For which field does the Eisenstein-criterion work?
I want to proof/show that a polynomial is reducible or irreducible in $Bbb Q[X]$ or $Bbb Z[X]$.
For $p=3$ the Eisenstein-Criterion works, but I'm not sure for which field.
All coefficients of the polynomials are in $Bbb Z$.
Wolfram Alpha says there are 5 zeros, two complex and three rationals.
But I thought it was irreducible in $Bbb Q$, because of Eisenstein.
Could someone maybe explain me it ?
P.S. Does the first coefficient necessarily need to be monic ?
$x^5-36x^4+6x^3+30x^2+24$
algebra-precalculus
asked Dec 4 '18 at 15:54
kratos88
496
|
show 1 more comment
I want to proof/show that a polynomial is reducible or irreducible in $Bbb Q[X]$ or $Bbb Z[X]$.
For $p=3$ the Eisenstein-Criterion works, but I'm not sure for which field.
All coefficients of the polynomials are in $Bbb Z$.
Wolfram Alpha says there are 5 zeros, two complex and three rationals.
But I thought it was irreducible in $Bbb Q$, because of Eisenstein.
Could someone maybe explain me it ?
P.S. Does the first coefficient necessarily need to be monic ?
$x^5-36x^4+6x^3+30x^2+24$
algebra-precalculus
asked Dec 4 '18 at 15:54
kratos88
496
1
Yes, if a polynomial is irreducible in $mathbb Z[X]$, it remains irreducible in $mathbb Q[X]$. This is a consequence of Gauss's lemma. Your original post fails to give the (degree five?) polynomial, so it's hard to guess where the disconnect occurs.
– hardmath
Dec 4 '18 at 15:58
Which polynomial do you mean?
– bellcircle
Dec 4 '18 at 16:01
@bellcircle: this one
– kratos88
Dec 4 '18 at 16:10
@hardmath: I thought, that it's irreducible in $Bbb Z$, but reducible in $Bbb Q$
– kratos88
Dec 4 '18 at 16:11
1
So, yes this is irreducible by Eisenstein's criterion with $p=3$. I don't think Wolfram Alpha is trying to tell you it has three rational roots, but instead that there are three real roots (and is showing you decimal approximations).
– hardmath
Dec 4 '18 at 16:21
|
show 1 more comment
I want to proof/show that a polynomial is reducible or irreducible in $Bbb Q[X]$ or $Bbb Z[X]$.
For $p=3$ the Eisenstein-Criterion works, but I'm not sure for which field.
All coefficients of the polynomials are in $Bbb Z$.
Wolfram Alpha says there are 5 zeros, two complex and three rationals.
But I thought it was irreducible in $Bbb Q$, because of Eisenstein.
Could someone maybe explain me it ?
P.S. Does the first coefficient necessarily need to be monic ?
$x^5-36x^4+6x^3+30x^2+24$
algebra-precalculus
asked Dec 4 '18 at 15:54
kratos88
496
I want to proof/show that a polynomial is reducible or irreducible in $Bbb Q[X]$ or $Bbb Z[X]$.
For $p=3$ the Eisenstein-Criterion works, but I'm not sure for which field.
All coefficients of the polynomials are in $Bbb Z$.
Wolfram Alpha says there are 5 zeros, two complex and three rationals.
But I thought it was irreducible in $Bbb Q$, because of Eisenstein.
Could someone maybe explain me it ?
P.S. Does the first coefficient necessarily need to be monic ?
$x^5-36x^4+6x^3+30x^2+24$
algebra-precalculus
algebra-precalculus
asked Dec 4 '18 at 15:54
kratos88
496
asked Dec 4 '18 at 15:54
kratos88
496
edited Dec 4 '18 at 16:09
asked Dec 4 '18 at 15:54
kratos88
496
asked Dec 4 '18 at 15:54
kratos88
496
asked Dec 4 '18 at 15:54
kratos88
496
496
1
Yes, if a polynomial is irreducible in $mathbb Z[X]$, it remains irreducible in $mathbb Q[X]$. This is a consequence of Gauss's lemma. Your original post fails to give the (degree five?) polynomial, so it's hard to guess where the disconnect occurs.
– hardmath
Dec 4 '18 at 15:58
Which polynomial do you mean?
– bellcircle
Dec 4 '18 at 16:01
@bellcircle: this one
– kratos88
Dec 4 '18 at 16:10
@hardmath: I thought, that it's irreducible in $Bbb Z$, but reducible in $Bbb Q$
– kratos88
Dec 4 '18 at 16:11
1
So, yes this is irreducible by Eisenstein's criterion with $p=3$. I don't think Wolfram Alpha is trying to tell you it has three rational roots, but instead that there are three real roots (and is showing you decimal approximations).
– hardmath
Dec 4 '18 at 16:21
|
show 1 more comment
1
Yes, if a polynomial is irreducible in $mathbb Z[X]$, it remains irreducible in $mathbb Q[X]$. This is a consequence of Gauss's lemma. Your original post fails to give the (degree five?) polynomial, so it's hard to guess where the disconnect occurs.
– hardmath
Dec 4 '18 at 15:58
Which polynomial do you mean?
– bellcircle
Dec 4 '18 at 16:01
@bellcircle: this one
– kratos88
Dec 4 '18 at 16:10
@hardmath: I thought, that it's irreducible in $Bbb Z$, but reducible in $Bbb Q$
– kratos88
Dec 4 '18 at 16:11
1
So, yes this is irreducible by Eisenstein's criterion with $p=3$. I don't think Wolfram Alpha is trying to tell you it has three rational roots, but instead that there are three real roots (and is showing you decimal approximations).
– hardmath
Dec 4 '18 at 16:21
1
1
Yes, if a polynomial is irreducible in $mathbb Z[X]$, it remains irreducible in $mathbb Q[X]$. This is a consequence of Gauss's lemma. Your original post fails to give the (degree five?) polynomial, so it's hard to guess where the disconnect occurs.
– hardmath
Dec 4 '18 at 15:58
Yes, if a polynomial is irreducible in $mathbb Z[X]$, it remains irreducible in $mathbb Q[X]$. This is a consequence of Gauss's lemma. Your original post fails to give the (degree five?) polynomial, so it's hard to guess where the disconnect occurs.
– hardmath
Dec 4 '18 at 15:58
Which polynomial do you mean?
– bellcircle
Dec 4 '18 at 16:01
Which polynomial do you mean?
– bellcircle
Dec 4 '18 at 16:01
@bellcircle: this one
– kratos88
Dec 4 '18 at 16:10
@bellcircle: this one
– kratos88
Dec 4 '18 at 16:10
@hardmath: I thought, that it's irreducible in $Bbb Z$, but reducible in $Bbb Q$
– kratos88
Dec 4 '18 at 16:11
@hardmath: I thought, that it's irreducible in $Bbb Z$, but reducible in $Bbb Q$
– kratos88
Dec 4 '18 at 16:11
1
1
So, yes this is irreducible by Eisenstein's criterion with $p=3$. I don't think Wolfram Alpha is trying to tell you it has three rational roots, but instead that there are three real roots (and is showing you decimal approximations).
– hardmath
Dec 4 '18 at 16:21
So, yes this is irreducible by Eisenstein's criterion with $p=3$. I don't think Wolfram Alpha is trying to tell you it has three rational roots, but instead that there are three real roots (and is showing you decimal approximations).
– hardmath
Dec 4 '18 at 16:21
|
show 1 more comment
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1
Yes, if a polynomial is irreducible in $mathbb Z[X]$, it remains irreducible in $mathbb Q[X]$. This is a consequence of Gauss's lemma. Your original post fails to give the (degree five?) polynomial, so it's hard to guess where the disconnect occurs.
– hardmath
Dec 4 '18 at 15:58
Which polynomial do you mean?
– bellcircle
Dec 4 '18 at 16:01
@bellcircle: this one
– kratos88
Dec 4 '18 at 16:10
@hardmath: I thought, that it's irreducible in $Bbb Z$, but reducible in $Bbb Q$
– kratos88
Dec 4 '18 at 16:11
1
So, yes this is irreducible by Eisenstein's criterion with $p=3$. I don't think Wolfram Alpha is trying to tell you it has three rational roots, but instead that there are three real roots (and is showing you decimal approximations).
– hardmath
Dec 4 '18 at 16:21