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.


1 Answer 1


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.

For more details, read the following:

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.

  • 1
    $\begingroup$ 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 ; ) ) $\endgroup$ Nov 4, 2018 at 20:26

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