# abstract algebra #2

• Jan 27th 2010, 02:50 PM
Deepu
abstract algebra #2
if p is prime and (a,b)=p, then (a^2,b^2)=?
• Jan 28th 2010, 01:20 AM
Sudharaka
Dear Deepu,

$(a,b)=p$

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

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

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

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

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

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

From A and B ; p=k

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

Hope this helps.
• Jan 28th 2010, 02:16 PM
Deepu
Thank you very much.
• Jan 28th 2010, 03:25 PM
Defunkt
Quote:

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

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

This implication is incorrect.

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

---

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

So, we now know that $p^2$ is a common divisor of $a^2,~b^2$. I'll leave it to you to show that it is the greatest common divisor.
• Jan 28th 2010, 09:19 PM
chiph588@
Suppose $a=p_1^{\alpha_1} p_2^{\alpha_2} \cdot\cdot\cdot p_n^{\alpha_n}$ and
$b=p_1^{\beta_1} p_2^{\beta_2} \cdot\cdot\cdot p_n^{\beta_n}$ where $\alpha_i,\beta_j \geq 0$.

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

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

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

Hence $(a^2,b^2) = (a,b)^2$.
• Jan 28th 2010, 10:52 PM
Sudharaka
Quote:

Originally Posted by Defunkt
This implication is incorrect.

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

---

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

So, we now know that $p^2$ is a common divisor of $a^2,~b^2$. 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.