Collinear Galois conjugatesAligned roots of irreducible polynomialsTschirnhaus TransformationRank of sum of Galois conjugates of a matrixLeast prime $p$ such that an irreducible polynomial of degree $n$ has no root modulo $p$?The holomorphic version of Galois theoryCan a product of conjugates be a Pisot number again?Which algebraic relations are possible between algebraic conjugates?Weyl Groups as Galois groupssmall Pisot numbers with real conjugatesGalois group and invariant polynomials
Collinear Galois conjugates
Aligned roots of irreducible polynomialsTschirnhaus TransformationRank of sum of Galois conjugates of a matrixLeast prime $p$ such that an irreducible polynomial of degree $n$ has no root modulo $p$?The holomorphic version of Galois theoryCan a product of conjugates be a Pisot number again?Which algebraic relations are possible between algebraic conjugates?Weyl Groups as Galois groupssmall Pisot numbers with real conjugatesGalois group and invariant polynomials
$begingroup$
This is inspired by this old question, which may provide a bit more background. But the two present questions seem somewhat more fundamental to me.
Let $p$ be an irreducible polynomial with integer coefficients.
- Is it possible that three of the roots of $p$ are collinear on a line in "general position", i.e. which is neither horizontal nor vertical nor through the origin?
Further, for $alphainmathbb R$, denote by $N_p(alpha)$ the number of zeros of $p$ with real part $alpha$.
- If $N_p(alpha)>1$, is it true that $N_p(alpha)$ is always a power of $2$?
cv.complex-variables polynomials galois-theory
asked Aug 12 at 19:57
WolfgangWolfgang
6,4234 gold badges29 silver badges74 bronze badges
$endgroup$
add a comment |
$begingroup$
This is inspired by this old question, which may provide a bit more background. But the two present questions seem somewhat more fundamental to me.
Let $p$ be an irreducible polynomial with integer coefficients.
- Is it possible that three of the roots of $p$ are collinear on a line in "general position", i.e. which is neither horizontal nor vertical nor through the origin?
Further, for $alphainmathbb R$, denote by $N_p(alpha)$ the number of zeros of $p$ with real part $alpha$.
- If $N_p(alpha)>1$, is it true that $N_p(alpha)$ is always a power of $2$?
cv.complex-variables polynomials galois-theory
asked Aug 12 at 19:57
WolfgangWolfgang
6,4234 gold badges29 silver badges74 bronze badges
$endgroup$
4
$begingroup$
You should probably exclude linear shifts here, as you did in the other question—in other words, replace "through the origin" with "through an integer". (Also, 1 is a power of 2....)
$endgroup$
– Greg Martin
Aug 12 at 20:10
$begingroup$
A somewhat related question from Math.SE where the answer shows that it is impossible for an irredecible polynomial from $BbbZ[x]$ to have three roots in an arithmetic progression.
$endgroup$
– Jyrki Lahtonen
Aug 13 at 7:03
add a comment |
$begingroup$
This is inspired by this old question, which may provide a bit more background. But the two present questions seem somewhat more fundamental to me.
Let $p$ be an irreducible polynomial with integer coefficients.
- Is it possible that three of the roots of $p$ are collinear on a line in "general position", i.e. which is neither horizontal nor vertical nor through the origin?
Further, for $alphainmathbb R$, denote by $N_p(alpha)$ the number of zeros of $p$ with real part $alpha$.
- If $N_p(alpha)>1$, is it true that $N_p(alpha)$ is always a power of $2$?
cv.complex-variables polynomials galois-theory
asked Aug 12 at 19:57
WolfgangWolfgang
6,4234 gold badges29 silver badges74 bronze badges
$endgroup$
This is inspired by this old question, which may provide a bit more background. But the two present questions seem somewhat more fundamental to me.
Let $p$ be an irreducible polynomial with integer coefficients.
- Is it possible that three of the roots of $p$ are collinear on a line in "general position", i.e. which is neither horizontal nor vertical nor through the origin?
Further, for $alphainmathbb R$, denote by $N_p(alpha)$ the number of zeros of $p$ with real part $alpha$.
- If $N_p(alpha)>1$, is it true that $N_p(alpha)$ is always a power of $2$?
cv.complex-variables polynomials galois-theory
cv.complex-variables polynomials galois-theory
asked Aug 12 at 19:57
WolfgangWolfgang
6,4234 gold badges29 silver badges74 bronze badges
asked Aug 12 at 19:57
WolfgangWolfgang
6,4234 gold badges29 silver badges74 bronze badges
asked Aug 12 at 19:57
WolfgangWolfgang
6,4234 gold badges29 silver badges74 bronze badges
asked Aug 12 at 19:57
WolfgangWolfgang
6,4234 gold badges29 silver badges74 bronze badges
asked Aug 12 at 19:57
WolfgangWolfgang
6,4234 gold badges29 silver badges74 bronze badges
6,4234 gold badges29 silver badges74 bronze badges
4
$begingroup$
You should probably exclude linear shifts here, as you did in the other question—in other words, replace "through the origin" with "through an integer". (Also, 1 is a power of 2....)
$endgroup$
– Greg Martin
Aug 12 at 20:10
$begingroup$
A somewhat related question from Math.SE where the answer shows that it is impossible for an irredecible polynomial from $BbbZ[x]$ to have three roots in an arithmetic progression.
$endgroup$
– Jyrki Lahtonen
Aug 13 at 7:03
add a comment |
4
$begingroup$
You should probably exclude linear shifts here, as you did in the other question—in other words, replace "through the origin" with "through an integer". (Also, 1 is a power of 2....)
$endgroup$
– Greg Martin
Aug 12 at 20:10
$begingroup$
A somewhat related question from Math.SE where the answer shows that it is impossible for an irredecible polynomial from $BbbZ[x]$ to have three roots in an arithmetic progression.
$endgroup$
– Jyrki Lahtonen
Aug 13 at 7:03
4
4
$begingroup$
You should probably exclude linear shifts here, as you did in the other question—in other words, replace "through the origin" with "through an integer". (Also, 1 is a power of 2....)
$endgroup$
– Greg Martin
Aug 12 at 20:10
$begingroup$
You should probably exclude linear shifts here, as you did in the other question—in other words, replace "through the origin" with "through an integer". (Also, 1 is a power of 2....)
$endgroup$
– Greg Martin
Aug 12 at 20:10
$begingroup$
A somewhat related question from Math.SE where the answer shows that it is impossible for an irredecible polynomial from $BbbZ[x]$ to have three roots in an arithmetic progression.
$endgroup$
– Jyrki Lahtonen
Aug 13 at 7:03
$begingroup$
A somewhat related question from Math.SE where the answer shows that it is impossible for an irredecible polynomial from $BbbZ[x]$ to have three roots in an arithmetic progression.
$endgroup$
– Jyrki Lahtonen
Aug 13 at 7:03
add a comment |
1 Answer
1
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oldest
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$begingroup$
The answer to Q1 is yes. For example, $p(x) = x^6 + 45x^4 + 122x^3 + 504x^2 + 1740x + 2213$ is a polynomial with three roots on the line $y = 2x+3$ (and the other three on the line $y = -2x-3$). Take your favorite irreducible cubic with real roots $f(x)$ (mine is $x^3 - 3x + 1$) and let $alpha_1$, $alpha_2$, $alpha_3$ be the roots. Now, choose rational numbers $m$ and $b$ - we'll make a degree 6 polynomial whose roots (thought of in the form $x + iy$) lie on the line $y = mx + b$. (I chose $m = 2$ and $b = 3$.) Let $p(x) = fleft(fracx-bi1+miright) fleft(fracx+bi1-miright)$. If $z$ is a root of $p(x)$ coming from the first factor, then $fracz-bi1+mi = alpha$ for one of the real roots $alpha$ of the cubic $f(x)$ and then $z = alpha + (m alpha + b)i$ lies on the line $y = mx+b$.
The answer to Q2 is no. You can take an irreducible cubic $f(x)$ with all roots real and negative. Then, $f(x^2)$ can be an irreducible polynomial with all imaginary roots and so $N_p(0) = 6$. (For example, you can take $f(x) = x^3 + 6x^2 + 9x + 3$. Then $f(x^2)$ is irreducible by Eisenstein's criterion.)
answered Aug 12 at 21:35
Jeremy RouseJeremy Rouse
12k2 gold badges46 silver badges72 bronze badges
$endgroup$
$begingroup$
Thank you for providing the construction for Q1. I had tried something similar which didn't work and then thought that, if ever a $rational$ such slope existed, $p$ would be reducible. For Q2, indeed. Combining with the roots of your favorite one, this solves my initial question, which I have just updated.
$endgroup$
– Wolfgang
Aug 13 at 10:52
add a comment |
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$begingroup$
The answer to Q1 is yes. For example, $p(x) = x^6 + 45x^4 + 122x^3 + 504x^2 + 1740x + 2213$ is a polynomial with three roots on the line $y = 2x+3$ (and the other three on the line $y = -2x-3$). Take your favorite irreducible cubic with real roots $f(x)$ (mine is $x^3 - 3x + 1$) and let $alpha_1$, $alpha_2$, $alpha_3$ be the roots. Now, choose rational numbers $m$ and $b$ - we'll make a degree 6 polynomial whose roots (thought of in the form $x + iy$) lie on the line $y = mx + b$. (I chose $m = 2$ and $b = 3$.) Let $p(x) = fleft(fracx-bi1+miright) fleft(fracx+bi1-miright)$. If $z$ is a root of $p(x)$ coming from the first factor, then $fracz-bi1+mi = alpha$ for one of the real roots $alpha$ of the cubic $f(x)$ and then $z = alpha + (m alpha + b)i$ lies on the line $y = mx+b$.
The answer to Q2 is no. You can take an irreducible cubic $f(x)$ with all roots real and negative. Then, $f(x^2)$ can be an irreducible polynomial with all imaginary roots and so $N_p(0) = 6$. (For example, you can take $f(x) = x^3 + 6x^2 + 9x + 3$. Then $f(x^2)$ is irreducible by Eisenstein's criterion.)
answered Aug 12 at 21:35
Jeremy RouseJeremy Rouse
12k2 gold badges46 silver badges72 bronze badges
$endgroup$
$begingroup$
Thank you for providing the construction for Q1. I had tried something similar which didn't work and then thought that, if ever a $rational$ such slope existed, $p$ would be reducible. For Q2, indeed. Combining with the roots of your favorite one, this solves my initial question, which I have just updated.
$endgroup$
– Wolfgang
Aug 13 at 10:52
add a comment |
$begingroup$
The answer to Q1 is yes. For example, $p(x) = x^6 + 45x^4 + 122x^3 + 504x^2 + 1740x + 2213$ is a polynomial with three roots on the line $y = 2x+3$ (and the other three on the line $y = -2x-3$). Take your favorite irreducible cubic with real roots $f(x)$ (mine is $x^3 - 3x + 1$) and let $alpha_1$, $alpha_2$, $alpha_3$ be the roots. Now, choose rational numbers $m$ and $b$ - we'll make a degree 6 polynomial whose roots (thought of in the form $x + iy$) lie on the line $y = mx + b$. (I chose $m = 2$ and $b = 3$.) Let $p(x) = fleft(fracx-bi1+miright) fleft(fracx+bi1-miright)$. If $z$ is a root of $p(x)$ coming from the first factor, then $fracz-bi1+mi = alpha$ for one of the real roots $alpha$ of the cubic $f(x)$ and then $z = alpha + (m alpha + b)i$ lies on the line $y = mx+b$.
The answer to Q2 is no. You can take an irreducible cubic $f(x)$ with all roots real and negative. Then, $f(x^2)$ can be an irreducible polynomial with all imaginary roots and so $N_p(0) = 6$. (For example, you can take $f(x) = x^3 + 6x^2 + 9x + 3$. Then $f(x^2)$ is irreducible by Eisenstein's criterion.)
answered Aug 12 at 21:35
Jeremy RouseJeremy Rouse
12k2 gold badges46 silver badges72 bronze badges
$endgroup$
$begingroup$
Thank you for providing the construction for Q1. I had tried something similar which didn't work and then thought that, if ever a $rational$ such slope existed, $p$ would be reducible. For Q2, indeed. Combining with the roots of your favorite one, this solves my initial question, which I have just updated.
$endgroup$
– Wolfgang
Aug 13 at 10:52
add a comment |
$begingroup$
The answer to Q1 is yes. For example, $p(x) = x^6 + 45x^4 + 122x^3 + 504x^2 + 1740x + 2213$ is a polynomial with three roots on the line $y = 2x+3$ (and the other three on the line $y = -2x-3$). Take your favorite irreducible cubic with real roots $f(x)$ (mine is $x^3 - 3x + 1$) and let $alpha_1$, $alpha_2$, $alpha_3$ be the roots. Now, choose rational numbers $m$ and $b$ - we'll make a degree 6 polynomial whose roots (thought of in the form $x + iy$) lie on the line $y = mx + b$. (I chose $m = 2$ and $b = 3$.) Let $p(x) = fleft(fracx-bi1+miright) fleft(fracx+bi1-miright)$. If $z$ is a root of $p(x)$ coming from the first factor, then $fracz-bi1+mi = alpha$ for one of the real roots $alpha$ of the cubic $f(x)$ and then $z = alpha + (m alpha + b)i$ lies on the line $y = mx+b$.
The answer to Q2 is no. You can take an irreducible cubic $f(x)$ with all roots real and negative. Then, $f(x^2)$ can be an irreducible polynomial with all imaginary roots and so $N_p(0) = 6$. (For example, you can take $f(x) = x^3 + 6x^2 + 9x + 3$. Then $f(x^2)$ is irreducible by Eisenstein's criterion.)
answered Aug 12 at 21:35
Jeremy RouseJeremy Rouse
12k2 gold badges46 silver badges72 bronze badges
$endgroup$
The answer to Q1 is yes. For example, $p(x) = x^6 + 45x^4 + 122x^3 + 504x^2 + 1740x + 2213$ is a polynomial with three roots on the line $y = 2x+3$ (and the other three on the line $y = -2x-3$). Take your favorite irreducible cubic with real roots $f(x)$ (mine is $x^3 - 3x + 1$) and let $alpha_1$, $alpha_2$, $alpha_3$ be the roots. Now, choose rational numbers $m$ and $b$ - we'll make a degree 6 polynomial whose roots (thought of in the form $x + iy$) lie on the line $y = mx + b$. (I chose $m = 2$ and $b = 3$.) Let $p(x) = fleft(fracx-bi1+miright) fleft(fracx+bi1-miright)$. If $z$ is a root of $p(x)$ coming from the first factor, then $fracz-bi1+mi = alpha$ for one of the real roots $alpha$ of the cubic $f(x)$ and then $z = alpha + (m alpha + b)i$ lies on the line $y = mx+b$.
The answer to Q2 is no. You can take an irreducible cubic $f(x)$ with all roots real and negative. Then, $f(x^2)$ can be an irreducible polynomial with all imaginary roots and so $N_p(0) = 6$. (For example, you can take $f(x) = x^3 + 6x^2 + 9x + 3$. Then $f(x^2)$ is irreducible by Eisenstein's criterion.)
answered Aug 12 at 21:35
Jeremy RouseJeremy Rouse
12k2 gold badges46 silver badges72 bronze badges
edited Aug 12 at 22:46
answered Aug 12 at 21:35
Jeremy RouseJeremy Rouse
12k2 gold badges46 silver badges72 bronze badges
answered Aug 12 at 21:35
Jeremy RouseJeremy Rouse
12k2 gold badges46 silver badges72 bronze badges
answered Aug 12 at 21:35
Jeremy RouseJeremy Rouse
12k2 gold badges46 silver badges72 bronze badges
12k2 gold badges46 silver badges72 bronze badges
$begingroup$
Thank you for providing the construction for Q1. I had tried something similar which didn't work and then thought that, if ever a $rational$ such slope existed, $p$ would be reducible. For Q2, indeed. Combining with the roots of your favorite one, this solves my initial question, which I have just updated.
$endgroup$
– Wolfgang
Aug 13 at 10:52
add a comment |
$begingroup$
Thank you for providing the construction for Q1. I had tried something similar which didn't work and then thought that, if ever a $rational$ such slope existed, $p$ would be reducible. For Q2, indeed. Combining with the roots of your favorite one, this solves my initial question, which I have just updated.
$endgroup$
– Wolfgang
Aug 13 at 10:52
$begingroup$
Thank you for providing the construction for Q1. I had tried something similar which didn't work and then thought that, if ever a $rational$ such slope existed, $p$ would be reducible. For Q2, indeed. Combining with the roots of your favorite one, this solves my initial question, which I have just updated.
$endgroup$
– Wolfgang
Aug 13 at 10:52
$begingroup$
Thank you for providing the construction for Q1. I had tried something similar which didn't work and then thought that, if ever a $rational$ such slope existed, $p$ would be reducible. For Q2, indeed. Combining with the roots of your favorite one, this solves my initial question, which I have just updated.
$endgroup$
– Wolfgang
Aug 13 at 10:52
add a comment |
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$begingroup$
You should probably exclude linear shifts here, as you did in the other question—in other words, replace "through the origin" with "through an integer". (Also, 1 is a power of 2....)
$endgroup$
– Greg Martin
Aug 12 at 20:10
$begingroup$
A somewhat related question from Math.SE where the answer shows that it is impossible for an irredecible polynomial from $BbbZ[x]$ to have three roots in an arithmetic progression.
$endgroup$
– Jyrki Lahtonen
Aug 13 at 7:03