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In a principal ideal ring:

$$\forall\, a,b \in \mathbb{Z},\quad a \wedge b = 1\ \text{ (notation for gcd)} \enspace \Longleftrightarrow \enspace (a) + (b)= \mathbb{Z}$$ where $(a)$ denotes the principal ideal generated by $a$. So it is indeed obvious that if $\ (n_1,\cdots , n_r)$ are pairwise coprime then they are setwise coprime:

$$ (n_i)+ (n_j)= \mathbb{Z} \quad \Longrightarrow\quad (n_1)+ \cdots + (n_r)= \mathbb{Z} $$ (the reasonning is as intuitive with the definition of greatest common divisor).

  • First useless detail, I nevertheless wanted to have an explicit Bézout relation: starting from relations $\ u\, n_i + v\, n_j=1$ for every couple $(i,j), i\neq j$, find a systematic way to obtain an expression of the form $$\ u_1\, n_1 + \cdots + u_r\, n_r=1$$

  • Secondly I was wondering whether an equivalence or just an implication relates the following properties:

  1. $ (n_1,\cdots , n_r)$ are pairwise coprime.
  2. $ \Big( \prod_{i\neq 1} n_i,\cdots ,\prod_{i\neq r} n_i \Big)$ are setwise coprime.

Edit: a relation as in the first $\ \bullet$ could be useful e.g. in the "kernel decomposition theorem". In addition to the comment and to the answer, an other way to obtain it is by induction to insert $1=u_1\, n_1 + \cdots + u_{r-1}\, n_{r-1}$ into $$\begin{split} 1 &= n_r - (n_r-1)\times 1= n_r - (n_r-1)\times (u_1\, n_1 + \cdots + u_{r-1}\, n_{r-1})\\ &= n_r + \tilde{u}_1\, n_1 + \cdots + \tilde{u}_{r-1}\, n_{r-1} \end{split}$$

Noix07
  • 3,499

1 Answers1

1

Let us prove the equivalence of the second $\enspace\bullet$

1. implies 2.: in fact this will also give a relation of the form mentionned in the first $\ \bullet$

"$(n_1,\cdots, n_r)$ pairwise coprime" gives by the Bézout identity ${r\choose 2}= \frac{r(r-1)}{2}$ equalities of the form $$ u_k n_i + v_k n_j = 1 \quad \text{for each pair}\ k:=(i,j),\ i\neq j$$

A first proof of the implication is to multiply all of these equalities and show that each of the $2^{{r\choose 2}}$ terms in the development of the product $\displaystyle \prod_{1\leq i<j\leq r} (u_k n_i + v_k n_j)$ into sums is of the form $\displaystyle \prod_{i\neq j} n_i $, i.e. contains at least $r-1$ of the $r$ factors $\{n_1,\cdots, n_r\}$.

In place of a formal proof, let us just picture ourselves the set of couples $(i,j)$ such that $1\leq i<j\leq r$ as the coordinates of ${r\choose 2}$ points which form a "triangle". (Or if they designate the components of a matrix, it would correspond to the strictly upper components).

Now for each factor $(u_k n_i + v_k n_j)$ of the product, choosing one of the summand correspond in our pictorial representation to projecting either on the absciss or ordinate (x or y axis). We keep this value and repeat this for each of the ${r\choose 2}$ points of the "triangle". The "worse" case where we have a few values as possible is obtained inductively by "deleting" the row or column after each step, which correspond to choosing the same factor as much as possible. What should be formally written is that "since it is an $r-1$ by $r-1$ triangle we will get at the end $r-1$ values".

As second proof is by induction on $r$. Assume the statement holds for $r-1$: "$(n_1,\cdots, n_{r-1})$ pairwise coprime" implies $\exists\ A_1,\cdots A_{r-1}\in \mathbb{N}$ s.t. $$ A_1\prod_{i\neq 1} n_i + \cdots + A_{r-1} \prod_{i\neq r-1} n_i = 1 $$ With the additional hypothesis the $n_r$ is pairwise coprime with the other $n_j$ we have $r-1$ relations of the form $$ u_j n_j + v_j n_r = 1 $$ which we "insert" in each of the summand: e.g. $$ A_j \prod_{\genfrac{}{}{0pt}{1}{1\leq i\leq r-1}{i\neq j}} n_i \times 1 = A_j \prod_{\genfrac{}{}{0pt}{1}{1\leq i\leq r-1}{i\neq j}} n_i \times ( u_j n_j + v_j n_r) = u_j\, A_j \prod_{\genfrac{}{}{0pt}{1}{1\leq i\leq r}{i\neq r}} n_i + v_j\, A_j \prod_{\genfrac{}{}{0pt}{1}{1\leq i\leq r}{i\neq j}} n_i $$

Conversely: assume that $$ A_1\prod_{i\neq 1} n_i + \cdots + A_r \prod_{i\neq r} n_i = 1 $$

Then $\forall\ k,\ 1 \leq k \leq r$ the rest of Euclidean division of $A_k\prod_{i\neq k} n_i$ by $n_k$ is 1: regroup all multiples of $n_k$ $$ A_k\prod_{i\neq k} n_i + B_k\, n_k = 1 $$ and for any $j\neq k$ $$ A_k \Big(\prod_{i\neq k,j} n_i \Big)\, n_j + B_k\, n_k = 1 \quad\text{i.e.}\quad n_j \wedge n_k = 1 $$

Noix07
  • 3,499