Description of the lattice that gcd and lcm form on a PID

567 Views Asked by At

I was reviewing my notes on principal ideal domains and I found this (to me) interesting property.

In any principal ideal domain $D$ every two elements have greatest common divisor $(a,b)$ and least common multiple. Therefore as internal operations over $D$ this two form a lattice.

Indeed, if I take $a \lor b = (a,b)$ as the supremum and $a \land b = [a,b]$ as the infimum it is clear that associativity and commutativity hold.

This gives some intuition on the properties that I have for these in the general context of an integral domain, for instance $a \lor 1 = 1,a \land 0 = 0$ and others.

However, I would like to know more about the properties of this lattice. To begin with I would like to know if it is distributive and complemented, since then I would have that it forms a Boolean algebra.

Other consequences:

I have realized also that in a principal ideal domain thanks to the Bézout's identity the traditional lattice ordered with the inclusion and with supremum and infimum given by $+,\cap$ has an interesting connection with this lattice:

$\langle a \rangle + \langle b \rangle = \langle (a,b) \rangle$

$\langle a \rangle \cap \langle b \rangle = \langle [a,b] \rangle$

$\langle a \rangle \cdot \langle b \rangle = \langle ab \rangle$

1

There are 1 best solutions below

2
On BEST ANSWER

Recall that a PID is a UFD. This is for $a,b,c \neq 0$ (distributivity obviously holds for $a,b$ or $c=0$) If $p$ is prime and $p^k$ divides $a$ and $b\lor c$, then it divides $b$ or it divides $c$, ans so it divides $a\land b$ or $a\land c$, therefore it divides $(a\land b)\lor (a\land c)$.

Conversely, if $p$ is prime and $p^k$ divides $(a\land b)\lor (a\land c)$, it divides (wlog) $a\land b$, so it divides $a$ and $b$, so it divides $b\lor c$, and so it divides $a\land (b\lor c)$.

Therefore $a\land (b\lor c) = (a\land b)\lor (a\land c)$

(I left a few details to you, of course) Therefore the lattice is distributive.

However, it is not complemented and is therefore not a boolean algebra : if $a,b\neq 0$, $ab\neq 0$ and $(ab)\subset (a)\cap (b)$.

It's a complete lattice because intersections of ideals are ideals. I don't know if much more can be said without further hypotheses