Is this plane curve irreducible?
$begingroup$
I want to define a plane curve in $mathbb{A}^2(mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)inmathbb{A}^2(mathbb{C})$, but my goal is for the plane curve to be irreducible.
How do I tell/prove that this curve is irreducible?
Thanks so much!
algebraic-curves
asked Dec 13 '18 at 12:09
mathgeenmathgeen
224
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add a comment |
$begingroup$
I want to define a plane curve in $mathbb{A}^2(mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)inmathbb{A}^2(mathbb{C})$, but my goal is for the plane curve to be irreducible.
How do I tell/prove that this curve is irreducible?
Thanks so much!
algebraic-curves
asked Dec 13 '18 at 12:09
mathgeenmathgeen
224
$endgroup$
add a comment |
$begingroup$
I want to define a plane curve in $mathbb{A}^2(mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)inmathbb{A}^2(mathbb{C})$, but my goal is for the plane curve to be irreducible.
How do I tell/prove that this curve is irreducible?
Thanks so much!
algebraic-curves
asked Dec 13 '18 at 12:09
mathgeenmathgeen
224
$endgroup$
I want to define a plane curve in $mathbb{A}^2(mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)inmathbb{A}^2(mathbb{C})$, but my goal is for the plane curve to be irreducible.
How do I tell/prove that this curve is irreducible?
Thanks so much!
algebraic-curves
algebraic-curves
asked Dec 13 '18 at 12:09
mathgeenmathgeen
224
asked Dec 13 '18 at 12:09
mathgeenmathgeen
224
edited Dec 13 '18 at 15:04
mathgeen
asked Dec 13 '18 at 12:09
mathgeenmathgeen
224
asked Dec 13 '18 at 12:09
mathgeenmathgeen
224
asked Dec 13 '18 at 12:09
mathgeenmathgeen
224
224
add a comment |
add a comment |
1 Answer
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I start from a change of cordinates:
begin{equation*}
begin{cases}
x^{prime}=x-1\
y^{prime}=y-1
end{cases}
end{equation*}
so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.
The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
begin{equation*}
varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
end{equation*}
because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.
answered Dec 13 '18 at 14:10
Armando j18eosArmando j18eos
2,63511328
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1 Answer
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1 Answer
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$begingroup$
I start from a change of cordinates:
begin{equation*}
begin{cases}
x^{prime}=x-1\
y^{prime}=y-1
end{cases}
end{equation*}
so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.
The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
begin{equation*}
varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
end{equation*}
because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.
answered Dec 13 '18 at 14:10
Armando j18eosArmando j18eos
2,63511328
$endgroup$
add a comment |
$begingroup$
I start from a change of cordinates:
begin{equation*}
begin{cases}
x^{prime}=x-1\
y^{prime}=y-1
end{cases}
end{equation*}
so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.
The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
begin{equation*}
varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
end{equation*}
because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.
answered Dec 13 '18 at 14:10
Armando j18eosArmando j18eos
2,63511328
$endgroup$
add a comment |
$begingroup$
I start from a change of cordinates:
begin{equation*}
begin{cases}
x^{prime}=x-1\
y^{prime}=y-1
end{cases}
end{equation*}
so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.
The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
begin{equation*}
varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
end{equation*}
because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.
answered Dec 13 '18 at 14:10
Armando j18eosArmando j18eos
2,63511328
$endgroup$
I start from a change of cordinates:
begin{equation*}
begin{cases}
x^{prime}=x-1\
y^{prime}=y-1
end{cases}
end{equation*}
so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.
The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
begin{equation*}
varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
end{equation*}
because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.
answered Dec 13 '18 at 14:10
Armando j18eosArmando j18eos
2,63511328
edited Dec 13 '18 at 17:33
answered Dec 13 '18 at 14:10
Armando j18eosArmando j18eos
2,63511328
answered Dec 13 '18 at 14:10
Armando j18eosArmando j18eos
2,63511328
answered Dec 13 '18 at 14:10
Armando j18eosArmando j18eos
2,63511328
2,63511328
add a comment |
add a comment |
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