2013/10/7 Jori Mantysalo
> On Mon, 7 Oct 2013, John Cremona wrote:
>
> sage: R. = ZZ[]; print (4*x^2-1).factor()
>> --**--**
>> ---
>> NotImplementedError Traceback (most recent call
>> last)
>>
>
> But tha
On Mon, 7 Oct 2013, John Cremona wrote:
sage: R. = ZZ[]; print (4*x^2-1).factor()
---
NotImplementedError Traceback (most recent call last)
But that is just what I would like to "implement", i.e. make
On 7 October 2013 08:15, Jori Mantysalo wrote:
> On Fri, 4 Oct 2013, Marco Streng wrote:
>
>> Just take the factorization over QQ, then for each factor, make it a
>> primitive integral polynomial, i.e., multiply by the lcm of the
>> denominators of the coefficients and divide by the gcd of the num
On Fri, 4 Oct 2013, Marco Streng wrote:
Just take the factorization over QQ, then for each factor, make it a
primitive integral polynomial, i.e., multiply by the lcm of the
denominators of the coefficients and divide by the gcd of the numerators of
the coefficients. Then you have a factorization
2013/10/4 Jori Mantysalo
> On Fri, 4 Oct 2013, Volker Braun wrote:
>
> If the integral polynomial is not monic then the roots need not be
>> integral:
>>
>> sage: R. = QQ[]
>> sage: (4*x^2-1).factor()
>> (4) * (x - 1/2) * (x + 1/2)
>>
>
(4*x^2-1) = (2*x-1)*(2*x+1)
ZZ[x] has unique factorization
On Fri, 4 Oct 2013, Volker Braun wrote:
If the integral polynomial is not monic then the roots need not be integral:
sage: R. = QQ[]
sage: (4*x^2-1).factor()
(4) * (x - 1/2) * (x + 1/2)
So this would not be factorizable in ZZ[x] but is factorizable in QQ[x]
Of course. Duh.
Anyways, is this
If the integral polynomial is not monic then the roots need not be integral:
sage: R. = QQ[]
sage: (4*x^2-1).factor()
(4) * (x - 1/2) * (x + 1/2)
So this would not be factorizable in ZZ[x] but is factorizable in QQ[x]
On Friday, October 4, 2013 12:44:41 PM UTC+1, Jori Mantysalo wrote:
>
> $SAG
