\section*{Comparable Infinite-Order Elements in Picard Groups of Total Rings of Fractions}

Consider a commutative ring \(R\) and its Picard group \(\mathrm{Pic}(R)\). For
two invertible \(R\)-modules \(M,N\), define \([M]\le [N]\) to mean that there
exists an injective \(R\)-linear map from \(M\) to \(N\). This makes
\(\mathrm{Pic}(R)\) a preordered group.

The order is trivial if \(\mathrm{Pic}(Q(R))\) is trivial, where \(Q\) denotes
the total ring of fractions. In particular, this holds if \(R\) is a domain,
because every invertible \(R\)-module \(M\) is isomorphic to an ideal, meaning
\([M]\le [R]\); see https://mathoverflow.net/a/375757/3332. Therefore, for
arbitrary \(M,N\),
\[
[M]=[N][M\otimes_R N^\vee]\le[N][R]=[N].
\]

Now focus on the case \(R=Q(R)\), i.e. every element of \(R\) is either a unit
or a zero divisor. This implies that invertible \(R\)-modules satisfy the
Schr\"oder--Bernstein property, making \(\mathrm{Pic}(R)\) a partially ordered
group. A proof due to Jun Koizumi is as follows: if \([M]\le[R]\) and
\([R]\le[M]\), equivalently \([M^\vee]\le[R]\), but \([R]\ne[M]\), then \(M\)
embeds as a proper ideal of \(R\). Such an ideal consists purely of zero
divisors, but then
\[
[R]=[M][M^\vee]\le[M][R]=[M],
\]
which contradicts the fact that \(1\in R\) is not annihilated by a nonzero
element of \(R\). Consequently, no element of finite order in
\(\mathrm{Pic}(R)\) is comparable to \([R]\), apart from \([R]\) itself.

\paragraph{Open problem.}
Could there exist an element \([M]\in\mathrm{Pic}(R)\) of infinite order that is
comparable to \([R]\)? Equivalently, could there be an invertible, proper ideal
of \(R\)?

Such an ideal necessarily consists of zero divisors since \(R=Q(R)\). If \(R\)
is Noetherian, the whole ideal is annihilated by some nonzero element of \(R\);
see https://math.stackexchange.com/a/4846756/12932. This is impossible since
\(R\) acts faithfully on any invertible \(R\)-module. In fact, one can show
that \(R\) is semi-local and therefore \(\mathrm{Pic}(R)=1\), if \(R\) is
Noetherian; see https://mathoverflow.net/a/499611/3332. On the other hand,
Koizumi's construction produces a total ring of fractions whose Picard group
contains an arbitrary Picard group \(\mathrm{Pic}(A)\) as a direct factor.

If an invertible ideal \(I\subsetneq R=Q(R)\) exists, then \(R_{\mathfrak{p}}\)
must fail to be a total ring of fractions for some prime ideal
\(\mathfrak{p}\subset R\). Indeed, each \(I_{\mathfrak{p}}\subseteq
R_{\mathfrak{p}}\) is free of rank \(1\), so equality must hold if every
non-zero-divisor in \(R_{\mathfrak{p}}\) is a unit, which then implies \(I=R\).
Similarly, since \(I\) is Zariski-locally free, \(R_x\) must fail to be a total
ring of fractions for some \(x\in R\).

Examples of such rings \(R\), including one similar to Koizumi's, appeared in
\emph{Misconceptions about \(K_X\)}:
https://freedommathdance.blogspot.com/2012/11/misconceptions-about-kx.html.
Another example is spelt out here:
https://www.perplexity.ai/search/732e9876-1d0d-4f63-bf27-6f19d284b5eb#1.

The embeddability order defines an ordered group because tensoring with any
flat module, in particular an invertible one, preserves injectivity. If
injectivity is replaced by surjectivity, flatness is not needed since tensoring
is always right exact, but the resulting order is uninteresting: any surjective
linear map between invertible modules is bijective. See the P.S. of
https://mathoverflow.net/q/511864/3332 for a proof.

\paragraph{Lean verification.}
https://github.com/google-deepmind/formal-conjectures/blob/main/FormalConjectures/Mathoverflow/507128.lean