First of all, thanks for any help! :)
I am thinking about the following question about possible denominators in linear combinations of integral polynomials:
Let $f, f_1,\dots,f_r \in \mathbb{Z}[X]$. Assume that $f$ lies in the ideal generated by $f_1,\dots,f_r$ over $\mathbb{Q}[X]$, that is, $f$ can be written as $$ f = \sum_{i=1}^r h_i f_i, $$ with $h_i \in \mathbb{Q}[X]$.
I am now interested in the possible prime factors of the denominators in the coefficients of the $h_i$. My feeling would be that these prime factors have to be related in some way to the prime factors of the coefficients of the $f_i$ (and of $f$), since all the denominators have to cancel out (as $f$ is integral).
For a very simple example, take $f = x$ and $f_1 = 2x - y, f_2 = 3y$. Then $$ f = \frac{1}{2} f_1 + \frac{1}{6} f_2, $$ and all prime factors of the denominators also appear as prime factors in the $f_i$. Moreover, the denominators can be bounded by the product of all coefficients in the $f_i$ (and $f$).
I guess my question is whether this is always the case, or whether anything else can be said about the possible denominators.
(For completeness, we assume that the representation $f = \sum_i h_i f_i$ is irredundant in the sense that it does not contain a syzygy of the $f_i$, as otherwise we could always add a syzygy multiplied with an arbitrary denominator.)