Translations of an irreducible plane curve

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Let $f(x,y)in mathbbC[x,y]$ be an irreducible polynomial. Then $f$ corresponds to a curve in $mathbbA^2$. For $(a,b)in mathbbZ^2$, consider $g = f(a+x,b+y)$ the translation of $f$ by $(a,b)$. Under what circumstances does $V(g) = V(f)$?

I can think of examples where this holds, for instance if $f=y-x$ then $(a,a)$ will work for any $a in mathbbZ$. However, I would like to conclude that lines with rational gradient is the only class of examples where this holds.

I think this boils down to looking at $f(x,y) = f(a+x,b+y)$, (looking for the intersection of the varieties) and showing that this cannot happen except when $a=b=0$. I think I have done this for when $f$ is a conic section but I'm struggling for higher degree polynomials. Does anyone have any suggestions, is this perhaps a standard problem in algebraic geometry that I have not found?

abstract-algebra algebraic-geometry ring-theory ideals curves

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asked Jul 30 at 15:34

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Let $f(x,y)in mathbbC[x,y]$ be an irreducible polynomial. Then $f$ corresponds to a curve in $mathbbA^2$. For $(a,b)in mathbbZ^2$, consider $g = f(a+x,b+y)$ the translation of $f$ by $(a,b)$. Under what circumstances does $V(g) = V(f)$?

I can think of examples where this holds, for instance if $f=y-x$ then $(a,a)$ will work for any $a in mathbbZ$. However, I would like to conclude that lines with rational gradient is the only class of examples where this holds.

I think this boils down to looking at $f(x,y) = f(a+x,b+y)$, (looking for the intersection of the varieties) and showing that this cannot happen except when $a=b=0$. I think I have done this for when $f$ is a conic section but I'm struggling for higher degree polynomials. Does anyone have any suggestions, is this perhaps a standard problem in algebraic geometry that I have not found?

abstract-algebra algebraic-geometry ring-theory ideals curves

share|cite|improve this question

asked Jul 30 at 15:34

Help please

22319

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $f(x,y)in mathbbC[x,y]$ be an irreducible polynomial. Then $f$ corresponds to a curve in $mathbbA^2$. For $(a,b)in mathbbZ^2$, consider $g = f(a+x,b+y)$ the translation of $f$ by $(a,b)$. Under what circumstances does $V(g) = V(f)$?

I can think of examples where this holds, for instance if $f=y-x$ then $(a,a)$ will work for any $a in mathbbZ$. However, I would like to conclude that lines with rational gradient is the only class of examples where this holds.

I think this boils down to looking at $f(x,y) = f(a+x,b+y)$, (looking for the intersection of the varieties) and showing that this cannot happen except when $a=b=0$. I think I have done this for when $f$ is a conic section but I'm struggling for higher degree polynomials. Does anyone have any suggestions, is this perhaps a standard problem in algebraic geometry that I have not found?

abstract-algebra algebraic-geometry ring-theory ideals curves

share|cite|improve this question

asked Jul 30 at 15:34

Help please

22319

Let $f(x,y)in mathbbC[x,y]$ be an irreducible polynomial. Then $f$ corresponds to a curve in $mathbbA^2$. For $(a,b)in mathbbZ^2$, consider $g = f(a+x,b+y)$ the translation of $f$ by $(a,b)$. Under what circumstances does $V(g) = V(f)$?

I can think of examples where this holds, for instance if $f=y-x$ then $(a,a)$ will work for any $a in mathbbZ$. However, I would like to conclude that lines with rational gradient is the only class of examples where this holds.

I think this boils down to looking at $f(x,y) = f(a+x,b+y)$, (looking for the intersection of the varieties) and showing that this cannot happen except when $a=b=0$. I think I have done this for when $f$ is a conic section but I'm struggling for higher degree polynomials. Does anyone have any suggestions, is this perhaps a standard problem in algebraic geometry that I have not found?

abstract-algebra algebraic-geometry ring-theory ideals curves

share|cite|improve this question

asked Jul 30 at 15:34

Help please

22319

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 15:34

Help please

22319

asked Jul 30 at 15:34

Help please

22319

asked Jul 30 at 15:34

Help please

22319

22319

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1 Answer
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Yes, your impression is correct. This can only happen when $V(f)$ is a line with slope $b/a$.

Let $P=(x_0,y_0)$ be an arbitrary point on $V(f)$. Let $L$ be the line with slope
$b/a$ through $P$. It follows that all the points $P_n=(x_0+na,y_0+nb)$, $ninBbbZ$, are on $V(f)$ (induction on $|n|$). Therefore $V(f)$ and $L$
intersect at infinitely many point. This is in violation of Bezout's theorem stating that, unless $V(f)=L$, the intersection $V(f)cap L$ has cardinality $deg f$ (counted projectively and with multiplicities).

---

Observe that the characteristic of the ground field plays a role here. In characteristic $p$ we have counterexamples. For example, Artin-Schreier curves $C:y^p-y=f(x)$ have the property that $(x_0,y_0)in C$ if and only $(x_0,y_0+1)in C$.

share|cite|improve this answer

edited Jul 31 at 6:58

answered Jul 30 at 15:46

Jyrki Lahtonen

105k12161355

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are uncomfortable with Bezout, you can turn the argument into one showing that a univariate polynomial has infinitely many zeros simply by plugging in a parametrization of $L$ into $f$. I gotta rush now. The local Old Farts' beach volleyball club is waiting for their scapegoat.
  – Jyrki Lahtonen
  Jul 30 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am happy with Bezout's. Thank you very much for the elegant solution, I hope you enjoy your volleyball!
  – Help please
  Jul 30 at 15:57
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Yes, your impression is correct. This can only happen when $V(f)$ is a line with slope $b/a$.

Let $P=(x_0,y_0)$ be an arbitrary point on $V(f)$. Let $L$ be the line with slope
$b/a$ through $P$. It follows that all the points $P_n=(x_0+na,y_0+nb)$, $ninBbbZ$, are on $V(f)$ (induction on $|n|$). Therefore $V(f)$ and $L$
intersect at infinitely many point. This is in violation of Bezout's theorem stating that, unless $V(f)=L$, the intersection $V(f)cap L$ has cardinality $deg f$ (counted projectively and with multiplicities).

---

Observe that the characteristic of the ground field plays a role here. In characteristic $p$ we have counterexamples. For example, Artin-Schreier curves $C:y^p-y=f(x)$ have the property that $(x_0,y_0)in C$ if and only $(x_0,y_0+1)in C$.

share|cite|improve this answer

edited Jul 31 at 6:58

answered Jul 30 at 15:46

Jyrki Lahtonen

105k12161355

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are uncomfortable with Bezout, you can turn the argument into one showing that a univariate polynomial has infinitely many zeros simply by plugging in a parametrization of $L$ into $f$. I gotta rush now. The local Old Farts' beach volleyball club is waiting for their scapegoat.
  – Jyrki Lahtonen
  Jul 30 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am happy with Bezout's. Thank you very much for the elegant solution, I hope you enjoy your volleyball!
  – Help please
  Jul 30 at 15:57
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Yes, your impression is correct. This can only happen when $V(f)$ is a line with slope $b/a$.

Let $P=(x_0,y_0)$ be an arbitrary point on $V(f)$. Let $L$ be the line with slope
$b/a$ through $P$. It follows that all the points $P_n=(x_0+na,y_0+nb)$, $ninBbbZ$, are on $V(f)$ (induction on $|n|$). Therefore $V(f)$ and $L$
intersect at infinitely many point. This is in violation of Bezout's theorem stating that, unless $V(f)=L$, the intersection $V(f)cap L$ has cardinality $deg f$ (counted projectively and with multiplicities).

---

Observe that the characteristic of the ground field plays a role here. In characteristic $p$ we have counterexamples. For example, Artin-Schreier curves $C:y^p-y=f(x)$ have the property that $(x_0,y_0)in C$ if and only $(x_0,y_0+1)in C$.

share|cite|improve this answer

edited Jul 31 at 6:58

answered Jul 30 at 15:46

Jyrki Lahtonen

105k12161355

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are uncomfortable with Bezout, you can turn the argument into one showing that a univariate polynomial has infinitely many zeros simply by plugging in a parametrization of $L$ into $f$. I gotta rush now. The local Old Farts' beach volleyball club is waiting for their scapegoat.
  – Jyrki Lahtonen
  Jul 30 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am happy with Bezout's. Thank you very much for the elegant solution, I hope you enjoy your volleyball!
  – Help please
  Jul 30 at 15:57
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Yes, your impression is correct. This can only happen when $V(f)$ is a line with slope $b/a$.

Let $P=(x_0,y_0)$ be an arbitrary point on $V(f)$. Let $L$ be the line with slope
$b/a$ through $P$. It follows that all the points $P_n=(x_0+na,y_0+nb)$, $ninBbbZ$, are on $V(f)$ (induction on $|n|$). Therefore $V(f)$ and $L$
intersect at infinitely many point. This is in violation of Bezout's theorem stating that, unless $V(f)=L$, the intersection $V(f)cap L$ has cardinality $deg f$ (counted projectively and with multiplicities).

---

Observe that the characteristic of the ground field plays a role here. In characteristic $p$ we have counterexamples. For example, Artin-Schreier curves $C:y^p-y=f(x)$ have the property that $(x_0,y_0)in C$ if and only $(x_0,y_0+1)in C$.

share|cite|improve this answer

edited Jul 31 at 6:58

answered Jul 30 at 15:46

Jyrki Lahtonen

105k12161355

Yes, your impression is correct. This can only happen when $V(f)$ is a line with slope $b/a$.

Let $P=(x_0,y_0)$ be an arbitrary point on $V(f)$. Let $L$ be the line with slope
$b/a$ through $P$. It follows that all the points $P_n=(x_0+na,y_0+nb)$, $ninBbbZ$, are on $V(f)$ (induction on $|n|$). Therefore $V(f)$ and $L$
intersect at infinitely many point. This is in violation of Bezout's theorem stating that, unless $V(f)=L$, the intersection $V(f)cap L$ has cardinality $deg f$ (counted projectively and with multiplicities).

---

Observe that the characteristic of the ground field plays a role here. In characteristic $p$ we have counterexamples. For example, Artin-Schreier curves $C:y^p-y=f(x)$ have the property that $(x_0,y_0)in C$ if and only $(x_0,y_0+1)in C$.

share|cite|improve this answer

edited Jul 31 at 6:58

answered Jul 30 at 15:46

Jyrki Lahtonen

105k12161355

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 6:58

edited Jul 31 at 6:58

edited Jul 31 at 6:58

answered Jul 30 at 15:46

Jyrki Lahtonen

105k12161355

answered Jul 30 at 15:46

Jyrki Lahtonen

105k12161355

answered Jul 30 at 15:46

Jyrki Lahtonen

105k12161355

105k12161355

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are uncomfortable with Bezout, you can turn the argument into one showing that a univariate polynomial has infinitely many zeros simply by plugging in a parametrization of $L$ into $f$. I gotta rush now. The local Old Farts' beach volleyball club is waiting for their scapegoat.
  – Jyrki Lahtonen
  Jul 30 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am happy with Bezout's. Thank you very much for the elegant solution, I hope you enjoy your volleyball!
  – Help please
  Jul 30 at 15:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are uncomfortable with Bezout, you can turn the argument into one showing that a univariate polynomial has infinitely many zeros simply by plugging in a parametrization of $L$ into $f$. I gotta rush now. The local Old Farts' beach volleyball club is waiting for their scapegoat.
  – Jyrki Lahtonen
  Jul 30 at 15:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am happy with Bezout's. Thank you very much for the elegant solution, I hope you enjoy your volleyball!
  – Help please
  Jul 30 at 15:57
  

  
  
  
  

If you are uncomfortable with Bezout, you can turn the argument into one showing that a univariate polynomial has infinitely many zeros simply by plugging in a parametrization of $L$ into $f$. I gotta rush now. The local Old Farts' beach volleyball club is waiting for their scapegoat.
– Jyrki Lahtonen
Jul 30 at 15:50

If you are uncomfortable with Bezout, you can turn the argument into one showing that a univariate polynomial has infinitely many zeros simply by plugging in a parametrization of $L$ into $f$. I gotta rush now. The local Old Farts' beach volleyball club is waiting for their scapegoat.
– Jyrki Lahtonen
Jul 30 at 15:50

I am happy with Bezout's. Thank you very much for the elegant solution, I hope you enjoy your volleyball!
– Help please
Jul 30 at 15:57

I am happy with Bezout's. Thank you very much for the elegant solution, I hope you enjoy your volleyball!
– Help please
Jul 30 at 15:57

add a comment | 

 

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