Proposition 2

(i) In $D(A)$ every non-empty set bounded above admits a least upper bound. More precisely, if $(a_i)$ is a non-empty family of elements of $I(A)$ which is bounded above, then

(ii) In $D(A)$ every non-empty set bounded below admits a greatest lower bound. More precisely, if $(a_i)$ is a non-empty family of elements of $I(A)$ which is bounded below, then (iii) The set $D(A)$ is a lattice.

Let $(a_i)$ be a non-empty family of elements of $I(A)$ which is bounded above. To say that a divisorial ideal $b$ bounds this family above amounts to saying that it is contained in all the $a_i$, that is that $b$ is contained in $\bigcap_i 6_i$. Hence $\bigcap_i a_i \neq (0)$ and $\bigcap_i 6_i$, is therefore a divisorial ideal, which shows (i).

Now let $(6_i)$ be a non-empty family of elements of $I(A)$ which is bounded below. To say that a divisorial ideal $b$ bounds this family below means that it contains all the $a_i$, that is (since $b$ is divisorial) that it contains all the $a_i$, or also that $b \supset \sum_i a_i$. This proves (ii).

Finally, to prove (iii) it is sufficient by (i) and (ii) to prove that, if $a, b$ are in $I(A)$, the set $\{a, b\}$ is bounded both above and below in $I(A)$; now it is bounded above by $a \cap b$ (which is distinct from $(0)$). It is bounded below by $a + b$, for $a + b \in I(A)$: if $d$ and $d'$ are non-zero elements of $A$ such that $da \subset A$ and $d'b \subset A$, then $dd'(a + b) \subset A$.