Plane Curve Examples
$begingroup$
teaching myself about plane curves, both affine and projective, and I was hoping to gain some exposition via some examples, if anyone can help me out.
First, I am thinking about irreducible plane curves and their tangent lines, and I was hoping to get an example of such a curve $C$ in complex affine 2-space $mathbb{A}^2(mathbb{C})$, say with two different tangent lines at the same point $(1,1)$.
Second, it would be great to get an example of a plane curve $C$ that is singular at the homogeneous coordinates $(1:0:0)$, $(0:1:0)$, and $(0:0:1)$ in complex projective 2-space $mathbb{P}^2(mathbb{C})$.
Thanks so much in advance. I have found that these concepts make a lot more sense when I have specific examples to look at, and sadly the material I am reading online is mostly text with few visual examples.
curves
asked Dec 12 '18 at 6:19
mathgeenmathgeen
224
$endgroup$
add a comment |
$begingroup$
teaching myself about plane curves, both affine and projective, and I was hoping to gain some exposition via some examples, if anyone can help me out.
First, I am thinking about irreducible plane curves and their tangent lines, and I was hoping to get an example of such a curve $C$ in complex affine 2-space $mathbb{A}^2(mathbb{C})$, say with two different tangent lines at the same point $(1,1)$.
Second, it would be great to get an example of a plane curve $C$ that is singular at the homogeneous coordinates $(1:0:0)$, $(0:1:0)$, and $(0:0:1)$ in complex projective 2-space $mathbb{P}^2(mathbb{C})$.
Thanks so much in advance. I have found that these concepts make a lot more sense when I have specific examples to look at, and sadly the material I am reading online is mostly text with few visual examples.
curves
asked Dec 12 '18 at 6:19
mathgeenmathgeen
224
$endgroup$
add a comment |
$begingroup$
teaching myself about plane curves, both affine and projective, and I was hoping to gain some exposition via some examples, if anyone can help me out.
First, I am thinking about irreducible plane curves and their tangent lines, and I was hoping to get an example of such a curve $C$ in complex affine 2-space $mathbb{A}^2(mathbb{C})$, say with two different tangent lines at the same point $(1,1)$.
Second, it would be great to get an example of a plane curve $C$ that is singular at the homogeneous coordinates $(1:0:0)$, $(0:1:0)$, and $(0:0:1)$ in complex projective 2-space $mathbb{P}^2(mathbb{C})$.
Thanks so much in advance. I have found that these concepts make a lot more sense when I have specific examples to look at, and sadly the material I am reading online is mostly text with few visual examples.
curves
asked Dec 12 '18 at 6:19
mathgeenmathgeen
224
$endgroup$
teaching myself about plane curves, both affine and projective, and I was hoping to gain some exposition via some examples, if anyone can help me out.
First, I am thinking about irreducible plane curves and their tangent lines, and I was hoping to get an example of such a curve $C$ in complex affine 2-space $mathbb{A}^2(mathbb{C})$, say with two different tangent lines at the same point $(1,1)$.
Second, it would be great to get an example of a plane curve $C$ that is singular at the homogeneous coordinates $(1:0:0)$, $(0:1:0)$, and $(0:0:1)$ in complex projective 2-space $mathbb{P}^2(mathbb{C})$.
Thanks so much in advance. I have found that these concepts make a lot more sense when I have specific examples to look at, and sadly the material I am reading online is mostly text with few visual examples.
curves
curves
asked Dec 12 '18 at 6:19
mathgeenmathgeen
224
asked Dec 12 '18 at 6:19
mathgeenmathgeen
224
edited Dec 12 '18 at 12:08
mathgeen
asked Dec 12 '18 at 6:19
mathgeenmathgeen
224
asked Dec 12 '18 at 6:19
mathgeenmathgeen
224
asked Dec 12 '18 at 6:19
mathgeenmathgeen
224
224
add a comment |
add a comment |
1 Answer
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Let $C={(x,y)inmathbb C:f(x,y)=0}$ defines an irreducible plane curve in $mathbb A^2_{mathbb C}$, $(1,1)in C$ implies that $f(x,y)=(x-1)p(x,y)+(y-1)q(x,y)$. $C$ is singular $(1,1)$ implies that both partial derivatives vanish, so we obtain $$begin{align}f_x(x,y)&=p(x,y)+(x-1)p_x(x,y)+(y-1)q_x(x,y) \ f_y(x,y)&=(x-1)p_y(x,y)+(y-1)q_y(x,y)+q(x,y)
end{align}$$
Evaluating at $(1,1)$ gives $p(1,1)=q(1,1)=0$, hence we can expand further $f(x,y)=a(x-1)^2+b(x-1)(y-1)+c(y-1)^2$ where $a,b,cinmathbb C[x,y]$. The simplest of such curve could be $f(x,y)=x(x-1)^2+(y-1)^2$, you could try to show that it is irreducible by contradiction.
About you second question, the idea is similar, suppose the curve is defined by $F(x,y,z)=0$, then you would need to solve $F_x(p_i)=F_y(p_i)=F_z(p_i)=0$ for $p_i$ the three points you specified. Now observe that if $F(x,y,z)=G(x,y)+H(y,z)+J(x,z)$ for homogeneous polynomials $G,H,J$ with same degree, then we can guarantee that $F$ vanishes at the three points if $H(0,1)=J(0,1)$, $G(1,0)=J(1,0)$, etc. This hinted that maybe we should look at polynomials like $F(x,y,z)=xy+yz+xz$. However this curve is not quite singular. The usual way of turning a curve singular is to raise the term of higher power, and turns out indeed $F(x,y,z)=x^2y^2+y^2z^2+x^2z^2$ works.
It is hard to visualize these curves because they live in at least $4$-(real) dimensional space, so sometimes it could be helpful to look at the real version of curves in hope of gaining some insights.
answered Dec 12 '18 at 11:02
lEmlEm
3,2621719
$endgroup$
1
$begingroup$
@mathgeen Feel free to post any questions here on this site, I will try to answer them.
$endgroup$
– lEm
Dec 13 '18 at 1:56
$begingroup$
@IEm could you give some insight into how to prove that $f(x,y)$ is irreducible in the first part?
$endgroup$
– mathgeen
Dec 13 '18 at 11:27
$begingroup$
@mathgeen The polynomial is of degree 3, so if it is reducible you can express it as a product of a degree 1 and a degree 2 polynomial, try to expand $(ax+by+c)(dx^2+ey^2+fxy+gx+hy+i)=f(x,y)$ and compare the coefficients.
$endgroup$
– lEm
Dec 14 '18 at 1:51
add a comment |
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$begingroup$
Let $C={(x,y)inmathbb C:f(x,y)=0}$ defines an irreducible plane curve in $mathbb A^2_{mathbb C}$, $(1,1)in C$ implies that $f(x,y)=(x-1)p(x,y)+(y-1)q(x,y)$. $C$ is singular $(1,1)$ implies that both partial derivatives vanish, so we obtain $$begin{align}f_x(x,y)&=p(x,y)+(x-1)p_x(x,y)+(y-1)q_x(x,y) \ f_y(x,y)&=(x-1)p_y(x,y)+(y-1)q_y(x,y)+q(x,y)
end{align}$$
Evaluating at $(1,1)$ gives $p(1,1)=q(1,1)=0$, hence we can expand further $f(x,y)=a(x-1)^2+b(x-1)(y-1)+c(y-1)^2$ where $a,b,cinmathbb C[x,y]$. The simplest of such curve could be $f(x,y)=x(x-1)^2+(y-1)^2$, you could try to show that it is irreducible by contradiction.
About you second question, the idea is similar, suppose the curve is defined by $F(x,y,z)=0$, then you would need to solve $F_x(p_i)=F_y(p_i)=F_z(p_i)=0$ for $p_i$ the three points you specified. Now observe that if $F(x,y,z)=G(x,y)+H(y,z)+J(x,z)$ for homogeneous polynomials $G,H,J$ with same degree, then we can guarantee that $F$ vanishes at the three points if $H(0,1)=J(0,1)$, $G(1,0)=J(1,0)$, etc. This hinted that maybe we should look at polynomials like $F(x,y,z)=xy+yz+xz$. However this curve is not quite singular. The usual way of turning a curve singular is to raise the term of higher power, and turns out indeed $F(x,y,z)=x^2y^2+y^2z^2+x^2z^2$ works.
It is hard to visualize these curves because they live in at least $4$-(real) dimensional space, so sometimes it could be helpful to look at the real version of curves in hope of gaining some insights.
answered Dec 12 '18 at 11:02
lEmlEm
3,2621719
$endgroup$
1
$begingroup$
@mathgeen Feel free to post any questions here on this site, I will try to answer them.
$endgroup$
– lEm
Dec 13 '18 at 1:56
$begingroup$
@IEm could you give some insight into how to prove that $f(x,y)$ is irreducible in the first part?
$endgroup$
– mathgeen
Dec 13 '18 at 11:27
$begingroup$
@mathgeen The polynomial is of degree 3, so if it is reducible you can express it as a product of a degree 1 and a degree 2 polynomial, try to expand $(ax+by+c)(dx^2+ey^2+fxy+gx+hy+i)=f(x,y)$ and compare the coefficients.
$endgroup$
– lEm
Dec 14 '18 at 1:51
add a comment |
$begingroup$
Let $C={(x,y)inmathbb C:f(x,y)=0}$ defines an irreducible plane curve in $mathbb A^2_{mathbb C}$, $(1,1)in C$ implies that $f(x,y)=(x-1)p(x,y)+(y-1)q(x,y)$. $C$ is singular $(1,1)$ implies that both partial derivatives vanish, so we obtain $$begin{align}f_x(x,y)&=p(x,y)+(x-1)p_x(x,y)+(y-1)q_x(x,y) \ f_y(x,y)&=(x-1)p_y(x,y)+(y-1)q_y(x,y)+q(x,y)
end{align}$$
Evaluating at $(1,1)$ gives $p(1,1)=q(1,1)=0$, hence we can expand further $f(x,y)=a(x-1)^2+b(x-1)(y-1)+c(y-1)^2$ where $a,b,cinmathbb C[x,y]$. The simplest of such curve could be $f(x,y)=x(x-1)^2+(y-1)^2$, you could try to show that it is irreducible by contradiction.
About you second question, the idea is similar, suppose the curve is defined by $F(x,y,z)=0$, then you would need to solve $F_x(p_i)=F_y(p_i)=F_z(p_i)=0$ for $p_i$ the three points you specified. Now observe that if $F(x,y,z)=G(x,y)+H(y,z)+J(x,z)$ for homogeneous polynomials $G,H,J$ with same degree, then we can guarantee that $F$ vanishes at the three points if $H(0,1)=J(0,1)$, $G(1,0)=J(1,0)$, etc. This hinted that maybe we should look at polynomials like $F(x,y,z)=xy+yz+xz$. However this curve is not quite singular. The usual way of turning a curve singular is to raise the term of higher power, and turns out indeed $F(x,y,z)=x^2y^2+y^2z^2+x^2z^2$ works.
It is hard to visualize these curves because they live in at least $4$-(real) dimensional space, so sometimes it could be helpful to look at the real version of curves in hope of gaining some insights.
answered Dec 12 '18 at 11:02
lEmlEm
3,2621719
$endgroup$
1
$begingroup$
@mathgeen Feel free to post any questions here on this site, I will try to answer them.
$endgroup$
– lEm
Dec 13 '18 at 1:56
$begingroup$
@IEm could you give some insight into how to prove that $f(x,y)$ is irreducible in the first part?
$endgroup$
– mathgeen
Dec 13 '18 at 11:27
$begingroup$
@mathgeen The polynomial is of degree 3, so if it is reducible you can express it as a product of a degree 1 and a degree 2 polynomial, try to expand $(ax+by+c)(dx^2+ey^2+fxy+gx+hy+i)=f(x,y)$ and compare the coefficients.
$endgroup$
– lEm
Dec 14 '18 at 1:51
add a comment |
$begingroup$
Let $C={(x,y)inmathbb C:f(x,y)=0}$ defines an irreducible plane curve in $mathbb A^2_{mathbb C}$, $(1,1)in C$ implies that $f(x,y)=(x-1)p(x,y)+(y-1)q(x,y)$. $C$ is singular $(1,1)$ implies that both partial derivatives vanish, so we obtain $$begin{align}f_x(x,y)&=p(x,y)+(x-1)p_x(x,y)+(y-1)q_x(x,y) \ f_y(x,y)&=(x-1)p_y(x,y)+(y-1)q_y(x,y)+q(x,y)
end{align}$$
Evaluating at $(1,1)$ gives $p(1,1)=q(1,1)=0$, hence we can expand further $f(x,y)=a(x-1)^2+b(x-1)(y-1)+c(y-1)^2$ where $a,b,cinmathbb C[x,y]$. The simplest of such curve could be $f(x,y)=x(x-1)^2+(y-1)^2$, you could try to show that it is irreducible by contradiction.
About you second question, the idea is similar, suppose the curve is defined by $F(x,y,z)=0$, then you would need to solve $F_x(p_i)=F_y(p_i)=F_z(p_i)=0$ for $p_i$ the three points you specified. Now observe that if $F(x,y,z)=G(x,y)+H(y,z)+J(x,z)$ for homogeneous polynomials $G,H,J$ with same degree, then we can guarantee that $F$ vanishes at the three points if $H(0,1)=J(0,1)$, $G(1,0)=J(1,0)$, etc. This hinted that maybe we should look at polynomials like $F(x,y,z)=xy+yz+xz$. However this curve is not quite singular. The usual way of turning a curve singular is to raise the term of higher power, and turns out indeed $F(x,y,z)=x^2y^2+y^2z^2+x^2z^2$ works.
It is hard to visualize these curves because they live in at least $4$-(real) dimensional space, so sometimes it could be helpful to look at the real version of curves in hope of gaining some insights.
answered Dec 12 '18 at 11:02
lEmlEm
3,2621719
$endgroup$
Let $C={(x,y)inmathbb C:f(x,y)=0}$ defines an irreducible plane curve in $mathbb A^2_{mathbb C}$, $(1,1)in C$ implies that $f(x,y)=(x-1)p(x,y)+(y-1)q(x,y)$. $C$ is singular $(1,1)$ implies that both partial derivatives vanish, so we obtain $$begin{align}f_x(x,y)&=p(x,y)+(x-1)p_x(x,y)+(y-1)q_x(x,y) \ f_y(x,y)&=(x-1)p_y(x,y)+(y-1)q_y(x,y)+q(x,y)
end{align}$$
Evaluating at $(1,1)$ gives $p(1,1)=q(1,1)=0$, hence we can expand further $f(x,y)=a(x-1)^2+b(x-1)(y-1)+c(y-1)^2$ where $a,b,cinmathbb C[x,y]$. The simplest of such curve could be $f(x,y)=x(x-1)^2+(y-1)^2$, you could try to show that it is irreducible by contradiction.
About you second question, the idea is similar, suppose the curve is defined by $F(x,y,z)=0$, then you would need to solve $F_x(p_i)=F_y(p_i)=F_z(p_i)=0$ for $p_i$ the three points you specified. Now observe that if $F(x,y,z)=G(x,y)+H(y,z)+J(x,z)$ for homogeneous polynomials $G,H,J$ with same degree, then we can guarantee that $F$ vanishes at the three points if $H(0,1)=J(0,1)$, $G(1,0)=J(1,0)$, etc. This hinted that maybe we should look at polynomials like $F(x,y,z)=xy+yz+xz$. However this curve is not quite singular. The usual way of turning a curve singular is to raise the term of higher power, and turns out indeed $F(x,y,z)=x^2y^2+y^2z^2+x^2z^2$ works.
It is hard to visualize these curves because they live in at least $4$-(real) dimensional space, so sometimes it could be helpful to look at the real version of curves in hope of gaining some insights.
answered Dec 12 '18 at 11:02
lEmlEm
3,2621719
answered Dec 12 '18 at 11:02
lEmlEm
3,2621719
answered Dec 12 '18 at 11:02
lEmlEm
3,2621719
answered Dec 12 '18 at 11:02
lEmlEm
3,2621719
3,2621719
1
$begingroup$
@mathgeen Feel free to post any questions here on this site, I will try to answer them.
$endgroup$
– lEm
Dec 13 '18 at 1:56
$begingroup$
@IEm could you give some insight into how to prove that $f(x,y)$ is irreducible in the first part?
$endgroup$
– mathgeen
Dec 13 '18 at 11:27
$begingroup$
@mathgeen The polynomial is of degree 3, so if it is reducible you can express it as a product of a degree 1 and a degree 2 polynomial, try to expand $(ax+by+c)(dx^2+ey^2+fxy+gx+hy+i)=f(x,y)$ and compare the coefficients.
$endgroup$
– lEm
Dec 14 '18 at 1:51
add a comment |
1
$begingroup$
@mathgeen Feel free to post any questions here on this site, I will try to answer them.
$endgroup$
– lEm
Dec 13 '18 at 1:56
$begingroup$
@IEm could you give some insight into how to prove that $f(x,y)$ is irreducible in the first part?
$endgroup$
– mathgeen
Dec 13 '18 at 11:27
$begingroup$
@mathgeen The polynomial is of degree 3, so if it is reducible you can express it as a product of a degree 1 and a degree 2 polynomial, try to expand $(ax+by+c)(dx^2+ey^2+fxy+gx+hy+i)=f(x,y)$ and compare the coefficients.
$endgroup$
– lEm
Dec 14 '18 at 1:51
1
1
$begingroup$
@mathgeen Feel free to post any questions here on this site, I will try to answer them.
$endgroup$
– lEm
Dec 13 '18 at 1:56
$begingroup$
@mathgeen Feel free to post any questions here on this site, I will try to answer them.
$endgroup$
– lEm
Dec 13 '18 at 1:56
$begingroup$
@IEm could you give some insight into how to prove that $f(x,y)$ is irreducible in the first part?
$endgroup$
– mathgeen
Dec 13 '18 at 11:27
$begingroup$
@IEm could you give some insight into how to prove that $f(x,y)$ is irreducible in the first part?
$endgroup$
– mathgeen
Dec 13 '18 at 11:27
$begingroup$
@mathgeen The polynomial is of degree 3, so if it is reducible you can express it as a product of a degree 1 and a degree 2 polynomial, try to expand $(ax+by+c)(dx^2+ey^2+fxy+gx+hy+i)=f(x,y)$ and compare the coefficients.
$endgroup$
– lEm
Dec 14 '18 at 1:51
$begingroup$
@mathgeen The polynomial is of degree 3, so if it is reducible you can express it as a product of a degree 1 and a degree 2 polynomial, try to expand $(ax+by+c)(dx^2+ey^2+fxy+gx+hy+i)=f(x,y)$ and compare the coefficients.
$endgroup$
– lEm
Dec 14 '18 at 1:51
add a comment |
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