0

Let $f_1$, $\ldots$, $f_m \colon \mathbb{Z}^n \to \mathbb{Z}$ affine functions ( an affine function $f$ is of the form $f(x_1, \ldots, x_n) = \sum_{j=1}^n a_j x_j + b $, with $a_j$, $b \in \mathbb{Z}$), such that for every $x \in \mathbb{Z}^n$ we have

$$\operatorname{gcd} (f_1(x), \ldots, f_m(x) ) = 1$$

Then there exist $(\alpha_1, \ldots, \alpha_m) \in \mathbb{Z}^m$ such that

$$\sum_{i=1}^m \alpha_i f_i = 1$$

Notes:

  1. The converse is clear.

  2. Here is some a particular case of this. Since all of the problems use the same idea of a linear combination, one is lead to believe that this is the underlying reason in general.

  3. It works for $A$ a PID rather than $\mathbb{Z}$. For $A = k$ a field, it is an earlier posted question .

  4. One could ask if it works or not for other rings, interesting in itself

Any feedback would be appreciated!

$\bf{Added:}$

The problem as stated is in fact Wrong. However, there are some cases where it seems to hold. For now, I will add the extra hypothesis that every $f_i$ has its coefficients relatively prime ( that is, it is not divisible by another integral affine function).

orangeskid
  • 53,909
  • What do you mean by $\sum\limits_{i=1}^m \alpha_if_i\equiv 1$? – ShyamalSayak Jul 22 '23 at 04:30
  • @ShyamalSayak: Oh, I mean identical $1$. But since it could be confused with congruence, I will write $=1$. Thanks for the feedback! – orangeskid Jul 22 '23 at 05:19
  • Is the constant $b$ meant to be $b_j$? And do you assume $gcd(a_i, b_j)=1$ as well as $gcd(a_i,a_j)=1$ for every pair? – aerile Jul 22 '23 at 22:25
  • @aerile: I mean that each $f_j$ is not an integral multiple of another integral form ( except $\pm$) – orangeskid Jul 23 '23 at 14:24
  • It seems that case such as $f_1=4x+1,f_2=4x+3$ or $f_1=3, f_2=3x+2$ should be excluded. (The first sentence of my previous comment was mistake.) – aerile Jul 24 '23 at 20:31
  • @aerile: Yes, I see... it seems the whole question is in fact wrong as stated. I should just delete it – orangeskid Jul 24 '23 at 20:34

0 Answers0