Hartshorne algebraic geometry IV Proposition 3.8.












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In Hartshorne's book for Algebraic Geometry, I am a little confused about the proof in Chapter IV Proposition 3.8.enter image description hereenter image description here
In particular, why does (a) imply (b)? If $psi$ is inseparable, why does every tangent line passes through $R$? In the other hand, if $psi$ is inseparable, he chooses $T$ be a nonramified point in $psi(X)$. As $T$ lives in $mathbb P^2$ rather than on $X-R$, what does he mean $L_P, L_Q$ are both in the plane spanned by $R$ and $L_T$? By the definition of projection, shall $R$ lie in the line corresponding to $T$?



I am confused because this is too formal rather than algebraic. Thanks for any suggestions










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  • If $psi$ is (purely) inseparable, then $dpsi = 0$. Let $pi : mathbb{P}^3 setminus R rightarrow mathbb{P}^2$. Then $dpsi(L_P) = 0 Leftrightarrow dpi(L_P) = 0$. I think it is more or less clear from here that the line $L_P$ has to pass through $R$. In the other case, the quoted text is assuming $mathbb{P}^2$ to be a hyperplane of $mathbb{P}^3$, not containing $R$.
    – random123
    Nov 19 at 8:30
















1














In Hartshorne's book for Algebraic Geometry, I am a little confused about the proof in Chapter IV Proposition 3.8.enter image description hereenter image description here
In particular, why does (a) imply (b)? If $psi$ is inseparable, why does every tangent line passes through $R$? In the other hand, if $psi$ is inseparable, he chooses $T$ be a nonramified point in $psi(X)$. As $T$ lives in $mathbb P^2$ rather than on $X-R$, what does he mean $L_P, L_Q$ are both in the plane spanned by $R$ and $L_T$? By the definition of projection, shall $R$ lie in the line corresponding to $T$?



I am confused because this is too formal rather than algebraic. Thanks for any suggestions










share|cite|improve this question






















  • If $psi$ is (purely) inseparable, then $dpsi = 0$. Let $pi : mathbb{P}^3 setminus R rightarrow mathbb{P}^2$. Then $dpsi(L_P) = 0 Leftrightarrow dpi(L_P) = 0$. I think it is more or less clear from here that the line $L_P$ has to pass through $R$. In the other case, the quoted text is assuming $mathbb{P}^2$ to be a hyperplane of $mathbb{P}^3$, not containing $R$.
    – random123
    Nov 19 at 8:30














1












1








1







In Hartshorne's book for Algebraic Geometry, I am a little confused about the proof in Chapter IV Proposition 3.8.enter image description hereenter image description here
In particular, why does (a) imply (b)? If $psi$ is inseparable, why does every tangent line passes through $R$? In the other hand, if $psi$ is inseparable, he chooses $T$ be a nonramified point in $psi(X)$. As $T$ lives in $mathbb P^2$ rather than on $X-R$, what does he mean $L_P, L_Q$ are both in the plane spanned by $R$ and $L_T$? By the definition of projection, shall $R$ lie in the line corresponding to $T$?



I am confused because this is too formal rather than algebraic. Thanks for any suggestions










share|cite|improve this question













In Hartshorne's book for Algebraic Geometry, I am a little confused about the proof in Chapter IV Proposition 3.8.enter image description hereenter image description here
In particular, why does (a) imply (b)? If $psi$ is inseparable, why does every tangent line passes through $R$? In the other hand, if $psi$ is inseparable, he chooses $T$ be a nonramified point in $psi(X)$. As $T$ lives in $mathbb P^2$ rather than on $X-R$, what does he mean $L_P, L_Q$ are both in the plane spanned by $R$ and $L_T$? By the definition of projection, shall $R$ lie in the line corresponding to $T$?



I am confused because this is too formal rather than algebraic. Thanks for any suggestions







algebraic-geometry algebraic-curves






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asked Nov 18 at 22:54









zzy

2,3571419




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  • If $psi$ is (purely) inseparable, then $dpsi = 0$. Let $pi : mathbb{P}^3 setminus R rightarrow mathbb{P}^2$. Then $dpsi(L_P) = 0 Leftrightarrow dpi(L_P) = 0$. I think it is more or less clear from here that the line $L_P$ has to pass through $R$. In the other case, the quoted text is assuming $mathbb{P}^2$ to be a hyperplane of $mathbb{P}^3$, not containing $R$.
    – random123
    Nov 19 at 8:30


















  • If $psi$ is (purely) inseparable, then $dpsi = 0$. Let $pi : mathbb{P}^3 setminus R rightarrow mathbb{P}^2$. Then $dpsi(L_P) = 0 Leftrightarrow dpi(L_P) = 0$. I think it is more or less clear from here that the line $L_P$ has to pass through $R$. In the other case, the quoted text is assuming $mathbb{P}^2$ to be a hyperplane of $mathbb{P}^3$, not containing $R$.
    – random123
    Nov 19 at 8:30
















If $psi$ is (purely) inseparable, then $dpsi = 0$. Let $pi : mathbb{P}^3 setminus R rightarrow mathbb{P}^2$. Then $dpsi(L_P) = 0 Leftrightarrow dpi(L_P) = 0$. I think it is more or less clear from here that the line $L_P$ has to pass through $R$. In the other case, the quoted text is assuming $mathbb{P}^2$ to be a hyperplane of $mathbb{P}^3$, not containing $R$.
– random123
Nov 19 at 8:30




If $psi$ is (purely) inseparable, then $dpsi = 0$. Let $pi : mathbb{P}^3 setminus R rightarrow mathbb{P}^2$. Then $dpsi(L_P) = 0 Leftrightarrow dpi(L_P) = 0$. I think it is more or less clear from here that the line $L_P$ has to pass through $R$. In the other case, the quoted text is assuming $mathbb{P}^2$ to be a hyperplane of $mathbb{P}^3$, not containing $R$.
– random123
Nov 19 at 8:30















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