# Showing that $(m)\cap (n)=(\operatorname{lcm}(m,n))$ and $(m)+(n)=(\gcd(m,n))$ for any $m,n\in\mathbb{Z}$

Suppose that $I$ and $J$ are ideals of $\mathbb{Z}$, with $I=(m)$ and $J=(n)$. This question has two parts:

1) Let r be the least common multiple of $m$ and $n$. Show that $I\cap J = (r)$.

2) Let $d=(m,n)$. Show that $I+J = (d)$.

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Remember that $(a) \subseteq (b)$ if and only if $b$ divides $a$.

1) $(\ell) = (m) \cap (n) \subseteq (m)$ and $(n)$. Thus $m$ and $n$ divide $\ell$ (so $\ell$ is common multiple). What if $k$ is divisible by $m$ and $n$? What would imply that $\ell$ divides $k$ so that $\ell$ is the least common multiple?

2) $(d)=(m,n)=(m)+(n)$. Then $(m)$ and $(n) \subseteq (d)$. Thus $d$ divides $m$ and $n$ (so $d$ is a common divisor). What if $k$ divides $m$ and $n$? What would imply that $k$ divides $d$ so that $d$ is the greatest common divisor?

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HINT $\$ It is straightforward if one employs the universal definitions of lcm and gcd:

$\begin{eqnarray}{} 1)\qquad\rm (m)\cap(n) = (k)\ &\iff&\rm\ [\ (m),(n)\supset (j) &\iff&\rm (k)\supset (j)\ ]\ \iff\ &\rm[\ m,n\ |\ j\ &\iff&\rm\ k\ |\ j\ ] \\ 2)\qquad\rm (m)+(n) = (k)\ &\iff&\rm\ [\ (j)\supset (m),(n)\ &\iff&\rm (j)\supset (k)\ ]\ \iff\ \ &\rm[\ j\ |\ m,n\ &\iff&\rm\ j\ |\ k\ ] \end{eqnarray}$

Notice how the above makes crystal-clear the innate duality between lcm and gcd.

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They're also respectively join and meet in the divisibility lattice (or coproduct and product, to make duality even more obvious). – Alexei Averchenko Nov 17 '11 at 6:51
There is something slightly misleading in this answer, in that it suggests that both identities of the question follow from the cited universal definitions, in any integral domain where all gcd's and lcm's exist (in particular in any UFD). However this is only true for the first (lcm) identity (indeed the principal ideal generated by lcm$(n,m)$ equals $(m)\cap(n)$ straight from the definition). But 2) should be read as that if $(m)+(n)$ is a principal ideal then its generator is $\gcd(n,m)$; however $(m)+(n)$ need not be so, and if not it will be strictly contained in $(\gcd(n,m))$. – Marc van Leeuwen Mar 17 '14 at 12:26
But of course the argument is perfectly valid in any principal ideal domain, like $\Bbb Z$ of the question, or even in any Bezout domain. – Marc van Leeuwen Mar 17 '14 at 12:27
@marc Try as I may, I cannot imagine how you could (mis)read the answer to think that it "suggest that..." That is quite far-fetched, esp. considering that the question is about $\,\Bbb Z\,$ and the answer makes no mention whatsoever of generalizations. – Bill Dubuque Mar 17 '14 at 12:51
Well, I guess the suggestion comes in part from the use of "universal" (which might suggest "not just in $\Bbb Z$", though maybe it means something else), and that under the link it says "the general definition of LCM in an arbitrary domain" (which indeed it is). If an argument uses only quite general definitions/properties, one might expect it to remain valid in that generality. It is legitimate to take a slick proof like this one, and ask oneself which hypotheses on the ring are actually being used. I just wanted to remark that $1$ of the $4$ $\iff$'s uses the Bezout domain hypothesis. – Marc van Leeuwen Mar 17 '14 at 13:14