Mixing
Roots of polynomial irreducible over the rationals
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If a polynomial is "irreducible over the rationals", does it mean that it has no rational roots?
I would say yes because otherwise I could divide out the linear factors (i.e. rational roots) but maybe I'm wrong?
polynomials roots factoring rational-numbers
asked Jan 15 at 14:35
Ecir HanaEcir Hana
417414
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add a comment |
$begingroup$
If a polynomial is "irreducible over the rationals", does it mean that it has no rational roots?
I would say yes because otherwise I could divide out the linear factors (i.e. rational roots) but maybe I'm wrong?
polynomials roots factoring rational-numbers
asked Jan 15 at 14:35
Ecir HanaEcir Hana
417414
$endgroup$
$begingroup$
If $f$ is irreducible over $Bbb Q$ of degree $nge 2$, then it has no rational root, yes. The converse is not true in general.
$endgroup$
– Dietrich Burde
Jan 15 at 14:37
add a comment |
$begingroup$
If a polynomial is "irreducible over the rationals", does it mean that it has no rational roots?
I would say yes because otherwise I could divide out the linear factors (i.e. rational roots) but maybe I'm wrong?
polynomials roots factoring rational-numbers
asked Jan 15 at 14:35
Ecir HanaEcir Hana
417414
$endgroup$
If a polynomial is "irreducible over the rationals", does it mean that it has no rational roots?
I would say yes because otherwise I could divide out the linear factors (i.e. rational roots) but maybe I'm wrong?
polynomials roots factoring rational-numbers
polynomials roots factoring rational-numbers
asked Jan 15 at 14:35
Ecir HanaEcir Hana
417414
asked Jan 15 at 14:35
Ecir HanaEcir Hana
417414
asked Jan 15 at 14:35
Ecir HanaEcir Hana
417414
asked Jan 15 at 14:35
Ecir HanaEcir Hana
417414
asked Jan 15 at 14:35
Ecir HanaEcir Hana
417414
417414
$begingroup$
If $f$ is irreducible over $Bbb Q$ of degree $nge 2$, then it has no rational root, yes. The converse is not true in general.
$endgroup$
– Dietrich Burde
Jan 15 at 14:37
add a comment |
$begingroup$
If $f$ is irreducible over $Bbb Q$ of degree $nge 2$, then it has no rational root, yes. The converse is not true in general.
$endgroup$
– Dietrich Burde
Jan 15 at 14:37
$begingroup$
If $f$ is irreducible over $Bbb Q$ of degree $nge 2$, then it has no rational root, yes. The converse is not true in general.
$endgroup$
– Dietrich Burde
Jan 15 at 14:37
$begingroup$
If $f$ is irreducible over $Bbb Q$ of degree $nge 2$, then it has no rational root, yes. The converse is not true in general.
$endgroup$
– Dietrich Burde
Jan 15 at 14:37
add a comment |
2 Answers
2
active
oldest
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Almost.
For example, the polynomial $X-frac23$ is irreducible and yet it has a rational root.
However, this (i.e., all linear polynomials) is the only exception, as otherwise your correct reasoning applies.
Note however, that the converse is not true: A polynomial may be recucible even if it does not have rational roots; consider e.g., $X^4-5X+6=(X^2-2)(X^2-2)$
answered Jan 15 at 14:38
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
add a comment |
$begingroup$
A polynomial over the rationals is a polynomial $f(x)in{Bbb Q}[x]$. If the polynomial is linear, $f(x)=ax+b$, $ane 0$, it is irreducible over the rationals. If the polynomial has degree $geq 2$, it is irreducible over the rationals provided that it has no root in the rationals.
answered Jan 15 at 14:40
WuestenfuxWuestenfux
4,7331413
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add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
Almost.
For example, the polynomial $X-frac23$ is irreducible and yet it has a rational root.
However, this (i.e., all linear polynomials) is the only exception, as otherwise your correct reasoning applies.
Note however, that the converse is not true: A polynomial may be recucible even if it does not have rational roots; consider e.g., $X^4-5X+6=(X^2-2)(X^2-2)$
answered Jan 15 at 14:38
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
add a comment |
$begingroup$
Almost.
For example, the polynomial $X-frac23$ is irreducible and yet it has a rational root.
However, this (i.e., all linear polynomials) is the only exception, as otherwise your correct reasoning applies.
Note however, that the converse is not true: A polynomial may be recucible even if it does not have rational roots; consider e.g., $X^4-5X+6=(X^2-2)(X^2-2)$
answered Jan 15 at 14:38
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
add a comment |
$begingroup$
Almost.
For example, the polynomial $X-frac23$ is irreducible and yet it has a rational root.
However, this (i.e., all linear polynomials) is the only exception, as otherwise your correct reasoning applies.
Note however, that the converse is not true: A polynomial may be recucible even if it does not have rational roots; consider e.g., $X^4-5X+6=(X^2-2)(X^2-2)$
answered Jan 15 at 14:38
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
Almost.
For example, the polynomial $X-frac23$ is irreducible and yet it has a rational root.
However, this (i.e., all linear polynomials) is the only exception, as otherwise your correct reasoning applies.
Note however, that the converse is not true: A polynomial may be recucible even if it does not have rational roots; consider e.g., $X^4-5X+6=(X^2-2)(X^2-2)$
answered Jan 15 at 14:38
Hagen von EitzenHagen von Eitzen
1643
answered Jan 15 at 14:38
Hagen von EitzenHagen von Eitzen
1643
answered Jan 15 at 14:38
Hagen von EitzenHagen von Eitzen
1643
answered Jan 15 at 14:38
Hagen von EitzenHagen von Eitzen
1643
1643
add a comment |
add a comment |
$begingroup$
A polynomial over the rationals is a polynomial $f(x)in{Bbb Q}[x]$. If the polynomial is linear, $f(x)=ax+b$, $ane 0$, it is irreducible over the rationals. If the polynomial has degree $geq 2$, it is irreducible over the rationals provided that it has no root in the rationals.
answered Jan 15 at 14:40
WuestenfuxWuestenfux
4,7331413
$endgroup$
add a comment |
$begingroup$
A polynomial over the rationals is a polynomial $f(x)in{Bbb Q}[x]$. If the polynomial is linear, $f(x)=ax+b$, $ane 0$, it is irreducible over the rationals. If the polynomial has degree $geq 2$, it is irreducible over the rationals provided that it has no root in the rationals.
answered Jan 15 at 14:40
WuestenfuxWuestenfux
4,7331413
$endgroup$
add a comment |
$begingroup$
A polynomial over the rationals is a polynomial $f(x)in{Bbb Q}[x]$. If the polynomial is linear, $f(x)=ax+b$, $ane 0$, it is irreducible over the rationals. If the polynomial has degree $geq 2$, it is irreducible over the rationals provided that it has no root in the rationals.
answered Jan 15 at 14:40
WuestenfuxWuestenfux
4,7331413
$endgroup$
A polynomial over the rationals is a polynomial $f(x)in{Bbb Q}[x]$. If the polynomial is linear, $f(x)=ax+b$, $ane 0$, it is irreducible over the rationals. If the polynomial has degree $geq 2$, it is irreducible over the rationals provided that it has no root in the rationals.
answered Jan 15 at 14:40
WuestenfuxWuestenfux
4,7331413
answered Jan 15 at 14:40
WuestenfuxWuestenfux
4,7331413
answered Jan 15 at 14:40
WuestenfuxWuestenfux
4,7331413
answered Jan 15 at 14:40
WuestenfuxWuestenfux
4,7331413
4,7331413
add a comment |
add a comment |
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$begingroup$
If $f$ is irreducible over $Bbb Q$ of degree $nge 2$, then it has no rational root, yes. The converse is not true in general.
$endgroup$
– Dietrich Burde
Jan 15 at 14:37