Simple prime/GCD proof question

[jeffreydk]

Hello, I'm working out of Hungerford's Abstract Algebra text and this proof has been bothering me because I think I know why it works and it's so simple but I can't figure out how you would show a rigorous proof of it...

If $$a=p_1^{r_1}p_2^{r_2}p_3^{r_3} \cdots p_k^{r_k}$$ and $$b=p_1^{s_1}p_2^{s_2}p_3^{s_3} \cdots p_k^{s_k}$$

where $$p_1,p_2, \ldots ,p_k$$ are distinct positive primes and each $$r_i,s_i \geq 0$$ ,

then prove that $$GCD(a,b)=p_1^{n_1}p_2^{n_2}p_3^{n_3} \cdots p_k^{n_k}$$, where for each $$i \text{, } n_i=\min(r_i,s_i)$$.

I had thought I might be able to show it through using the definition of GCD as a linear combination--where the GCD, d, is the smallest positive element in the set

$$S=\big\{ d=am+bn \text{ } \vert \text{ } m,n \in \mathbb{Z} \big\}$$

and therefore I could use that to show that the GCD(a,b) must be the minimum of each $$r_i,s_i$$. But that just isn't working out and seems like it's making the proof too complicated anyway. I apologize that I don't have more of a solution worked out--any hint or help would be greatly appreciated. Thank you.

 

[tiny-tim]

it's the greatest!

If $$a=p_1^{r_1}p_2^{r_2}p_3^{r_3} \cdots p_k^{r_k}$$ and $$b=p_1^{s_1}p_2^{s_2}p_3^{s_3} \cdots p_k^{s_k}$$

where $$p_1,p_2, \ldots ,p_k$$ are distinct positive primes and each $$r_i,s_i \geq 0$$ ,

then prove that $$GCD(a,b)=p_1^{n_1}p_2^{n_2}p_3^{n_3} \cdots p_k^{n_k}$$, where for each $$i \text{, } n_i=\min(r_i,s_i)$$.

Hi jeffreydk!

Well, it obviously is a divisor of both … and if you multiply it by anything else, it won't be, and therefore …

it's the greatest! ​

 

[jeffreydk]

ohh hah yes of course, thanks.