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Suppose $a,b$ be two positive integers such that $a = {p_{1}}^{r_{1}}{p_{2}}^{r_{2}}...{p_{n}}^{r_{n}}$ and $b = {p_{1}}^{s_{1}}{p_{2}}^{s_{2}}...{p_{n}}^{s_{n}}$, where $p_{i}$ are distinct primes and $r_{i} , s_{i} \in \mathbb N_{0}$ for $i = 1,2,...,n$ where $\mathbb N_{0} = \mathbb N \cup \{0\}$. Then show that $d = {p_{1}}^{t_{1}}{p_{2}}^{t_{2}}...{p_{n}}^{t_{n}}$ is the $gcd (a,b)$ where $t_{i} = \min \{r_{i} , s_{i}\}$ for $i = 1,2,...,n$.

It is easy to see that $d|a$ and $d|b$.But I find difficulty to show the second part i.e. if $x$ be any common divisor of $a$ and $b$.Then $x|d$.How can I show this?Please help me.

Thank you in advance.

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    It boils down to showing that if $q_i(x)$ is the exponent of $p_i$ in the prime factorization of $x$ and $y$ divides $z$ then $a_i(y) \leq a_i(z)$. – 2017-01-10
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    Notice if $k_i > t_i = \min(r_i, s_i)$ then $p^{k_i}$ does *not* divide both a and b. – 2017-01-10
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    Let N be a common divisor. If $N = \prod q_i^{k_i}$ divides *both* a and b then none of the prime $q$ can be different than any of $p_i$. And none of the $k_i$ can be greater than $t_i$. So $N = \prod p_i^{k_i}$ where {$p_i$} are *some* the the primes and $k_i \le t_i$ so $N \le d$. So $d$ is the greatest common divisor. – 2017-01-10
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    I want to show that if $x$ does not divide $d$ then $x$ fails to divide at least one of $a$ or $b$.But I can't do it.Please help me. – 2017-01-10
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    If x does not divide d then either i) x has a prime factor that d does not. Neither a nor b have that prime factor so x does not ddivide either a nor b. or ii) x has a prime factor $p_i$ whose power is larger than $t_i$. If it's larger than $t_i$, it is larger than $r_i$ or larger than $s_i$. If it is larger than $r_i$ then $x$ does not divide $a$. If it is larger than $s_i$ than $x$ does not divide $b$. – 2017-01-10

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Let $x$ be a common divisor.

Let $q$ be prime and $q|x$ then $q|a$ and $q|b$. The only prime factors of $a$ and $b$ are $p_i$. So $q = p_i$ for some $i$.

Let $k$ be the highest power of $q=p_i$ that divides $x$. That is $q^k|x$ but $q^{k+1} \not \mid x$. Then $q^k|a$ and $q^k|b$. If $k > r_i$ then $p_i^k\not \mid a$. If $k > s_i$ then $p_i^k \not \mid b$. So $k \le t_i = \min(r_i,s_i)$.

Let $x = \prod q_l^{k_l}$ be the prime factorization. We have shown that each $q_l = $ some $p_i$ and that each $k_l \le t_i$. So we know $\prod q_l^{k_l}|\prod p_i^{t_i}$. So $x|d$.

So $d$ is gcd.

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    Fantastic answer.This is something I am looking for.Thank you. – 2017-01-10