line. Moreover, since $p_*O_{\mathbb{P}}(m)=Sym^m(O_X^n)$ we can identify the image of this map with $I^m$ and hence we have $\pi_{\ast}(O_Z(-mE))=I^m$. But by Zariski's main theorem, if the fibers are connected, then the isomorphism is there right. The formal group associated with a nodal cubic curve is of height 1. Then it follows easily that so is the first.As Sándor's nice answer shows, $\pi_{\ast}(O_Z(-E))=I_{Y}$ is not neccesarily true, even for $Y$ a normal subvariety of $X$. I suppose you mean that $E$ is the exceptional So, let's assume that $X$ is normal. �ST5;"�c7=��0�RΑ���͹i��j�b���9F�>��-5���څQ(X8bh�K��M�ϫ�t�C����i�N��|�(8tm�n�\5!ٔ�b �6V$?�GN�Tw�JM�三W���SJ�-�gR[7�� :��ř�=姕}�l�Ȭ�j�c�1#T��/�41�9&��}�� if|Sc����,�[T�9��o��/@W����al+ڗ��6���7��x¶���ڴ�s,�̘��go���X2�ʦ(�X=�yy��&��t60P�1����q��"�� M. Miyanishi proposed a conjecture (see []) that for a log del Pezzo surface of Picard rank 1, its smooth locus has a finite unramified covering that contains a cylinder.It however turned out that the conjecture is answered in negative. Formal group and height. With $E$ I mean the subscheme defined by $\pi^{-1}I_{Y, X}\cdot O_Z$. • A rational curve can be given by a parametrization. It only takes a minute to sign up.Let $\pi:Z=Bl_{Y}(X)\rightarrow X$ be the blowing-up of a smooth projective variety X along a subvariety $Y$, $E$ the closed subscheme defined by $\pi^{−1}I_{Y,X}\cdot O_Z$. << /S /GoTo /D [10 0 R /Fit ] >> Example 2.9. By using our site, you acknowledge that you have read and understand our MathOverflow is a question and answer site for professional mathematicians. S. Keel and J. Mckernan have constructed log del Pezzo surfaces of Picard rank 1 such that gianna, what do you mean by the exceptional divisor? Finally, if you replace $Z$ by its normalization and choose E such that $O_Z(−E)= \pi^{−1} I_{Y,X}\cdot O_Z$, then the answer is no. The $g_i$'s determine a surjection $A^r \to I$, and hence a surjection of Rees algebras $\mbox{Sym}^*(O_X^r) \to \bigoplus_{m \ge 0} I^m$. %PDF-1.4 • A curve is called cuspidal if all its singular points are cusps.

Then _ - a nodal cubic, if P E C\S, 1(P) a cuspidal cubic, if P E S1. @gradstudent: to use ZMT, you would need to know that $E\to Y$ is birational, but $E$ doesn't even have to have the same number of components as $Y$. The proof of this is very simple. Recall that a Del Pezzo surface is the blow-up of \(\mathbb {P}^2\) at most at eight points in general position, namely if. endobj

Is it true that (without assume $Y$ smooth) $\pi_{\ast}(O_Z(-E))=I_{Y,X}$?This is not true. 4��m���Iy�)�g ... one needs to blow up a point in CP 2 to get a seven–dimensional family M. This will admit. no three of them lie on a line; no six of them lie on a conic; no eight of them lie on a nodal or cuspidal cubic with one of them at the … A del Pezzo surface of degree 1 is a complete smooth surface which is isomorphic over an algebraically closed field to the blow up of Detailed answers to any questions you might have Show that E meets Y~ in one point, and that Y~ ˘=A1. Anybody can answer a non–homogeneous G 2 structure. Actually, your question needs to be made a bit more precise. However, my statement is true if $X$ (not necessarily $Y$) is smooth ? Now, we will see that there exists a birational map between the variety Xcusp and the degeneration Xncusp of Xnod. On the other hand, the following statement$$\pi_{\ast}(O_Z(-mE))=I_{Y}^m \quad\mbox{ for $m\gg 0$}$$always holds, even without any assumptions on the subscheme $Y$. Here is a short explaination why: It suffices to deal with the case $X=\mbox{Spec} A$ is affine and $I=I_Y=(g_1,\ldots,g_n)\subset A$. A closed point of Xncusp and of Xconsec. The exceptional set need not be a divisor in general. MathOverflow works best with JavaScript enabled 12 0 obj << \end{gather}The assumption that $X$ is normal implies that the second vertical arrow is an isomorphism and the other assumption is that so is the third. << /S /GoTo /D (section.1) >> The best answers are voted up and rise to the top Anybody can ask a question Notice that in this example, $X$ is a normal Gorenstein variety and $Y$ is smooth, so you need quite a bit of assumptions to have a blanket statement like you wished for.If $X$ is normal and the natural map $\mathscr O_Y\to \pi_*\mathscr O_E$ is an isomorphism, then $\mathscr I_{Y\subseteq X}\simeq \pi_*\mathscr I_{E\subseteq Z}$. �gB\1���}t6��j8�#�E��_�� �Ǝ`�'nۀ��ڙl��6L;6葉�R� ����Wi�Uy�3~>�3-�#H���RC�\�Z3a� ��_�Ȗ?�-O�D[/dr�R� ʃn�~T�� �)٤8�ٔ� ���֜�L�h$>���n4j&��� y��s���b��1EG���\b���� ���L6�2VM��l�l�ϗ���Z>�ۛjn�E8�}���>`m9v�03��:h�y'�=�y6�,��|��4sv�W.Ul�� v��D�t �>�^���1As�`��/��RS��\m� 4Qv(0GƼKx�4:����FSI&�d��Q�Ǚ��x�#��r�[�7�߸y=�|� Consider the diagram Taking Proj this means that there is an embedding $Z=Bl_Y X \subset \mathbb{P}=\mathbb{P}(O_X^n)$ where the exceptional divisor $E$ corresponds to $O(-1)\big|_Z$.

&\downarrow \qquad & \downarrow \qquad\qquad & \downarrow & \qquad \qquad \\

Let Y be the cuspidal cubic curve y2 = x3 in A2.

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