Characteristic of a field and algebraic curves
In what way the characteristic of a field influences the curves defined over that field? For example, let $C$ be the curve defined by $X^3+Y^3+Z^3=0$ over an algebraically closed field $K$. What happen to the curve if $char(K)=3$ or $char(K)ne 3$?
Thank you!
algebraic-geometry
asked Nov 26 '18 at 7:52
mip
334
add a comment |
In what way the characteristic of a field influences the curves defined over that field? For example, let $C$ be the curve defined by $X^3+Y^3+Z^3=0$ over an algebraically closed field $K$. What happen to the curve if $char(K)=3$ or $char(K)ne 3$?
Thank you!
algebraic-geometry
asked Nov 26 '18 at 7:52
mip
334
add a comment |
In what way the characteristic of a field influences the curves defined over that field? For example, let $C$ be the curve defined by $X^3+Y^3+Z^3=0$ over an algebraically closed field $K$. What happen to the curve if $char(K)=3$ or $char(K)ne 3$?
Thank you!
algebraic-geometry
asked Nov 26 '18 at 7:52
mip
334
In what way the characteristic of a field influences the curves defined over that field? For example, let $C$ be the curve defined by $X^3+Y^3+Z^3=0$ over an algebraically closed field $K$. What happen to the curve if $char(K)=3$ or $char(K)ne 3$?
Thank you!
algebraic-geometry
algebraic-geometry
asked Nov 26 '18 at 7:52
mip
334
asked Nov 26 '18 at 7:52
mip
334
asked Nov 26 '18 at 7:52
mip
334
asked Nov 26 '18 at 7:52
mip
334
asked Nov 26 '18 at 7:52
mip
334
334
add a comment |
add a comment |
1 Answer
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Over characteristic $p=3$, $X^3+Y^3+Z^3 = (X+Y+Z)^3$.
answered Nov 26 '18 at 7:54
Wuestenfux
3,5791411
Can you explain why is that true?
– mip
Nov 26 '18 at 8:12
1
@mip Why not just multiply out the cube and see what happens?
– Mark Bennet
Nov 26 '18 at 8:16
Oh, I see! Besides $X^3, Y^3$ and $Z^3$, all other terms from the expansion have a coefficient divisible by 3, which mean that they are $0$. Thank you!!
– mip
Nov 26 '18 at 8:23
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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Over characteristic $p=3$, $X^3+Y^3+Z^3 = (X+Y+Z)^3$.
answered Nov 26 '18 at 7:54
Wuestenfux
3,5791411
Can you explain why is that true?
– mip
Nov 26 '18 at 8:12
1
@mip Why not just multiply out the cube and see what happens?
– Mark Bennet
Nov 26 '18 at 8:16
Oh, I see! Besides $X^3, Y^3$ and $Z^3$, all other terms from the expansion have a coefficient divisible by 3, which mean that they are $0$. Thank you!!
– mip
Nov 26 '18 at 8:23
add a comment |
Over characteristic $p=3$, $X^3+Y^3+Z^3 = (X+Y+Z)^3$.
answered Nov 26 '18 at 7:54
Wuestenfux
3,5791411
Can you explain why is that true?
– mip
Nov 26 '18 at 8:12
1
@mip Why not just multiply out the cube and see what happens?
– Mark Bennet
Nov 26 '18 at 8:16
Oh, I see! Besides $X^3, Y^3$ and $Z^3$, all other terms from the expansion have a coefficient divisible by 3, which mean that they are $0$. Thank you!!
– mip
Nov 26 '18 at 8:23
add a comment |
Over characteristic $p=3$, $X^3+Y^3+Z^3 = (X+Y+Z)^3$.
answered Nov 26 '18 at 7:54
Wuestenfux
3,5791411
Over characteristic $p=3$, $X^3+Y^3+Z^3 = (X+Y+Z)^3$.
answered Nov 26 '18 at 7:54
Wuestenfux
3,5791411
answered Nov 26 '18 at 7:54
Wuestenfux
3,5791411
answered Nov 26 '18 at 7:54
Wuestenfux
3,5791411
answered Nov 26 '18 at 7:54
Wuestenfux
3,5791411
3,5791411
Can you explain why is that true?
– mip
Nov 26 '18 at 8:12
1
@mip Why not just multiply out the cube and see what happens?
– Mark Bennet
Nov 26 '18 at 8:16
Oh, I see! Besides $X^3, Y^3$ and $Z^3$, all other terms from the expansion have a coefficient divisible by 3, which mean that they are $0$. Thank you!!
– mip
Nov 26 '18 at 8:23
add a comment |
Can you explain why is that true?
– mip
Nov 26 '18 at 8:12
1
@mip Why not just multiply out the cube and see what happens?
– Mark Bennet
Nov 26 '18 at 8:16
Oh, I see! Besides $X^3, Y^3$ and $Z^3$, all other terms from the expansion have a coefficient divisible by 3, which mean that they are $0$. Thank you!!
– mip
Nov 26 '18 at 8:23
Can you explain why is that true?
– mip
Nov 26 '18 at 8:12
Can you explain why is that true?
– mip
Nov 26 '18 at 8:12
1
1
@mip Why not just multiply out the cube and see what happens?
– Mark Bennet
Nov 26 '18 at 8:16
@mip Why not just multiply out the cube and see what happens?
– Mark Bennet
Nov 26 '18 at 8:16
Oh, I see! Besides $X^3, Y^3$ and $Z^3$, all other terms from the expansion have a coefficient divisible by 3, which mean that they are $0$. Thank you!!
– mip
Nov 26 '18 at 8:23
Oh, I see! Besides $X^3, Y^3$ and $Z^3$, all other terms from the expansion have a coefficient divisible by 3, which mean that they are $0$. Thank you!!
– mip
Nov 26 '18 at 8:23
add a comment |
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