Re: Divisibility

In article <170220060854019832%anniel@xxxxxxxxxxxxxxxxxxxxx>,
A N Niel <anniel@xxxxxxxxxxxxxxxxxxxxx> wrote:
In article <1140166128.266290.90220@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<jonhak@xxxxxxxxx> wrote:

If p(x) is a polynom over a field F with characteristic 0,
show that p«(x) divides p(x) if and only if p(x)=c*(x-x0)^n.

I have problems showing the implication in the direction
p«(x)|p(x) => p(x)=c*(x-x0)^n

I have tried to start out with a polynomial on general form.
p(x)=cn*x^n + ... + c0

derive it, match coefficients and come to the conclusion that
p(x)=p«(x)(ax+b) only possible if c0=c1=...cn-1=b=0.

but this isn't working out for me.

Any help would be appriciated.

Thanks.


Compare the degrees of p and p' to see that p/p' has degree 1.
Change variables to reduce to the case p/p' = cx.
Now can you proceed by comparing coefficients?

In article <1140254793.566742.253640@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
jonhak@xxxxxxxxx wrote:
I'm sorry, but i don't get it.

First, it would be nice if you quoted the relevant parts of the
message to which you are replying so that others might know what
the context of your comments is.

Next, I think your approach works fine, but you made one small
oversight. If you look at Mr Niel's comment, you can see what
you missed. Here is that approach expanded:

Suppose we change variables so that p/p' = cx. Let

n
--- k
p(x) = > a x
--- k
k=0

Then
n
--- k
cx p'(x) = > ck a x
--- k
k=1

which means that a_k = ck a_k or that (ck - 1)a_k = 0 for all
k. So for all k except 1/c, a_k = 0. When k = 1/c, a_k can
be anything. It appears you missed that one coefficient need
not be 0.

Here is a second method:

Suppose p/p' = ax+b. Take reciprocals and integrate

log(p(x))

|\ p'(x)
= | ----- dx
\| p(x)

|\ dx
= | ----
\| ax+b

1
= - log(ax+b) + log(c)
a

Thus, p(x) = c(ax+b)^{1/a}. Since p is a polynomial, we must have
that 1/a = n is an integer, so we get p(x) = ca^n (x - (-b/a))^n.

Rob Johnson <rob@xxxxxxxxxxxxxx>
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