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.
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
asked Nov 18 at 22:54
zzy
2,2231319
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up vote
1
down vote
favorite
In Hartshorne's book for Algebraic Geometry, I am a little confused about the proof in Chapter IV Proposition 3.8.
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
asked Nov 18 at 22:54
zzy
2,2231319
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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Hartshorne's book for Algebraic Geometry, I am a little confused about the proof in Chapter IV Proposition 3.8.
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
asked Nov 18 at 22:54
zzy
2,2231319
In Hartshorne's book for Algebraic Geometry, I am a little confused about the proof in Chapter IV Proposition 3.8.
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
algebraic-geometry algebraic-curves
asked Nov 18 at 22:54
zzy
2,2231319
asked Nov 18 at 22:54
zzy
2,2231319
asked Nov 18 at 22:54
zzy
2,2231319
asked Nov 18 at 22:54
zzy
2,2231319
asked Nov 18 at 22:54
zzy
2,2231319
2,2231319
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
add a comment |
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
add a comment |
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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