1. abstract algebra #3

If p is prime and p|a^n, is it true that p^n|a^n? Justify your answer.

Thank you.

2. Yes indeed. If $\displaystyle p|a^n \Rightarrow p|a$. You can show this inductively:

It follows from the prime-property: $\displaystyle p|xy \Rightarrow p|x$ or $\displaystyle p|y$. Take $\displaystyle x= a, y= a^{n-1}$. If $\displaystyle p|a$ you're done if $\displaystyle p|a^{n-1}$ you go on with $\displaystyle x= a, y= a^{n-2}$. Eventually you'll obtain $\displaystyle p|a$

Conclusion: $\displaystyle p^n|a^n$

3. Originally Posted by Dinkydoe
Yes indeed. If $\displaystyle p|a^n \Rightarrow p|a$. You can show this inductively:

It follows from the prime-property: $\displaystyle p|xy \Rightarrow p|x$ or $\displaystyle p|y$. Take $\displaystyle x= a, y= a^{n-1}$. If $\displaystyle p|a$ you're done if $\displaystyle p|a^{n-1}$ you go on with $\displaystyle x= a, y= a^{n-2}$. Eventually you'll obtain $\displaystyle p|a$

Conclusion: $\displaystyle p^n|a^n$
thank you. you showed p|a. but what about p^n.

4. Originally Posted by Deepu
thank you. you showed p|a. but what about p^n.
$\displaystyle p\mid a\implies a=kp$ so $\displaystyle a^n=k^np^n$...so