A projective variety is irreducible if and only if it does not contain a line
$begingroup$
Let $X$ be a proper closed subset of $mathbb P^2_{mathbb C}$. Is it true that $X$ is irreducible if and only if it does not contain a line $ax+by+cz = 0$? I have never seen or thought about irreducibility this way.
This is claimed in these notes on elliptic curves in the proof that an elliptic curve is irreducible.
algebraic-geometry
asked Dec 16 '18 at 22:49
D_SD_S
13.9k61553
$endgroup$
add a comment |
$begingroup$
Let $X$ be a proper closed subset of $mathbb P^2_{mathbb C}$. Is it true that $X$ is irreducible if and only if it does not contain a line $ax+by+cz = 0$? I have never seen or thought about irreducibility this way.
This is claimed in these notes on elliptic curves in the proof that an elliptic curve is irreducible.
algebraic-geometry
asked Dec 16 '18 at 22:49
D_SD_S
13.9k61553
$endgroup$
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
add a comment |
$begingroup$
Let $X$ be a proper closed subset of $mathbb P^2_{mathbb C}$. Is it true that $X$ is irreducible if and only if it does not contain a line $ax+by+cz = 0$? I have never seen or thought about irreducibility this way.
This is claimed in these notes on elliptic curves in the proof that an elliptic curve is irreducible.
algebraic-geometry
asked Dec 16 '18 at 22:49
D_SD_S
13.9k61553
$endgroup$
Let $X$ be a proper closed subset of $mathbb P^2_{mathbb C}$. Is it true that $X$ is irreducible if and only if it does not contain a line $ax+by+cz = 0$? I have never seen or thought about irreducibility this way.
This is claimed in these notes on elliptic curves in the proof that an elliptic curve is irreducible.
algebraic-geometry
algebraic-geometry
asked Dec 16 '18 at 22:49
D_SD_S
13.9k61553
asked Dec 16 '18 at 22:49
D_SD_S
13.9k61553
asked Dec 16 '18 at 22:49
D_SD_S
13.9k61553
asked Dec 16 '18 at 22:49
D_SD_S
13.9k61553
asked Dec 16 '18 at 22:49
D_SD_S
13.9k61553
13.9k61553
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
add a comment |
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
4
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
add a comment |
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$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18