Exercise 1.3.1 (Condition for $(a,b) = 1$)

Question

With $a$ and $b$ as in (1.6) what conditions on the exponents must be satisfied if $(a,b) = 1$.

Richard Ganaye · Accepted Answer

\begin{proof} Write $A$ the set of prime factor of $ab$. Then  $a = \prod\limits_{p\in A} p^{\alpha(p)}, \ b = \prod\limits_{p\in A} p^{\beta(p)}$, where $\alpha(p) \geq 0, \beta(p) \geq 0$ for all $p \in A$. Then
\begin{align*}
a \wedge b = 1 &\iff \forall p \in A,\ \alpha(p) \beta(p) = 0\
&\iff \forall p \in A,\  \alpha(p) = 0 	ext{ or }  \beta(p) = 0.
\end{align*}
Indeed, if $a \wedge b = 1$, $p \mid a$ implies $p 
mid b$, so $\alpha(p) 
e 0 \Rightarrow \beta(p) = 0$.

Conversely, if $ \alpha(p) = 0 	ext{ or }  \beta(p) = 0$ for all $p$, then no prime $p$ divides both $a$ and $b$, so $a \wedge b = 1$.
\end{proof}

Exercise 1.3.1 (Condition for $(a,b) = 1$)

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Exercise 1.3.1 (Condition for $(a,b) = 1$)

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