Mixing
Every rational Polynomial is a product of the content and a primitive Polynomial
0
$begingroup$
If $fin mathbb{Q}[X]setminus {0}$ then $, f=cont(f)cdot f_1$ with $f_1 in mathbb{Z}[X]$ being a primitive Polynomial.
Why is that the case?
abstract-algebra polynomials irreducible-polynomials
asked Jan 11 at 17:16
KingDingelingKingDingeling
1497
$endgroup$
add a comment |
0
$begingroup$
If $fin mathbb{Q}[X]setminus {0}$ then $, f=cont(f)cdot f_1$ with $f_1 in mathbb{Z}[X]$ being a primitive Polynomial.
Why is that the case?
abstract-algebra polynomials irreducible-polynomials
asked Jan 11 at 17:16
KingDingelingKingDingeling
1497
$endgroup$
$begingroup$
See wikipedia.
$endgroup$
– Dietrich Burde
Jan 11 at 17:24
1
$begingroup$
I believe you want $f(x)in mathbb{Z}[x]$, not $mathbb{Q}[x]$. In that case, you can just factor out the gcd of the coefficients.
$endgroup$
– David Hill
Jan 11 at 18:43
add a comment |
0
0
0
$begingroup$
If $fin mathbb{Q}[X]setminus {0}$ then $, f=cont(f)cdot f_1$ with $f_1 in mathbb{Z}[X]$ being a primitive Polynomial.
Why is that the case?
abstract-algebra polynomials irreducible-polynomials
asked Jan 11 at 17:16
KingDingelingKingDingeling
1497
$endgroup$
If $fin mathbb{Q}[X]setminus {0}$ then $, f=cont(f)cdot f_1$ with $f_1 in mathbb{Z}[X]$ being a primitive Polynomial.
Why is that the case?
abstract-algebra polynomials irreducible-polynomials
abstract-algebra polynomials irreducible-polynomials
asked Jan 11 at 17:16
KingDingelingKingDingeling
1497
asked Jan 11 at 17:16
KingDingelingKingDingeling
1497
asked Jan 11 at 17:16
KingDingelingKingDingeling
1497
asked Jan 11 at 17:16
KingDingelingKingDingeling
1497
asked Jan 11 at 17:16
KingDingelingKingDingeling
1497
1497
$begingroup$
See wikipedia.
$endgroup$
– Dietrich Burde
Jan 11 at 17:24
1
$begingroup$
I believe you want $f(x)in mathbb{Z}[x]$, not $mathbb{Q}[x]$. In that case, you can just factor out the gcd of the coefficients.
$endgroup$
– David Hill
Jan 11 at 18:43
add a comment |
$begingroup$
See wikipedia.
$endgroup$
– Dietrich Burde
Jan 11 at 17:24
1
$begingroup$
I believe you want $f(x)in mathbb{Z}[x]$, not $mathbb{Q}[x]$. In that case, you can just factor out the gcd of the coefficients.
$endgroup$
– David Hill
Jan 11 at 18:43
$begingroup$
See wikipedia.
$endgroup$
– Dietrich Burde
Jan 11 at 17:24
$begingroup$
See wikipedia.
$endgroup$
– Dietrich Burde
Jan 11 at 17:24
1
1
$begingroup$
I believe you want $f(x)in mathbb{Z}[x]$, not $mathbb{Q}[x]$. In that case, you can just factor out the gcd of the coefficients.
$endgroup$
– David Hill
Jan 11 at 18:43
$begingroup$
I believe you want $f(x)in mathbb{Z}[x]$, not $mathbb{Q}[x]$. In that case, you can just factor out the gcd of the coefficients.
$endgroup$
– David Hill
Jan 11 at 18:43
add a comment |
0
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$begingroup$
See wikipedia.
$endgroup$
– Dietrich Burde
Jan 11 at 17:24
1
$begingroup$
I believe you want $f(x)in mathbb{Z}[x]$, not $mathbb{Q}[x]$. In that case, you can just factor out the gcd of the coefficients.
$endgroup$
– David Hill
Jan 11 at 18:43