Exercise 1.3.21 ($[a,b,c](ab,bc,ca) = |abc|$)

Proof. Let $a,b,c$ be integers.

• Suppose first that $a,b,c$ are positive integers. We may write

  $$a=\underset{p\in A}{\prod }{p}^{\alpha (p)},b=\underset{p\in A}{\prod }{p}^{\beta (p)},c=\underset{p\in A}{\prod }{p}^{\gamma (p)},$$

  where $0\le \alpha (p),0\le \beta (p),0\le \gamma (p)$.

  Then

  $$\begin{array}{llll}\hfill (a\vee b\vee c)(\mathit{ab}\wedge \mathit{bc}\wedge \mathit{ca})& =\underset{p\in A}{\prod }{p}^{\max <!--\; FUNCTION\; APPLICATION\; -->(\alpha (p),\beta (p),\gamma (p))+\min <!--\; FUNCTION\; APPLICATION\; -->(\alpha (p)+\beta (p),\beta (p)+\gamma (p),\gamma (p)+\alpha (p))}.& \hfill & \end{array}$$

  For any integers $u,v,w$,

  $$\max <!--\; FUNCTION\; APPLICATION\; -->(u,v,w)+\min <!--\; FUNCTION\; APPLICATION\; -->(u+v,v+w,w+u)=u+v+w.$$

  Indeed, we may assume without loss of generality that $u\le v\le w$. Then $u+v\le v+w,u+v\le w+u$, thus $\min <!--\; FUNCTION\; APPLICATION\; -->(u+v,v+w,w+u)=u+v$, and

  $$\max <!--\; FUNCTION\; APPLICATION\; -->(u,v,w)+\min <!--\; FUNCTION\; APPLICATION\; -->(u+v,v+w,w+u)=w+(u+v)=u+v+w.$$

  (Same result for the six permutations of $u,v,w$.)

  Therefore

  $$\begin{array}{llll}\hfill (a\vee b\vee c)(\mathit{ab}\wedge \mathit{bc}\wedge \mathit{ca})& =\underset{p\in A}{\prod }{p}^{\alpha (p)+\beta (p)+\gamma (p)}& \hfill & \\ \hfill & =\mathit{abc}.& \hfill & \end{array}$$
• If $a=0$ or $b=0$ or $c=0$, then $a\vee b\vee c=0$, thus $(a\vee b\vee c)(\mathit{ab}\wedge \mathit{bc}\wedge \mathit{ca})=0=|\mathit{abc}|$.

• Suppose now that $a,b,c\in \mathbb{Z}$. Then $a\vee b\vee c=|a|\vee |b|\vee |c|$ and $\mathit{ab}\wedge \mathit{bc}\wedge \mathit{ca}=|\mathit{ab}|\wedge |\mathit{bc}|\wedge |\mathit{ca}|$, thus

  $$\begin{array}{llll}\hfill (a\vee b\vee c)(\mathit{ab}\wedge \mathit{bc}\wedge \mathit{ca})& =(|a|\vee |b|\vee |c)(|a||b|\wedge |b||c|\wedge |c||a|)& \hfill & \\ \hfill & =|a||b||c|=|\mathit{abc}|.& \hfill & \end{array}$$

For all $(a,b,c)\in {\mathbb{Z}}^3$,

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