For a line $L$ and an algebraic curve $C$ of an irreducible polynomial, prove $C cap L$ contains at most d...
0
$begingroup$
Artin Algebra Chapter 11
This has been answered here. My questions are about the solution of Brian Bi:
- By stronger, does he mean that $C ne L$ and $f$ is irreducible $implies l nmid f?$ If so, here is my proof. Is it correct?
Since $f$ is irreducible, $l$ divides $f$ if and only if $frac f l =: k$ is a nonzero constant. Then
$$C={f=0}={kl=0}={l=0}=L$$
- Is $a=0=b$ ruled out by definition of linear?
If so, is this in book, or is it understood because $a=0=b$ gives either $emptyset$ or $mathbb C^2$? If not, how do we prove $|C cap L| le d$ in the case where $c=0$? I was able to show just $C cap L = C$.
- Why is it that $g$ has degree $le d$?
I think this has something to do with:
- irreducibility of $f$ and
- what it means for $d$ to be the degree of $f$.
I think of writing $f=e_0(y)+e_1(y)x^1 + ... + e_m(y)x^m$. If each monomial has degree $le d$, then because $y=px+q$ has degree 1, $g$ has degree $le d$, being made up of monomials in one variable that have degree $le d$. Therefore,
- it's irrelevant that $f$ is irreducible, and
- the definition that $d$ is the degree of $f$ is that all its monomials have degree $le d$, and one of them has degree $d$?
abstract-algebra algebraic-geometry polynomials irreducible-polynomials algebraic-curves
$endgroup$
add a comment |
0
$begingroup$
Artin Algebra Chapter 11
This has been answered here. My questions are about the solution of Brian Bi:
- By stronger, does he mean that $C ne L$ and $f$ is irreducible $implies l nmid f?$ If so, here is my proof. Is it correct?
Since $f$ is irreducible, $l$ divides $f$ if and only if $frac f l =: k$ is a nonzero constant. Then
$$C={f=0}={kl=0}={l=0}=L$$
- Is $a=0=b$ ruled out by definition of linear?
If so, is this in book, or is it understood because $a=0=b$ gives either $emptyset$ or $mathbb C^2$? If not, how do we prove $|C cap L| le d$ in the case where $c=0$? I was able to show just $C cap L = C$.
- Why is it that $g$ has degree $le d$?
I think this has something to do with:
- irreducibility of $f$ and
- what it means for $d$ to be the degree of $f$.
I think of writing $f=e_0(y)+e_1(y)x^1 + ... + e_m(y)x^m$. If each monomial has degree $le d$, then because $y=px+q$ has degree 1, $g$ has degree $le d$, being made up of monomials in one variable that have degree $le d$. Therefore,
- it's irrelevant that $f$ is irreducible, and
- the definition that $d$ is the degree of $f$ is that all its monomials have degree $le d$, and one of them has degree $d$?
abstract-algebra algebraic-geometry polynomials irreducible-polynomials algebraic-curves
$endgroup$
3
$begingroup$
If f is not irreducible, then it might happen that L is a component of C (i.e. L is a factor of f), without being all of C. In this case there are infinitely many points in the intersection. This is the stronger thing. L divides f does not jmply that C is L. The third equality is wrong - kl = 0 holds if l is zero OR k is zero (a union of sets). Yes $a = b = 0$ is ruled out in order for it to be a line (one dimensional affine subspace). The degree of f is in terms of monomials like $x^ay^b$, which has degree $a + b$.
$endgroup$
– Lorenzo
Dec 30 '18 at 8:08
$begingroup$
@Lorenzo 1. I am thinking. 2. Thank you! 3. So we know the degree of the monomials that make up $f$. Then what's the degree of $f$? Highest of the degrees?
$endgroup$
– user198044
Dec 30 '18 at 9:15
$begingroup$
@Lorenzo For 1, $k$ is a nonzero constant because $f$ is irreducible, correct?
$endgroup$
– user198044
Dec 30 '18 at 9:16
$begingroup$
@Lorenzo Also for 1, my understanding was that the exercise asks us to show $f$ is irreducible and $C ne L implies |C cap L| le d$ and what Brian Bi tries to show is that $f$ is irreducible and $l nmid f implies |C cap L| le d$. Do you mean it's actually that Brian Bi is showing that $l nmid f implies |C cap L| le d$ and therefore the "stronger" is that "$C ne L$ and $f$ is irreducible$ implies l nmid f?$" I think so, so I'll edit that now.
$endgroup$
– user198044
Dec 30 '18 at 9:18
$begingroup$
If $f$ is irreducible (and of degree $> 1$), then there is no $k$ so that $f= kl$. If $f$ is irreducible and we write $f = kl$, then $k$ or $l$ is a constant. Yes the degree of $f$ is the highest of the degrees. Probably reading a book like Fultons' algebraic curves will help clarify this stuff: math.lsa.umich.edu/~wfulton/CurveBook.pdf
$endgroup$
– Lorenzo
Dec 30 '18 at 23:23
add a comment |
0
0
0
$begingroup$
Artin Algebra Chapter 11
This has been answered here. My questions are about the solution of Brian Bi:
- By stronger, does he mean that $C ne L$ and $f$ is irreducible $implies l nmid f?$ If so, here is my proof. Is it correct?
Since $f$ is irreducible, $l$ divides $f$ if and only if $frac f l =: k$ is a nonzero constant. Then
$$C={f=0}={kl=0}={l=0}=L$$
- Is $a=0=b$ ruled out by definition of linear?
If so, is this in book, or is it understood because $a=0=b$ gives either $emptyset$ or $mathbb C^2$? If not, how do we prove $|C cap L| le d$ in the case where $c=0$? I was able to show just $C cap L = C$.
- Why is it that $g$ has degree $le d$?
I think this has something to do with:
- irreducibility of $f$ and
- what it means for $d$ to be the degree of $f$.
I think of writing $f=e_0(y)+e_1(y)x^1 + ... + e_m(y)x^m$. If each monomial has degree $le d$, then because $y=px+q$ has degree 1, $g$ has degree $le d$, being made up of monomials in one variable that have degree $le d$. Therefore,
- it's irrelevant that $f$ is irreducible, and
- the definition that $d$ is the degree of $f$ is that all its monomials have degree $le d$, and one of them has degree $d$?
abstract-algebra algebraic-geometry polynomials irreducible-polynomials algebraic-curves
$endgroup$
Artin Algebra Chapter 11
This has been answered here. My questions are about the solution of Brian Bi:
- By stronger, does he mean that $C ne L$ and $f$ is irreducible $implies l nmid f?$ If so, here is my proof. Is it correct?
Since $f$ is irreducible, $l$ divides $f$ if and only if $frac f l =: k$ is a nonzero constant. Then
$$C={f=0}={kl=0}={l=0}=L$$
- Is $a=0=b$ ruled out by definition of linear?
If so, is this in book, or is it understood because $a=0=b$ gives either $emptyset$ or $mathbb C^2$? If not, how do we prove $|C cap L| le d$ in the case where $c=0$? I was able to show just $C cap L = C$.
- Why is it that $g$ has degree $le d$?
I think this has something to do with:
- irreducibility of $f$ and
- what it means for $d$ to be the degree of $f$.
I think of writing $f=e_0(y)+e_1(y)x^1 + ... + e_m(y)x^m$. If each monomial has degree $le d$, then because $y=px+q$ has degree 1, $g$ has degree $le d$, being made up of monomials in one variable that have degree $le d$. Therefore,
- it's irrelevant that $f$ is irreducible, and
- the definition that $d$ is the degree of $f$ is that all its monomials have degree $le d$, and one of them has degree $d$?
abstract-algebra algebraic-geometry polynomials irreducible-polynomials algebraic-curves
abstract-algebra algebraic-geometry polynomials irreducible-polynomials algebraic-curves
edited Dec 30 '18 at 9:22
asked Dec 30 '18 at 7:42
user198044
3
$begingroup$
If f is not irreducible, then it might happen that L is a component of C (i.e. L is a factor of f), without being all of C. In this case there are infinitely many points in the intersection. This is the stronger thing. L divides f does not jmply that C is L. The third equality is wrong - kl = 0 holds if l is zero OR k is zero (a union of sets). Yes $a = b = 0$ is ruled out in order for it to be a line (one dimensional affine subspace). The degree of f is in terms of monomials like $x^ay^b$, which has degree $a + b$.
$endgroup$
– Lorenzo
Dec 30 '18 at 8:08
$begingroup$
@Lorenzo 1. I am thinking. 2. Thank you! 3. So we know the degree of the monomials that make up $f$. Then what's the degree of $f$? Highest of the degrees?
$endgroup$
– user198044
Dec 30 '18 at 9:15
$begingroup$
@Lorenzo For 1, $k$ is a nonzero constant because $f$ is irreducible, correct?
$endgroup$
– user198044
Dec 30 '18 at 9:16
$begingroup$
@Lorenzo Also for 1, my understanding was that the exercise asks us to show $f$ is irreducible and $C ne L implies |C cap L| le d$ and what Brian Bi tries to show is that $f$ is irreducible and $l nmid f implies |C cap L| le d$. Do you mean it's actually that Brian Bi is showing that $l nmid f implies |C cap L| le d$ and therefore the "stronger" is that "$C ne L$ and $f$ is irreducible$ implies l nmid f?$" I think so, so I'll edit that now.
$endgroup$
– user198044
Dec 30 '18 at 9:18
$begingroup$
If $f$ is irreducible (and of degree $> 1$), then there is no $k$ so that $f= kl$. If $f$ is irreducible and we write $f = kl$, then $k$ or $l$ is a constant. Yes the degree of $f$ is the highest of the degrees. Probably reading a book like Fultons' algebraic curves will help clarify this stuff: math.lsa.umich.edu/~wfulton/CurveBook.pdf
$endgroup$
– Lorenzo
Dec 30 '18 at 23:23
add a comment |
3
$begingroup$
If f is not irreducible, then it might happen that L is a component of C (i.e. L is a factor of f), without being all of C. In this case there are infinitely many points in the intersection. This is the stronger thing. L divides f does not jmply that C is L. The third equality is wrong - kl = 0 holds if l is zero OR k is zero (a union of sets). Yes $a = b = 0$ is ruled out in order for it to be a line (one dimensional affine subspace). The degree of f is in terms of monomials like $x^ay^b$, which has degree $a + b$.
$endgroup$
– Lorenzo
Dec 30 '18 at 8:08
$begingroup$
@Lorenzo 1. I am thinking. 2. Thank you! 3. So we know the degree of the monomials that make up $f$. Then what's the degree of $f$? Highest of the degrees?
$endgroup$
– user198044
Dec 30 '18 at 9:15
$begingroup$
@Lorenzo For 1, $k$ is a nonzero constant because $f$ is irreducible, correct?
$endgroup$
– user198044
Dec 30 '18 at 9:16
$begingroup$
@Lorenzo Also for 1, my understanding was that the exercise asks us to show $f$ is irreducible and $C ne L implies |C cap L| le d$ and what Brian Bi tries to show is that $f$ is irreducible and $l nmid f implies |C cap L| le d$. Do you mean it's actually that Brian Bi is showing that $l nmid f implies |C cap L| le d$ and therefore the "stronger" is that "$C ne L$ and $f$ is irreducible$ implies l nmid f?$" I think so, so I'll edit that now.
$endgroup$
– user198044
Dec 30 '18 at 9:18
$begingroup$
If $f$ is irreducible (and of degree $> 1$), then there is no $k$ so that $f= kl$. If $f$ is irreducible and we write $f = kl$, then $k$ or $l$ is a constant. Yes the degree of $f$ is the highest of the degrees. Probably reading a book like Fultons' algebraic curves will help clarify this stuff: math.lsa.umich.edu/~wfulton/CurveBook.pdf
$endgroup$
– Lorenzo
Dec 30 '18 at 23:23
3
3
$begingroup$
If f is not irreducible, then it might happen that L is a component of C (i.e. L is a factor of f), without being all of C. In this case there are infinitely many points in the intersection. This is the stronger thing. L divides f does not jmply that C is L. The third equality is wrong - kl = 0 holds if l is zero OR k is zero (a union of sets). Yes $a = b = 0$ is ruled out in order for it to be a line (one dimensional affine subspace). The degree of f is in terms of monomials like $x^ay^b$, which has degree $a + b$.
$endgroup$
– Lorenzo
Dec 30 '18 at 8:08
$begingroup$
If f is not irreducible, then it might happen that L is a component of C (i.e. L is a factor of f), without being all of C. In this case there are infinitely many points in the intersection. This is the stronger thing. L divides f does not jmply that C is L. The third equality is wrong - kl = 0 holds if l is zero OR k is zero (a union of sets). Yes $a = b = 0$ is ruled out in order for it to be a line (one dimensional affine subspace). The degree of f is in terms of monomials like $x^ay^b$, which has degree $a + b$.
$endgroup$
– Lorenzo
Dec 30 '18 at 8:08
$begingroup$
@Lorenzo 1. I am thinking. 2. Thank you! 3. So we know the degree of the monomials that make up $f$. Then what's the degree of $f$? Highest of the degrees?
$endgroup$
– user198044
Dec 30 '18 at 9:15
$begingroup$
@Lorenzo 1. I am thinking. 2. Thank you! 3. So we know the degree of the monomials that make up $f$. Then what's the degree of $f$? Highest of the degrees?
$endgroup$
– user198044
Dec 30 '18 at 9:15
$begingroup$
@Lorenzo For 1, $k$ is a nonzero constant because $f$ is irreducible, correct?
$endgroup$
– user198044
Dec 30 '18 at 9:16
$begingroup$
@Lorenzo For 1, $k$ is a nonzero constant because $f$ is irreducible, correct?
$endgroup$
– user198044
Dec 30 '18 at 9:16
$begingroup$
@Lorenzo Also for 1, my understanding was that the exercise asks us to show $f$ is irreducible and $C ne L implies |C cap L| le d$ and what Brian Bi tries to show is that $f$ is irreducible and $l nmid f implies |C cap L| le d$. Do you mean it's actually that Brian Bi is showing that $l nmid f implies |C cap L| le d$ and therefore the "stronger" is that "$C ne L$ and $f$ is irreducible$ implies l nmid f?$" I think so, so I'll edit that now.
$endgroup$
– user198044
Dec 30 '18 at 9:18
$begingroup$
@Lorenzo Also for 1, my understanding was that the exercise asks us to show $f$ is irreducible and $C ne L implies |C cap L| le d$ and what Brian Bi tries to show is that $f$ is irreducible and $l nmid f implies |C cap L| le d$. Do you mean it's actually that Brian Bi is showing that $l nmid f implies |C cap L| le d$ and therefore the "stronger" is that "$C ne L$ and $f$ is irreducible$ implies l nmid f?$" I think so, so I'll edit that now.
$endgroup$
– user198044
Dec 30 '18 at 9:18
$begingroup$
If $f$ is irreducible (and of degree $> 1$), then there is no $k$ so that $f= kl$. If $f$ is irreducible and we write $f = kl$, then $k$ or $l$ is a constant. Yes the degree of $f$ is the highest of the degrees. Probably reading a book like Fultons' algebraic curves will help clarify this stuff: math.lsa.umich.edu/~wfulton/CurveBook.pdf
$endgroup$
– Lorenzo
Dec 30 '18 at 23:23
$begingroup$
If $f$ is irreducible (and of degree $> 1$), then there is no $k$ so that $f= kl$. If $f$ is irreducible and we write $f = kl$, then $k$ or $l$ is a constant. Yes the degree of $f$ is the highest of the degrees. Probably reading a book like Fultons' algebraic curves will help clarify this stuff: math.lsa.umich.edu/~wfulton/CurveBook.pdf
$endgroup$
– Lorenzo
Dec 30 '18 at 23:23
add a comment |
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3
$begingroup$
If f is not irreducible, then it might happen that L is a component of C (i.e. L is a factor of f), without being all of C. In this case there are infinitely many points in the intersection. This is the stronger thing. L divides f does not jmply that C is L. The third equality is wrong - kl = 0 holds if l is zero OR k is zero (a union of sets). Yes $a = b = 0$ is ruled out in order for it to be a line (one dimensional affine subspace). The degree of f is in terms of monomials like $x^ay^b$, which has degree $a + b$.
$endgroup$
– Lorenzo
Dec 30 '18 at 8:08
$begingroup$
@Lorenzo 1. I am thinking. 2. Thank you! 3. So we know the degree of the monomials that make up $f$. Then what's the degree of $f$? Highest of the degrees?
$endgroup$
– user198044
Dec 30 '18 at 9:15
$begingroup$
@Lorenzo For 1, $k$ is a nonzero constant because $f$ is irreducible, correct?
$endgroup$
– user198044
Dec 30 '18 at 9:16
$begingroup$
@Lorenzo Also for 1, my understanding was that the exercise asks us to show $f$ is irreducible and $C ne L implies |C cap L| le d$ and what Brian Bi tries to show is that $f$ is irreducible and $l nmid f implies |C cap L| le d$. Do you mean it's actually that Brian Bi is showing that $l nmid f implies |C cap L| le d$ and therefore the "stronger" is that "$C ne L$ and $f$ is irreducible$ implies l nmid f?$" I think so, so I'll edit that now.
$endgroup$
– user198044
Dec 30 '18 at 9:18
$begingroup$
If $f$ is irreducible (and of degree $> 1$), then there is no $k$ so that $f= kl$. If $f$ is irreducible and we write $f = kl$, then $k$ or $l$ is a constant. Yes the degree of $f$ is the highest of the degrees. Probably reading a book like Fultons' algebraic curves will help clarify this stuff: math.lsa.umich.edu/~wfulton/CurveBook.pdf
$endgroup$
– Lorenzo
Dec 30 '18 at 23:23