**Proof.**
If $A$ has weak dimension $\leq 1$, then the resolution $0 \to I \to A \to A/I \to 0$ shows that every ideal $I$ is flat by Lemma 15.63.2. Hence (1) $\Rightarrow $ (2).

Assume (4). Let $M$ be an $A$-module. Choose a surjection $F \to M$ where $F$ is a free $A$-module. Then $\mathop{\mathrm{Ker}}(F \to M)$ is flat by assumption, and we see that $M$ has tor dimension $\leq 1$ by Lemma 15.63.6. Hence (4) $\Rightarrow $ (1).

Every ideal is the union of the finitely generated ideals contained in it. Hence (3) implies (2) by Algebra, Lemma 10.38.3. Thus (3) $\Leftrightarrow $ (2).

Assume (2). Suppose that $N \subset M$ with $M$ a flat $A$-module. We will prove that $N$ is flat. We can write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ with each $M_ i$ finite free, see Algebra, Theorem 10.80.4. Setting $N_ i \subset M_ i$ the inverse image of $N$ we see that $N = \mathop{\mathrm{colim}}\nolimits N_ i$. By Algebra, Lemma 10.38.3. it suffices to prove $N_ i$ is flat and we reduce to the case $M = R^{\oplus n}$. In this case the module $N$ has a finite filtration by the submodules $R^{\oplus j} \cap N$ whose subquotients are ideals. By (2) these ideals are flat and hence $N$ is flat by Algebra, Lemma 10.38.13. Thus (2) $\Rightarrow $ (4).

Assume $A$ satisfies (1) and let $\mathfrak p \subset A$ be a prime ideal. By Lemmas 15.96.14 and 15.96.4 we see that $A_\mathfrak p$ satisfies (1). We will show $A$ is a valuation ring if $A$ is a local ring satisfying (3). Let $f \in \mathfrak m$ be a nonzero element. Then $(f)$ is a flat nonzero module generated by one element. Hence it is a free $A$-module by Algebra, Lemma 10.77.5. It follows that $f$ is a nonzerodivisor and $A$ is a domain. If $I \subset A$ is a finitely generated ideal, then we similarly see that $I$ is a finite free $A$-module, hence (by considering the rank) free of rank $1$ and $I$ is a principal ideal. Thus $A$ is a valuation ring by Algebra, Lemma 10.49.15. Thus (1) $\Rightarrow $ (5).

Assume (5). Let $I \subset A$ be a finitely generated ideal. Then $I_\mathfrak p \subset A_\mathfrak p$ is a finitely generated ideal in a valuation ring, hence principal (Algebra, Lemma 10.49.15), hence flat. Thus $I$ is flat by Algebra, Lemma 10.38.18. Thus (5) $\Rightarrow $ (3). This finishes the proof of the lemma.
$\square$

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