0

Let $R$ be an UFD and $a,b\in R\setminus\left\{ 0\right\}$. Prove that $ab=\operatorname{lcm}(a,b)\gcd(a,b)$.

Truong
  • 4,162
  • It's correct modulo multiplication by units. – Derek Holt Nov 27 '13 at 14:18
  • Can you explain more precisely to me? – Truong Nov 27 '13 at 14:20
  • Write down the factorisation of $a$ and $b$ as products of powers of irreducibles, and then use that to work out the lcm and gcd. For example, in ${\mathbb Z}$, $a=2^3.3.5^2.11$, $b=2^2.3^4.5.7^2$, the lcm is $2^3.3^4.5^2.7^2.11$ and the gcd is $2^2.3.5$. It's basically the same argument for any UFD. – Derek Holt Nov 27 '13 at 14:24
  • I know that in a UFD $R$, every finite sets of nonzero elements of $R$ have $gcd$. Hence $a,b$ have $gcd$ but why $ab=lcm(a,b).gcd(a,b)$ (modulo multiplication by units)? – Truong Nov 27 '13 at 14:37
  • 1
    Your comment suggests me that you are in a GCD domain (that is, an integral domain where every two non-zero elements have a $\gcd$), not in an UFD. –  Nov 27 '13 at 19:37

1 Answers1

0

Please be aware that commenters and readers like to be interacted with socially. When you copy a question verbatim without writing any of your own thoughts or even a greeting or any formatting to indicate you're posting a quote, you invoke the image of someone anonymously exploiting others to do their homework for them for free. Even if this isn't homework, please see:

http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question/

In a UFD this is more concrete than the general situation (a GCD domain). This is because all elements can be thought of as multisets of primes (each prime that divides a given element appears in this multiset with multiplicity equal to its exponent in the factorization). The gcd is a kind of "intersection" whereas the lcm is a kind of "union," and our law is a sort of spiritual realization of the combinatorial fact $|A|+|B|=|A\cap B|+|A\cup B|$ (interpreted correctly for multisets).

In particular, in a UFD one can prove the following explicit formulas:

$$\gcd(a,b)=\prod_{\pi\mid a,b}\pi^{\large\min\{v_{\pi}(a),v_\pi(b)\}}$$

$$\operatorname{lcm}(a,b)=\prod_{\pi\mid a,b}\pi^{\large\max\{v_\pi(a),v_\pi(b)\}}$$

where $v_\pi(\cdot)$ represents the power of $\pi$ in a factorization and $\pi$ runs over irreducibles. The desired law $ab=\operatorname{lcm}(a,b)\gcd(a,b)$ boils down to $e_1+e_2=\min\{e_1,e_2\}+\max\{e_1,e_2\}$ in $\Bbb N$ (which can be seen clearly by assume $e_1\le e_2$ without loss of generality).

More generally, in a GCD-domain, where gcds are defined (up to units) but factorization needn't be unique, the law still holds. Firstly, we define gcd and lcm by "universal properties," and only up to units. In particular, let $R$ be a GCD domain, and consider the multiplicative monoid $(R\setminus0)/R^\times$ (i.e. the nonzero elements modulo units as an algebraic structure with multiplication and identity).

Define the gcd of $a$ and $b$ as $(a,b)$ and their $\rm lcm$ as $[a,b]$, defined uniquely by

$$d\mid a,b\iff d\mid (a,b), \qquad a,b\mid m\iff [a,b]\mid m.$$

If $[a,b]$ exists then $ab/[a,b]\in R$ satisfies the universal property of $(a,b)$ via

$$d\mid a,b\iff db,ad\mid ab\iff a,b\mid ab/d\iff [a,b]\mid ab/d\iff d\mid ab/[a,b].$$

This is explained in Bill Dubuque's answer here.

Community
  • 1
anon
  • 151,657