Proving that lcm(a,b)gcd(a,b)=ab

[Mr Davis 97]

Homework Statement

Let $a\neq 0, b\neq 0$ be integers and let $(a),(b)$ be the ideals they generate in $\mathbb{Z}$. Show that $\operatorname{gcd}(a,b)\operatorname{lcm}(a,b) =ab$

Homework Equations

The Attempt at a Solution

Suppose that I already know from the second isomorphism theorem that
$$\frac{(a)}{(\gcd(a,b))} \cong \frac{(\rm{lcm}(a,b))}{(b)}.$$ How can I get from this the result $\operatorname{gcd}(a,b)\operatorname{lcm}(a,b) =ab$?

 

[andrewkirk]

There's no need to use isomorphism theorems. Basic set theory will suffice.

Let the prime factors of $ab$ be $p_1,...,p_n$ and let $a=\prod_{k=1}^n p_k{}^{r_k}$ and $b=\prod_{k=1}^n p_k{}^{s_k}$.

Now write each of gcd($a,b$), lcm($a,b$) and $ab$ in that prime-factorised product form (you'll need to use min and max functions). It should be easy to prove the required identity from there, given that $p^{\min(\alpha,\beta)}p^{\max(\alpha,\beta)} = p^{\alpha+\beta}$.

 

[fresh_42]

There's no need to use isomorphism theorems.

... but I sounds as an elegant idea.

Suppose that I already know from the second isomorphism theorem ...

Consideration of $\mathbb{Z}_a\; , \;\mathbb{Z}_b\; , \;\mathbb{Z}_a + \mathbb{Z}_b\; , \;\mathbb{Z}_a \cap \mathbb{Z}_b$ should do.

 

[Mr Davis 97]

... but I sounds as an elegant idea.

Consideration of $\mathbb{Z}_a\; , \;\mathbb{Z}_b\; , \;\mathbb{Z}_a + \mathbb{Z}_b\; , \;\mathbb{Z}_a \cap \mathbb{Z}_b$ should do.

How can I consider these rings such as $\mathbb{Z}_a$ and $\mathbb{Z}_b$ if they are not principal ideals of $\mathbb{Z}$, which they must be by hypothesis?

 

[fresh_42]

How can I consider these rings such as $\mathbb{Z}_a$ and $\mathbb{Z }_b$ if they are not principal ideals of $\mathbb{Z}$, which they must be by hypothesis?

Sorry, my fault. I meant $a\mathbb{Z}\; , \;b\mathbb{Z}\; , \;(a+b)\mathbb{Z}$ or $(ab)\mathbb{Z}$ and $a\mathbb{Z} \cap b\mathbb{Z}$ of course, the ideals, not the quotients.

 

[Mr Davis 97]

Sorry, my fault. I meant $a\mathbb{Z}\; , \;b\mathbb{Z}\; , \;(a+b)\mathbb{Z}$ or $(ab)\mathbb{Z}$ and $a\mathbb{Z} \cap b\mathbb{Z}$ of course, the ideals, not the quotients.

I'm not seeing how consideration of these implies the result though

 

[fresh_42]

I'm not seeing how consideration of these implies the result though

Yes, you completely confused me, as we need the quotients for the finite number of elements to represent the formula.
... guess, I should do it first on my own ...

 

[Mr Davis 97]

Yes, you completely confused me, as we need the quotients for the finite number of elements to represent the formula.
... guess, I should do it first on my own ...

I think maybe we just look at subgroup indices

 

[fresh_42]

I must admit, that I haven't found a nice way, although I still think there is one. But my attempts all ended up in diagram chasing. To use isomorphism theorems, one probably needs to show that $(a\mathbb{Z}\cap b\mathbb{Z})/ ab\mathbb{Z} \cong \mathbb{Z}/(a\mathbb{Z}+b\mathbb{Z})$. Maybe you have an idea how to show that fast. I even tried to use the four, five and nine lemmas.