# $\gcd(a(X),\,a'(X))$ w.r.t. squarefree decomposition

We are considering polynomials over a field $$\mathbb{F}$$. For $$a \in \mathbb{F}[X]$$, we have a squarefree decomposition $$a = \prod_{i=1}^k a_i^i$$ where $$\gcd(a_i,\,a_j) = 1$$ for $$i \neq j$$ and the $$a_i$$ are squarefree. Existence is clear; simply group together all prime factors of the same multiplicity.

At this webpage the following is said:

If $$a = \prod_{i=1}^k a_i^i$$ is the square-free factorization of $$a$$, then it is clear that $$a' = a \sum_{i=1}^k\frac{i\cdot a_i'}{a_i}$$ and hence that $$c = \gcd(a, a')$$ is $$c = \prod_{i=2}^k a_i^{i-1}$$

I have some trouble verifying this - obviously, the given polynomial divides $$\gcd(a,\,a')$$, but how do I show divisibility in the other direction?
Writing $$a'$$ as $$a' = \left( \prod_{i=2}^k a_i^{i-1} \right) \sum_{j=1}^k j \cdot a_j'\, \prod_{i=1,\,i\neq j}^k a_i$$

how can I show that there isn't some other prime factor $$p\,\vert\,a$$, such that

$$p \,\mid\, \sum_{j=1}^k j \cdot a_j'\, \prod_{i=1,\,i\neq j}^k a_i$$ i.e., that $$\gcd(a,\,a')$$ doesn't contain more prime factors?

I guess my main problem is that I want to know about the divisibility of a sum by some prime, but prime numbers intrinsically only directly relate to products, not sums...

I'm sorry if this question has some obviously easy answer I failed to see, I'm rather uninspired as of late.

Short trick, notice that $$p$$ would divide all the summands except at most one. Hence we are reduce to $$p$$ dividing one of the summands.
Just notice that if $$p$$ divides $$a$$, then $$p$$ divides some of the $$a_i$$. Let's say $$p$$ divides $$a_{i_0}$$. The only summand of $$\sum_{j=1}^k j \cdot a_j'\, \prod_{i=1,\,i\neq j}^k a_i$$ which do not have $$a_{i_0}$$ is $$i_0 \cdot a_{i_0}'\, \prod_{i=1,\,i\neq i_0}^k a_i$$. Thus $$p$$ divides the sum if and only if it divides $$i_0 \cdot a_{i_0}'\, \prod_{i=1,\,i\neq i_0}^k a_i$$. Even more, since the $$a_i$$ are pairwise coprime, $$p$$ does not divide $$\prod_{i=1,\,i\neq i_0}^k a_i$$ and so $$p$$ divides the sum if and only if it divides $$a_{i_0}'$$.
Finally, this means that $$p$$ divides the sum if and only if $$p$$ divides both $$a_{i_0}$$ and $$a_{i_0}'$$. By an argument similar to the one above (with the factorization of $$a_{i_0}$$), this happens if and only if $$p$$ divides $$p'$$, but this is not possible as $$p'$$ has less degree.
• Great, thank you! I was going into that direction but failed to make the crucial final thought :) I guess I would prefer a different explanation for $p$ not dividing both $a_{i_0}$ and $a_{i_0}'$: $a_{i_0}$ is squarefree, and hence $\gcd(a_{i_0},\,a_{i_0}')$ is trivial (wait, I just figured you gave the proof for that ; ) ) Nov 4, 2018 at 20:26