On Thursday, January 5, 2017 at 1:08:22 PM UTC+2, ecir...@gmail.com wrote:
>
> No, not a typo but an oversight from me.
>
> Is a polynomial "irreducible" if `factor_list(Poly(...))` is exactly of
> length 1?
>
It is irreducible if there is only one factor *and* its exponent is 1. The
square of an irreducible polynomial is not irreducible.
>
>
> On Thursday, January 5, 2017 at 12:23:57 AM UTC+1, Aaron Meurer wrote:
>>
>> Is there a typo below? Those polynomials aren't minimal polynomials
>> because they aren't irrreducible.
>>
>> Aaron Meurer
>>
>> On Wed, Jan 4, 2017 at 9:25 AM <ecir...@gmail.com> wrote:
>>
>>> I have two algebraic numbers defined by minimal polynomials:
>>>
>>> x, z = symbols('x, z')
>>> p = Poly((x-1)*(x-2)*(x-3)*(x-4))
>>> q = Poly((x-5)*(x-6)*(x-7)*(x-8))
>>>
>>>
>>>
>>> and I would like to compute the sum of these numbers. I [found](
>>> http://math.stackexchange.com/a/155132/327863) that I need to "`z=x+y`
>>> is a root of the resultant of `P(x)` and `Q(z−x)` (where we take this
>>> resultant by regarding `Q` as a polynomial in only `x`)".
>>>
>>> I'm totally new to all this algebraic numbers thing so I don't quite
>>> understand the advice but I tried:
>>>
>>> resultant(p, q.subs(x, z-x))
>>>
>>> but then I got stuck. Please, could someone explain to me:
>>>
>>> - I would like to see the steps that lead to the computation of desired
>>> minimal polynomial with the help of resultant. I think I defined the two
>>> numbers properly but I don't know how to express that `Q(z−x)`.
>>>
>>> - As it stands now, the `resultant` function returns bivariate
>>> polynomial but I would've assumed that the resulting minimal polynomial
>>> should be univariate. How do I get rid of the second variable?
>>>
>>> - The above link also says "`P(x) = Q(y) = 0`". Does it mean that `p`
>>> and `q` can't be both `Poly((x-1)*(x-2)*(x-3)*(x-4))`? What if I would like
>>> to add the same two numbers?
>>>
>>> Thank you very much in advance!
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
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