symmetric monoidal (∞,1)-category of spectra
A ring $R$ is a principal ideal domain if
it is an integral domain;
every ideal in $R$ is a principal ideal.
Often pid is used as an abbreviation of “principal ideal domain”.
any field
the ring of integers
a polynomial ring with coefficients in a field (in fact, for any commutative ring $R$, the ring $R[x]$ is a pid if and only if $R$ is a field)
a discrete valuation ring (for example, a ring of formal power series over a field)
the ring of entire holomorphic functions on $\mathbb{C}$
If $R$ is a pid, then any submodule $M$ of a free module $F$ over $R$ is also free. (For the converse statement, see here.)
By freeness of $F$, there exists an isomorphism $F \cong \sum_{j \in J} R_j$, a coproduct of copies $R_j$ of $R$ (as a module over the ring $R$) indexed over a set $J$, which we assume well-ordered using the axiom of choice. Define submodules of $F$:
Any element of $M \cap F_{\leq j}$ can be written uniquely as $(x, r)$ where $x \in F_{\lt j}$ and $r \in R_j$. Define a homomorphism
by $p_j((x, r)) = r$. The kernel of $p_j$ is $M \cap F_{\lt j}$, and we have an exact sequence
where $\im p_j$ is a submodule (i.e., an ideal) of $R$, hence generated by a single element $r_j$. Let $K \subseteq J$ consist of those $j$ such that $r_j \neq 0$, and for $k \in K$, choose $m_k$ such that $p_k(m_k) = r_k$. We claim that $\{m_k: k \in K\}$ forms a basis for $M$.
First we prove linear independence of $\{m_k\}$. Suppose $\sum_{i=1}^{n} a_i m_{k_i} = 0$, with $k_1 \lt k_2 \lt \ldots \lt k_n$. Applying $p_{k_n}$, we get $a_n p_{k_n}(m_{k_n}) = a_n r_{k_n} = 0$. Since $r_{k_n} \neq 0$, we have $a_n = 0$ (since we are working over a domain). The assertion now follows by induction.
Now we prove that the $m_k$ generate $M$. Assume otherwise, and let $j \in J$ be the least $j$ such that some $m \in M \cap F_{\leq j}$ cannot be written as a linear combination of the $m_k$, for $k \in K$. If $j \notin K$, then $m \in M \cap F_{\lt j}$, so that $m \in M \cap F_{\leq i}$ for some $i \lt j$, but this contradicts minimality of $j$. Therefore $j \in K$. Now, we have $p_j(m) = r \cdot r_j$ for some $r$; put $m' = m - r m_j$. Clearly
and so $m' \in M \cap F_{\lt j}$. Thus $m' \in M \cap F_{\leq i}$ for some $i \lt j$. At the same time, $m'$ cannot be written as a linear combination of the $m_k$; again, this contradicts minimality of $j$. Thus the $m_k$ generate $M$, as claimed.
Since the integers $\mathbb{Z}$ for a pid, and abelian groups are the same as $\mathbb{Z}$-modules, we have
(Dedekind) A subgroup of a free abelian group is also free abelian.
The analog of this statement for possibly non-abelian groups is the Nielsen-Schreier theorem.
Also, since projective modules are retracts of free modules, we have
Projective modules over a pid are free. In particular, submodules of projective modules are projective.
A finitely generated torsionfree module $M$ over a pid $R$ is free.
Let $S$ be a finite set of generators of $M$, and let $T \subseteq S$ be a maximal subset of linearly independent elements. (Unless $M = 0$, then $T$ has at least one element, because $M$ is torsionfree.) We claim that $M$ can be embedded as a submodule of the free module $F$ generated by $T$ (which in turn is the span of $T$ as a submodule $F \subseteq M$). By Theorem 1, it follows that $M$ is free.
Let $x_1, \ldots, x_n$ be the elements of $T$. It follows from maximality of $T$ that for any $m \in S - T$, there is a linear relation
with $r_m \neq 0$. For each $m$ in the complement $S - T$, pick such an $r_m$, and form $r = \prod_{m \in S - T} r_m$. Then the image of the scalar multiplication $\lambda_r \colon M \to M$ factors through $F \subseteq M$, and $M \to \lambda_r(M)$ is monic because $M$ is torsionfree. This completes the proof.
Let $R$ be a pid. Then an $R$-module $M$ is torsionfree if and only if it is flat.
Suppose $M$ is flat. Let $K$ be the field of fractions of $R$; since $R$ is a domain, we have a monic $R$-module map $R \to K$. By flatness, we have an induced monomorphism $M \cong R \otimes_R M \to K \otimes_R M$. For any nonzero $r \in R$, the naturality square
commutes, and since the map $1 \otimes \lambda_r$ is multiplication by a non-zero scalar on a vector space, it follows that this map and therefore also $\lambda_r$ is monic, i.e., $M$ is torsionfree.
In the other direction, suppose $M$ is torsionfree. Any module is the filtered colimit over the system of finitely generated submodules and inclusions between them; in this case all the finitely generated submodules of $M$ are torsion-free and hence free, by Proposition 1. Thus $M$ is a filtered colimit of free modules; it is therefore flat by a standard result proved here.
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