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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 $I+J = (d)$.

2 Answers 2

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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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    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.2014-03-17