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
1
active
oldest
votes
$begingroup$
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
3,3021819
$endgroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036311%2fplane-curve-examples%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
3,3021819
$endgroup$
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
3,3021819
$endgroup$
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
3,3021819
$endgroup$
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
3,3021819
answered Dec 12 '18 at 11:02
lEmlEm
3,3021819
answered Dec 12 '18 at 11:02
lEmlEm
3,3021819
answered Dec 12 '18 at 11:02
lEmlEm
3,3021819
3,3021819
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036311%2fplane-curve-examples%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
