The Stacks project

Proof. Suppose $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ Q) \subset \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ Q)$ for all finitely presented modules $Q$. If $Q$ is an arbitrary module, write $Q = \mathop{\mathrm{colim}}\nolimits _{i \in I} Q_ i$ as a colimit of a directed system of finitely presented modules $Q_ i$. Then $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_{Q_ i}) \subset \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_{Q_ i})$ for all $i$. Since taking directed colimits is exact and commutes with tensor product, it follows that $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ Q) \subset \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ Q)$. $\square$

Proof. Recall that $N'$ is $M' \oplus N$ modulo the submodule consisting of elements $(g(x), -f(x))$ for $x \in M$. From the construction of $N'$ we have a short exact sequence

Since tensoring commutes with taking pushouts, we have such a short exact sequence

for every $R$-module $Q$. So $f'$ is universally injective if and only if $\mathop{\mathrm{Ker}}(f \otimes \text{id}_ Q ) \subset \mathop{\mathrm{Ker}}(g \otimes \text{id}_ Q)$ for every $Q$, if and only if $g$ dominates $f$. $\square$

Proof. Consider the pushout of $f$ and $g$ as in the statement of Lemma 10.88.4. From the construction of the pushout it follows that $\mathop{\mathrm{Coker}}(f') = \mathop{\mathrm{Coker}}(f)$, so $\mathop{\mathrm{Coker}}(f')$ is of finite presentation. Then by Lemma 10.82.4, $f'$ is universally injective if and only if

splits. This is the case if and only if there is a map $h' : N' \to M'$ such that $h' \circ f' = \text{id}_{M'}$. From the universal property of the pushout, the existence of such an $h'$ is equivalent to $g$ factoring through $f$. $\square$

Proof. First we prove the equivalence of (1) and (2). Suppose (1) holds and let $i \in I$. Corresponding to the map $f_ i: M_ i \to M$, we can choose $g: M_ i \to Q$ as in (1). Since $M_ i$ and $Q$ are of finite presentation, so is $\mathop{\mathrm{Coker}}(g)$. Then by Lemma 10.88.5, $f_ i : M_ i \to M$ factors through $g: M_ i \to Q$, say $f_ i = h \circ g$ for some $h: Q \to M$. Then since $Q$ is finitely presented, $h$ factors through $M_ j \to M$ for some $j \geq i$, say $h = f_ j \circ h'$ for some $h': Q \to M_ j$. In total we have a commutative diagram

Thus $f_{ij}$ dominates $g$. But $g$ dominates $f_ i$, so $f_{ij}$ dominates $f_ i$.

Conversely, suppose (2) holds. Let $P$ be of finite presentation and $f: P \to M$ a module map. Then $f$ factors through $f_ i: M_ i \to M$ for some $i \in I$, say $f = f_ i \circ g'$ for some $g': P \to M_ i$. Choose by (2) a $j \geq i$ such that $f_{ij}$ dominates $f_ i$. We have a commutative diagram

From the diagram and the fact that $f_{ij}$ dominates $f_ i$, we find that $f$ and $f_{ij} \circ g'$ dominate each other. Hence taking $g = f_{ij} \circ g' : P \to M_ j$ works.

Next we prove (2) is equivalent to (3). Let $i \in I$. It is always true that $f_ i$ dominates $f_{ik}$ for $k \geq i$, since $f_ i$ factors through $f_{ik}$. If (2) holds, choose $j \geq i$ such that $f_{ij}$ dominates $f_ i$. Then since domination is a transitive relation, $f_{ij}$ dominates $f_{ik}$ for $k \geq i$. All $M_ i$ are of finite presentation, so $\mathop{\mathrm{Coker}}(f_{ik})$ is of finite presentation for $k \geq i$. By Lemma 10.88.5, $f_{ij}$ factors through $f_{ik}$ for all $k \geq i$. Thus (2) implies (3). On the other hand, if (3) holds then for any $R$-module $N$, $f_{ij} \otimes _ R \text{id}_ N$ factors through $f_{ik} \otimes _ R \text{id}_ N$ for $k \geq i$. So $\mathop{\mathrm{Ker}}(f_{ik} \otimes _ R \text{id}_ N) \subset \mathop{\mathrm{Ker}}(f_{ij} \otimes _ R \text{id}_ N)$ for $k \geq i$. But $\mathop{\mathrm{Ker}}(f_ i \otimes _ R \text{id}_ N: M_ i \otimes _ R N \to M \otimes _ R N)$ is the union of $\mathop{\mathrm{Ker}}(f_{ik} \otimes _ R \text{id}_ N)$ for $k \geq i$. Thus $\mathop{\mathrm{Ker}}(f_ i \otimes _ R \text{id}_ N) \subset \mathop{\mathrm{Ker}}(f_{ij} \otimes _ R \text{id}_ N)$ for any $R$-module $N$, which by definition means $f_{ij}$ dominates $f_ i$.

It is trivial that (3) implies (4) implies (5). We show (5) implies (3). Let $N = \prod _{s \in I} M_ s$. If (5) holds, then given $i \in I$ choose $j \geq i$ such that

for all $k \geq j$. Passing the product over $s \in I$ outside of the $\mathop{\mathrm{Hom}}\nolimits $'s and looking at the maps on each component of the product, this says

for all $k \geq j$ and $s \in I$. Taking $s = j$ we have

for all $k \geq j$. Since $f_{ij}$ is the image of $\text{id} \in \mathop{\mathrm{Hom}}\nolimits (M_ j, M_ j)$ under $\mathop{\mathrm{Hom}}\nolimits (M_ j, M_ j) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ j)$, this shows that for any $k \geq j$ there is $h \in \mathop{\mathrm{Hom}}\nolimits (M_ k, M_ j)$ such that $f_{ij} = h \circ f_{ik}$. If $j \geq k$ then we can take $h = f_{kj}$. Hence (3) holds. $\square$

Proof. Write $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ and $N = \mathop{\mathrm{colim}}\nolimits _{j \in J} N_ j$ as directed colimits of finitely presented $R$-modules. Denote $f_{ii'} : M_ i \to M_{i'}$ and $g_{jj'} : N_ j \to N_{j'}$ the transition maps. Then $M_ i \otimes _ R N_ j$ is a finitely presented $R$-module (see Lemma 10.12.14), and $M \otimes _ R N = \mathop{\mathrm{colim}}\nolimits _{(i, j) \in I \times J} M_ i \otimes _ R M_ j$. Pick $(i, j) \in I \times J$. By the definition of a Mittag-Leffler module we have Proposition 10.88.6 (3) for both systems. In other words there exist $i' \geq i$ and $j' \geq j$ such that for every choice of $i'' \geq i$ and $j'' \geq j$ there exist maps $a : M_{i''} \to M_{i'}$ and $b : M_{j''} \to M_{j'}$ such that $f_{ii'} = a \circ f_{ii''}$ and $g_{jj'} = b \circ g_{jj''}$. Then it is clear that $a \otimes b : M_{i''} \otimes _ R N_{j''} \to M_{i'} \otimes _ R N_{j'}$ serves the same purpose for the system $(M_ i \otimes _ R N_ j, f_{ii'} \otimes g_{jj'})$. Thus by the characterization Proposition 10.88.6 (3) we conclude that $M \otimes _ R N$ is Mittag-Leffler. $\square$

Proof. Since the condition is clear weaker than condition (1) of Proposition 10.88.6 we see that a Mittag-Leffler module satisfies the condition. Conversely, suppose that $M$ satisfies the condition and that $f : P \to M$ is an $R$-module map from a finitely presented $R$-module $P$ into $M$. Choose a surjection $F \to P$ where $F$ is a finite free $R$-module. By assumption we can find a map $F \to Q$ where $Q$ is a finitely presented $R$-module such that $F \to Q$ and $F \to M$ dominate each other. In particular, the kernel of $F \to Q$ contains the kernel of $F \to P$, hence we obtain an $R$-module map $g : P \to Q$ such that $F \to Q$ is equal to the composition $F \to P \to Q$. Let $N$ be any $R$-module and consider the commutative diagram

By assumption the kernels of $F \otimes _ R N \to Q \otimes _ R N$ and $F \otimes _ R N \to M \otimes _ R N$ are equal. Hence, as $F \otimes _ R N \to P \otimes _ R N$ is surjective, also the kernels of $P \otimes _ R N \to Q \otimes _ R N$ and $P \otimes _ R N \to M \otimes _ R N$ are equal. $\square$

Proof. Assume $M$ is a Mittag-Leffler module over $S$. Write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ as a directed colimit of finitely presented $S$-modules $M_ i$. As $M$ is Mittag-Leffler over $S$ there exists for each $i$ an index $j \geq i$ such that for all $k \geq j$ there is a factorization $f_{ij} = h \circ f_{ik}$ (where $h$ depends on $i$, the choice of $j$ and $k$). Note that by Lemma 10.36.23 the modules $M_ i$ are also finitely presented as $R$-modules. Moreover, all the maps $f_{ij}, f_{ik}, h$ are maps of $R$-modules. Thus we see that the system $(M_ i, f_{ij})$ satisfies the same condition when viewed as a system of $R$-modules. Thus $M$ is Mittag-Leffler as an $R$-module. $\square$

Proof. One implication follows from Lemma 10.88.11. To prove the other, assume $M$ is Mittag-Leffler as an $R$-module. Write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ as a directed colimit of finitely presented $S$-modules. As $I$ is finitely generated, the ring $S$ is finite and finitely presented as an $R$-algebra, hence the modules $M_ i$ are finitely presented as $R$-modules, see Lemma 10.36.23. Next, let $N$ be any $S$-module. Note that for each $i$ we have $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N) = \mathop{\mathrm{Hom}}\nolimits _ S(M_ i, N)$ as $R \to S$ is surjective. Hence the condition that the inverse system $(\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N))_ i$ satisfies Mittag-Leffler, implies that the system $(\mathop{\mathrm{Hom}}\nolimits _ S(M_ i, N))_ i$ satisfies Mittag-Leffler. Thus $M$ is Mittag-Leffler over $S$ by definition. $\square$