Exercise 1.20

Proof. (a) Let $a=𝜀\underset{p}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{p}^{a(p)},b={𝜀}^{\prime }\underset{p}{\prod }{p}^{b(p)},𝜀,{𝜀}^{\prime }=\pm 1$, and

Then

(i) $m\ge 0$.

(ii) As $a(p)\le \max <!--\; FUNCTION\; APPLICATION\; -->(a(p),b(p))$, $p^{a(p)}\mid p^{\max <!--\; FUNCTION\; APPLICATION\; -->(a(p),b(p))}$, so $a\mid m$. Similarly, $b\mid m$.

(iii) If $\mu ={𝜀}^{″}\underset{p}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{p}^{c(p)}$ is a common multiple of $a$ and $b$, then for all primes $p$, $a(p)\le c(p),b(p)\le c(p)$, so $\max <!--\; FUNCTION\; APPLICATION\; -->(a(p),b(p))\le c(p)$, so $m\mid \mu$. Since $m$ verifies the characterization of lcm, we obtain

Therefore ${\mathrm{ord}}_p[a,b]=\max <!--\; FUNCTION\; APPLICATION\; -->({\mathrm{ord}}_p(a),{\mathrm{ord}}_p(b))$. (b) Similarly, we prove that

As $\max <!--\; FUNCTION\; APPLICATION\; -->(a,b)+\min <!--\; FUNCTION\; APPLICATION\; -->(a,b)=a+b$, we obtain

Second proof (without decompositions in primes):

Let $d=a\wedge b$. If $d=0$, then $a=b=0$ and $(a\vee b)(a\wedge b)=\mathit{ab}$.

Suppose now that $d\ne 0$. There exist integers $a^{\prime },b^{\prime }$ such that

Let $m=d{a}^{\prime }{b}^{\prime }$. Then $a=d{a}^{\prime }\mid m$ and $b=d{b}^{\prime }\mid m$. If $\mu$ is a common multiple of $a$ and $b$, then $d\mid \mu$, and $a^{\prime }\mid \mu ∕d,b^{\prime }\mid \mu ∕d$. As $a^{\prime }\wedge b^{\prime }=1$, $a^{\prime }{b}^{\prime }\mid \mu ∕d$ (see Ex.1.9). Thus $m=d{a}^{\prime }{b}^{\prime }\mid \mu$.

$|m|$ verifies the characterization of lcm (Ex. 1.19), so $a\vee b=|m|=|d{a}^{\prime }{b}^{\prime }|=|\mathit{ab}|∕d$. Conclusion : $(a\vee b)(a\wedge b)=|\mathit{ab}|$. (c) Let $\delta \in \mathbb{Z}$. If $\delta \mid a,\delta \mid b$, then $\delta \mid a+b$ and $\delta \mid a\vee b$.

Conversely, suppose that $\delta \mid a+b,\delta \mid a\vee b$.

Let $a^{\prime },b^{\prime }\in \mathbb{Z}$ such that $a=d{a}^{\prime },b=d{b}^{\prime },a^{\prime }\wedge b^{\prime }=1$. Then $a\vee b=d{a}^{\prime }{b}^{\prime }$, so

Multiplying the first relation by $b^{\prime }$ and similarly by $a^{\prime }$, we obtain :

Since $\delta \mid d{a}^{\prime }{b}^{\prime }$, we obtain :

As ${a{}^{\prime }}^2\wedge {b{}^{\prime }}^2=1$, $\delta \mid d({a{}^{\prime }}^2\wedge {b{}^{\prime }}^2)=d$, so $\delta \mid a,\delta \mid b$.

The set of divisors of $a,b$ is the same that the set of divisors of $a+b,a\vee b$, so

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