# abstract algebra #2

• Jan 27th 2010, 01:50 PM
Deepu
abstract algebra #2
if p is prime and (a,b)=p, then (a^2,b^2)=?
• Jan 28th 2010, 12:20 AM
Sudharaka
Dear Deepu,

$\displaystyle (a,b)=p$

Take $\displaystyle (a^2,b^2)=k$

Now since, $\displaystyle k\mid{a^2}$ and $\displaystyle k\mid{b^2}$

Therefore, $\displaystyle k\mid{a}$ and $\displaystyle k\mid{b}$

Then since the gcd of "a" and "b" is "p" ; $\displaystyle k\leq{p}$ ---------------A

Also since $\displaystyle p=(a,b)\Rightarrow{p\mid{a}}$ and $\displaystyle p\mid{b}\Rightarrow{p\mid{a^2}}$ and $\displaystyle p\mid{b^2}$

Then since the gcd of $\displaystyle a^2$ and $\displaystyle b^2$ is "k" ; $\displaystyle p\leq{k}$ ----------------B

From A and B ; p=k

Therefore $\displaystyle (a^2,b^2)=p$

Hope this helps.
• Jan 28th 2010, 01:16 PM
Deepu
Thank you very much.
• Jan 28th 2010, 02:25 PM
Defunkt
Quote:

Originally Posted by Sudharaka
...
Now since, $\displaystyle k\mid{a^2}$ and $\displaystyle k\mid{b^2}$

Therefore, $\displaystyle k\mid{a}$ and $\displaystyle k\mid{b}$

This implication is incorrect.

Consider, for example, $\displaystyle a=6,b=9 \Rightarrow a^2=36, ~b^2=81 \Rightarrow 9 | 36, ~ 9 | 81$ but 9 obviously does not divide 6.

---

Let $\displaystyle (a,b)=p \Rightarrow p|a, ~ p|b \Rightarrow p^2|a^2,~p^2|b^2$

So, we now know that $\displaystyle p^2$ is a common divisor of $\displaystyle a^2,~b^2$. I'll leave it to you to show that it is the greatest common divisor.
• Jan 28th 2010, 08:19 PM
chiph588@
Suppose $\displaystyle a=p_1^{\alpha_1} p_2^{\alpha_2} \cdot\cdot\cdot p_n^{\alpha_n}$ and
$\displaystyle b=p_1^{\beta_1} p_2^{\beta_2} \cdot\cdot\cdot p_n^{\beta_n}$ where $\displaystyle \alpha_i,\beta_j \geq 0$.

Set $\displaystyle k=(a,b)$. Then $\displaystyle k=p_1^{\min{\alpha_1,\beta_1}} p_2^{\min{\alpha_2,\beta_2}} \cdot\cdot\cdot p_n^{\min{\alpha_n,\beta_n}}$.

Now $\displaystyle a^2=p_1^{2\alpha_1} p_2^{2\alpha_2} \cdot\cdot\cdot p_n^{2\alpha_n}$ and $\displaystyle b^2=p_1^{2\beta_1} p_2^{2\beta_2} \cdot\cdot\cdot p_n^{2\beta_n}$.

So $\displaystyle (a^2,b^2) = p_1^{\min{2\alpha_1,2\beta_1}} p_2^{\min{2\alpha_2,2\beta_2}} \cdot\cdot\cdot p_n^{\min{2\alpha_n,2\beta_n}} = p_1^{2\min{\alpha_1,\beta_1}} p_2^{2\min{\alpha_2,\beta_2}} \cdot\cdot\cdot p_n^{2\min{\alpha_n,\beta_n}} = k^2$.

Hence $\displaystyle (a^2,b^2) = (a,b)^2$.
• Jan 28th 2010, 09:52 PM
Sudharaka
Quote:

Originally Posted by Defunkt
This implication is incorrect.

Consider, for example, $\displaystyle a=6,b=9 \Rightarrow a^2=36, ~b^2=81 \Rightarrow 9 | 36, ~ 9 | 81$ but 9 obviously does not divide 6.

---

Let $\displaystyle (a,b)=p \Rightarrow p|a, ~ p|b \Rightarrow p^2|a^2,~p^2|b^2$

So, we now know that $\displaystyle p^2$ is a common divisor of $\displaystyle a^2,~b^2$. I'll leave it to you to show that it is the greatest common divisor.

Dear Defunkt,

Thank you for showing my mistake. I tried to solve the problem again using the suggestion you had given, but still haven't had any luck.