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From: quasi on 8 Aug 2010 04:10
On Sat, 07 Aug 2010 07:55:53 EDT, "Roman B. Binder"
<rbinder(a)netvision.net.il> wrote:

>How are You !

Fine thanks.

>Nowadays I am trying to find sufficient
>condition not allowing F(x)/H(x) completely.

As in the past, your English is difficult for me to understand.

I'll guess that you are trying to develop some kind of polynomial
irreducibility test.

Thus, given a polynomial F(x), you want to determine whether it has a
polynomial factor H(x). Yes?

But what kind of coefficients are your polynomials assumed to have?
Integers?

>It looks once coefficients of the oldest therm are
>of gcd=1

Ok, so it looks like you meant integer coefficients.

But what does "the oldest of them" mean?

>and once also absolute terms are of gcd=1

And what are "absolute terms"?

>so there is the deal: when we suppose some polynomial
>G(x) to complete the division;

So now you are assuming F(x) = G(x) H(x).

>let F(x) is n degree;
>H(x) of n-k degree so G(x) of k degree and should be:
>(fn)=[h(n-k)](gk)

The statement "(fn)=[h(n-k)](gk)" is not meaningful to me. I think
perhaps you are just repeating the assumption on the degrees:

deg(f) = n
deg(g) = k
deg(h) = n-k

>and (f0)=(h0)(g0);

True; f(x) = g(x) h(x) implies f(0) = g(0) h(0).

>then once:
>{(fn);[h(n-k)]} of gcd=1

I don't know what you mean by the above.

>and [(f0);(h0)] of gcd=1....(**)
>so such division for primitive polynomials F(x) and H(x)
>is not possible. ( it looks even once the one of these
>two simple condition fails such division could not be
>completed: (suppose we can vary (fn);(f0);[h(n-k);(h0); to be the most simple numbers eg. 1 or 2 but one of (**) conditions still holds ?

I think what you're saying is this: The assumption f(x) = g(x) h(x)
implies f(a) = g(a) h(a) for all integers a. So if for some integer a,
h(a) doesn't divide f(a), it would follow that the polynomial h(x)
doesn't divide f(x). But if you are trying to determine the
irreducibility of f(x), the polynomial h(x) is an _unknown_
polynomial, so you can't directly test whether h(a) divides f(a). On
the other hand, if h(x) is somehow specified, then you can just use
polynomial long division to determine if h(x) divides f(x) -- no need
to test using selected values of x.

So I don't really follow your idea. Perhaps you can illustrate it
using a specific example? Still, unless I can understand it fairly
easily, I may not have time to work on it.

quasi
From: Roman B. Binder on 9 Aug 2010 11:31
Hi, and Thanks too !
> On Sat, 07 Aug 2010 07:55:53 EDT, "Roman B. Binder"
> <rbinder(a)netvision.net.il> wrote:
>
> >How are You !
>
> Fine thanks.
>
> >Nowadays I am trying to find sufficient
> >condition not allowing F(x)/H(x) completely.
>
> As in the past, your English is difficult for me to
> understand.
>
> I'll guess that you are trying to develop some kind
> of polynomial
> irreducibility test.
Exactly not the test of irreducibility but some sufficient condition, which not allows for whole
division of F(x) by H(x).
( H(x) of smaller degree than F(x) of course )
> Thus, given a polynomial F(x), you want to determine
> whether it has a
> polynomial factor H(x). Yes?
Yes.
> But what kind of coefficients are your polynomials
> assumed to have?
> Integers?
Yes.
> >It looks once coefficients of the oldest therm are
> >of gcd=1
>
> Ok, so it looks like you meant integer coefficients.
>
> But what does "the oldest of them" mean?
for F(x)=(an)x^n +[a(n-1)]x^(n-1)+..(ak)^k..+(a1)x +(a0)
there used to be to call (an) the oldest coefficient
and (a0) as pure value as absolute term: Sorry I used
the oldest but I meant the oldest coefficient...
> >and once also absolute terms are of gcd=1
>
> And what are "absolute terms"?
> >so there is the deal: when we suppose some
> polynomial
> >G(x) to complete the division;
>
> So now you are assuming F(x) = G(x) H(x).
>
> >let F(x) is n degree;
> >H(x) of n-k degree so G(x) of k degree and should
> be:
> >(fn)=[h(n-k)](gk)
>
> The statement "(fn)=[h(n-k)](gk)" is not meaningful
> to me. I think
> perhaps you are just repeating the assumption on the
> degrees:
>
> deg(f) = n
> deg(g) = k
> deg(h) = n-k
Yes It was better to use Your statement...
> >and (f0)=(h0)(g0);
>
> True; f(x) = g(x) h(x) implies f(0) = g(0) h(0).
>
> >then once:
> >{(fn);[h(n-k)]} of gcd=1
Here just I recall coefficients: (an) = (fn) etc.
> I don't know what you mean by the above.
>
> >and [(f0);(h0)] of gcd=1....(**)
> >so such division for primitive polynomials F(x) and
> H(x)
> >is not possible. ( it looks even once the one of
> these
> >two simple condition fails such division could not
> be
> >completed: (suppose we can vary
> (fn);(f0);[h(n-k);(h0); to be the most simple numbers
> eg. 1 or 2 but one of (**) conditions still holds ?
>
> I think what you're saying is this: The assumption
> f(x) = g(x) h(x)
> implies f(a) = g(a) h(a) for all integers a. So if
> for some integer a,
> h(a) doesn't divide f(a), it would follow that the
> polynomial h(x)
> doesn't divide f(x). But if you are trying to
> determine the
> irreducibility of f(x), the polynomial h(x) is an
> _unknown_
> polynomial, so you can't directly test whether h(a)
> divides f(a). On
> the other hand, if h(x) is somehow specified, then
> you can just use
> polynomial long division to determine if h(x) divides
> f(x) -- no need
> to test using selected values of x.
>
> So I don't really follow your idea. Perhaps you can
> illustrate it
> using a specific example? Still, unless I can
> understand it fairly
> easily, I may not have time to work on it.
>
> quasi
O.K. One more time Thank You !
My first example is like this:
F(x) = 3^(2u-1) a^2 x^2 -3^u ab^2 x +b^4
H(x) = bx +3^(2u-1) a^2.............................(I.T)
where a;b;3 and also x should be of gcd=1
also we can take as the most simple a;b parameters
numbers a=2 once b=1 or inverse:
now for supposed polynomial G(x):
(h1)*(g1) = (f2)..................................(C.1.)
1*(g1) =3^(2u-1) a^2 can be true for (g1)=3^(2u-1) a^2
but (h0)*(g0) = (f0)..............................(C.2.)
3^(2u-1) 2^2 *(g0) = 1 can not be true...
Somehow similar once:
F(x) = 3^(2u-1) a^2 x^2 -3^u ab^2 x +b^4
H(x) = 3^u bx +a^2................................(II.T)
(h1)*(g1) = (f2)..................................(C.1.)
3^u *(g1)=3^(2u-1) a^2 can be true for (g1)=3^(u-1) a^2
but (h0)*(g0) = (f0)..............................(C.2.)
a^2 *(g0) = 2^2 (g0) = b^4 = 1 can not be true
here for supposed polynomial G(x):
Obviously not true for any bigger a;b of gcd,
what will involve more prime numbers and will done
impossible the two of upper conditions: (C.1.) and (C.2.)
What do You really think of these?
So on I can imagine for to find only some part of common
factors in F(x) and H(x) but not H(x) of such two
types (I.T) or (II.T) to be just factor of F(x)
Thanks for Your attention!

Best Regards
Ro-Bin

P.S. I can still remember one of Your comments:
" it could be more elementary but so much less easy
for to find... certain proof"
From: quasi on 10 Aug 2010 02:59
On Mon, 09 Aug 2010 15:31:38 EDT, "Roman B. Binder"
<rbinder(a)netvision.net.il> wrote:

>Hi, and Thanks too !
>> On Sat, 07 Aug 2010 07:55:53 EDT, "Roman B. Binder"
>> <rbinder(a)netvision.net.il> wrote:
>>
>> >How are You !
>>
>> Fine thanks.
>>
>> >Nowadays I am trying to find sufficient
>> >condition not allowing F(x)/H(x) completely.
>>
>> As in the past, your English is difficult for me to
>> understand.
>>
>> I'll guess that you are trying to develop some kind
>> of polynomial
>> irreducibility test.
>Exactly not the test of irreducibility but some sufficient condition, which not allows for whole
>division of F(x) by H(x).
>( H(x) of smaller degree than F(x) of course )
>> Thus, given a polynomial F(x), you want to determine
>> whether it has a
>> polynomial factor H(x). Yes?
>Yes.
>> But what kind of coefficients are your polynomials
>> assumed to have?
>> Integers?
>Yes.
>> >It looks once coefficients of the oldest therm are
>> >of gcd=1
>>
>> Ok, so it looks like you meant integer coefficients.
>>
>> But what does "the oldest of them" mean?
>for F(x)=(an)x^n +[a(n-1)]x^(n-1)+..(ak)^k..+(a1)x +(a0)
>there used to be to call (an) the oldest coefficient
>and (a0) as pure value as absolute term: Sorry I used
>the oldest but I meant the oldest coefficient...
>> >and once also absolute terms are of gcd=1
>>
>> And what are "absolute terms"?
>> >so there is the deal: when we suppose some
>> polynomial
>> >G(x) to complete the division;
>>
>> So now you are assuming F(x) = G(x) H(x).
>>
>> >let F(x) is n degree;
>> >H(x) of n-k degree so G(x) of k degree and should
>> be:
>> >(fn)=[h(n-k)](gk)
>>
>> The statement "(fn)=[h(n-k)](gk)" is not meaningful
>> to me. I think
>> perhaps you are just repeating the assumption on the
>> degrees:
>>
>> deg(f) = n
>> deg(g) = k
>> deg(h) = n-k
>Yes It was better to use Your statement...
>> >and (f0)=(h0)(g0);
>>
>> True; f(x) = g(x) h(x) implies f(0) = g(0) h(0).
>>
>> >then once:
>> >{(fn);[h(n-k)]} of gcd=1
>Here just I recall coefficients: (an) = (fn) etc.
>> I don't know what you mean by the above.
>>
>> >and [(f0);(h0)] of gcd=1....(**)
>> >so such division for primitive polynomials F(x) and
>> H(x)
>> >is not possible. ( it looks even once the one of
>> these
>> >two simple condition fails such division could not
>> be
>> >completed: (suppose we can vary
>> (fn);(f0);[h(n-k);(h0); to be the most simple numbers
>> eg. 1 or 2 but one of (**) conditions still holds ?
>>
>> I think what you're saying is this: The assumption
>> f(x) = g(x) h(x)
>> implies f(a) = g(a) h(a) for all integers a. So if
>> for some integer a,
>> h(a) doesn't divide f(a), it would follow that the
>> polynomial h(x)
>> doesn't divide f(x). But if you are trying to
>> determine the
>> irreducibility of f(x), the polynomial h(x) is an
>> _unknown_
>> polynomial, so you can't directly test whether h(a)
>> divides f(a). On
>> the other hand, if h(x) is somehow specified, then
>> you can just use
>> polynomial long division to determine if h(x) divides
>> f(x) -- no need
>> to test using selected values of x.
>>
>> So I don't really follow your idea. Perhaps you can
>> illustrate it
>> using a specific example? Still, unless I can
>> understand it fairly
>> easily, I may not have time to work on it.
>>
>> quasi
>O.K. One more time Thank You !
>My first example is like this:
>F(x) = 3^(2u-1) a^2 x^2 -3^u ab^2 x +b^4

Too many unknowns -- I was expecting only one unknown, namely x.

>H(x) = bx +3^(2u-1) a^2.............................(I.T)

So H is an assumed linear factor of F?

But why are you assuming that H has leading coefficient equal to b?
And why the strange form for the constant term of H?

Also, isn't b already used in F? And now that I examine F more
carefully, I see that b is used in more than one term of F. Very
strange.

At this point, I have no idea what you are doing.

How about using an irreducible polynomial F(x) in one variable x, with
no obscure parameters -- just _constant_ integer coefficients. Then
let's see how your method proceeds to show that F is irreducible.

quasi
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