Prove Proposition (2.21).
Proposition (2.21)
Let $a = p_1^{e_1} × p_2^{e_2} × ⋯ × p_k^{e_k}$ and $b= p_1^{f_1} × p_2^{f_2} × ⋯ × p_k^{f_k}$ be the decompositions of $a$ and $b$ and $e_j ≥ 0$ and $f_j ≥ 0$. Then the gcd is given by $$\gcd (a, b) = p_1^{\min(e_1, f_1)} × p_2^{\min(e_2, f_2)} × ⋯ × p_k^{\min(e_k, f_k)}$$
The $\gcd(a,b)$ is a divisor of both $a$ and $b$. That means, for some integers $j,k$
$$ j \times \gcd(a,b) = a \quad \land \quad k \times \gcd(a,b) = b $$
Some, or all, of the prime factors of $a$ are in $\gcd(a,b)$. In addition, $\gcd(a,b)$ contains no primes that are not in $a$ because then $\gcd(a,b)$ would not be a factor of $a$. Similarly, some, or all, of the prime factors of $b$ are in $\gcd(a,b)$, and $\gcd(a,b)$ contains no primes that are not in $b$.
Also, $\gcd(a,b)$ contains only primes that are in both $a$ and $b$. If it contained a prime that was in $a$ but not $b$, then $\gcd(a,b)$ would not be a factor of $b$. Similarly, if it contained a prime that was in $b$ but not in $a$, then $\gcd(a,b)$ would not be a factor of $a$.
If a prime $p$ occurs $\alpha \ge 0$ times in $a$ and $\beta \ge 0$ times in $b$, where $\alpha \ne \beta$ then it occurs $\min(\alpha, \beta)$ in $\gcd(a,b)$. If it occurred more frequently then $\gcd(a,b)$ would not be a divisor of both $a$ and $b$. If it occurred fewer times, then $\gcd(a,b)$ would not be the greatest common divisor.
This is equivalent to Proposition (2.21).