# Direct Limits of Rings

Let $((A_i), (\beta_{ij}))$ be a directed system of rings. Regard them as a directed system of abelian groups (i.e. ℤ-modules) and take the direct limit A.

Proposition 1.

The abelian group A has a natural structure of a commutative ring.

Note

General philosophy of the direct limit: “if something happens at index i and another thing happens at index j, then by picking k greater than i and j, both things may be assumed to happen at the same index”. The cautious reader is advised to fill in the gaps in the following proof.

Sketch of Proof

For each $i\in J$, let $\epsilon_i : A_i \to A$ be the canonical map.

Given $a,b \in A$, by proposition 2 here there exists $j\in J$ and $a', b' \in A_j$ such that $a = \epsilon_j(a')$, $b = \epsilon_j(b')$. Now define multiplication in A by $a\times b := \epsilon_j(a'b')$. This does not depend on our choice of j and $a',b' \in A_j$. Indeed, if we have another index $i\in J$ and $a_1, b_1 \in A_i$ such that $a = \epsilon_i(a_1)$, $b = \epsilon_i(b_1)$, by proposition 2 here again pick $k\in J$ greater than i and j such that $\beta_{ik}(a_1) = \beta_{jk}(a'), \ \beta_{ik}(b_1) = \beta_{jk}(b') \implies \beta_{ik}(a_1 b_1) = \beta_{jk}(a'b').$

This gives us the desired equality $\epsilon_i(a_1 b_1) = \epsilon_k(\beta_{ik}(a_1 b_1)) = \epsilon_k (\beta_{jk}(a'b')) = \epsilon_j(a'b').$

Clearly product in A is commutative. To show associativity, given $a,b,c\in A$, pick $j\in J$ and $a',b',c' \in A_j$ such that $a = \epsilon_j(a')$, $b = \epsilon_j(b')$ and $c = \epsilon_j(c')$. Then $a(bc) = \epsilon_j(a'(b'c')) = \epsilon_j((a'b')c') = (ab)c$.

To define $1\in A$, we pick any index $i\in J$ and set $1_A := \epsilon_i(1_{A_i})$. ♦

Exercise A

1. Prove that $1_A$ in the above proof is well-defined, and $1_A \times a = a$ for all $a\in A$.

2. Prove that the resulting ring A with the canonical $\epsilon_i : A_i \to A$ gives the direct limit of $A_i$ in the category of rings.

3. Prove that if the direct limit of rings $A_i$ is zero, then $A_i = 0$ for some i. [ Hint: a ring is zero if and only if 1 = 0. ]

4. Suppose $(B_i)_{i\in I}$ is an arbitrary collection of A-algebras. For each finite subset $L = \{i_1, \ldots, i_n\} \subseteq I$, define $B_L := B_{i_1} \otimes_A \ldots \otimes_A B_{i_n}$. Define a directed system of $B_L$ over the directed set of all finite subsets of I, ordered by inclusion $L\subseteq L'$. The tensor product of $B_i$ over A is defined to be the direct limit of this system. Prove that this gives the coproduct of $(B_i)_{i\in I}$ in the category A-algebras.

Note

Since direct limits are denoted by $\varinjlim$, we will write $\varprojlim$ for the earlier limits and call them inverse limits. For most cases of interest, inverse limits will be taken over J such that $J^{op}$ is directed. Even over directed sets, taking the inverse limit is not exact. A useful criterion for determining exactness is given by the Mittag-Lefler condition, which we will not cover (for now). # Taking Stock

We have seen many constructions which commute and some which do not. In the following examples, $(M_i)$ is an arbitrary collection of modules; M and N are modules, B is an A-algebra and $S\subseteq A$ is a multiplicative subset.

Case 1 (proposition 1 here): $S^{-1} (\oplus_i M_i) = \oplus_i S^{-1} M_i$ but $S^{-1}(\prod_i M_i) \ne \prod_i S^{-1}M_i$ in general.

Case 2 (exercise A here): $\mathfrak a (\oplus_i M_i) = \oplus_i \mathfrak a M_i$ but $\prod_i \mathfrak a M_i \ne \mathfrak a (\prod_i M_i)$.

Case 3 (corollary 1 here): a direct sum of projective modules is projective. A direct product of projective modules is not projective in general, but a counter-example is not too easy to construct.

Case 4 (exercise B here): for any collection $(N_i)$ of submodules of M, we have $S^{-1}(\sum_i N_i) = \sum_i S^{-1}N_i$ but $S^{-1}(\cap_i N_i) \ne \cap_i S^{-1} N_i$ in general.

Case 5 (exercise A here): if M is a flat A-module, then $M^B$ is a flat B-module.

Case 6 (proposition 1 here): we have $(\oplus_i M_i) \otimes_A N \cong \oplus_i (M_i \otimes_A N)$.

Case 7: more generally, we have $(\mathrm{colim}_{i\in J} M_i) \otimes_A N \cong \mathrm{colim}_{i\in J} (M_i \otimes_A N)$ if $(M_i)_{i\in J}$ is a diagram of A-modules of type J.

Case 8: hence we have an isomorphism $(\mathrm{colim}_{i\in J} M_i)^B \cong \mathrm{colim}_{i\in J}(M_i)^B$ of B-modules; in particular $S^{-1}(\mathrm{colim}_{i\in J} M_i) \cong \mathrm{colim}_{i\in J} (S^{-1} M_i)$.

Case 9 (proposition 3 here): if $f: M_1 \to M_2$ is surjective, then $f\otimes_A 1_N : M_1 \otimes_A N \to M_2 \otimes_A N$ is surjective; however, if f is injective $f\otimes_A 1_N$ is not injective in general.

Exercise B

Prove that there is always a canonical map between $(\prod_i M_i) \otimes_A N$ and $\prod_i (M_i \otimes_A N)$. Find an example where the map is not an isomorphism.

If M is a projective A-module, must $M^B = B\otimes_A M$ be a projective B-module? # Duality Principle

Remembering all the above relations may seem like a pain: in general if we have n constructions, we have about $O(n^2)$ relations to learn. It turns out most of these constructions can be classified as either “left-adjoint-like” or “right-adjoint-like”, which saves us a whole lot of effort in remembering them.

In the following table, constructions on the same side tend to commute or have consistent properties. Constructions on different sides may commute under specific additional conditions (e.g. finiteness, noetherianness).

 Left-adjoint-like Right-adjoint-like Sum of submodules Intersection of submodules Coproducts Products Right-exact functors Left-exact functors Pushouts Pullbacks, fibre products Direct sum of modules Direct product of modules Tensor products of modules Hom modules HomA(-, M) HomA(M, -) Colimits / Direct limits Limits / Inverse limits Injective maps Surjective maps Quotient modules Submodules Induced modules $M\mapsto M^B$. (Coinduced modules) Projective / free modules (Injective modules) Localization Multiplying ideal by module: $\mathfrak a M$ Do consider this table as a very rough guide. For example, if $0 \to N \to M \to P \to 0$ is a short exact sequence of A-modules, we do not get a right-exact sequence $\mathfrak a N \to \mathfrak a M \to \mathfrak a P \to 0$. [ Take $0\to 2\mathbb Z \to \mathbb Z \to \mathbb Z / 2\mathbb Z \to 0$ and $\mathfrak a = 2\mathbb Z$. ]

Also note that the terms in brackets have not been defined yet.

### Further Examples

1. We have $\mathrm{Hom}_A(\oplus_i N_i, M) \cong \prod_i \mathrm{Hom}_A(N_i, M)$.

2. The functor $\mathrm{Hom}_A(-, M)$ takes a right-exact sequence to a left-exact sequence.

3. Recall that the colimit of a diagram of A-modules was constructed by taking a quotient of the direct sum of these modules. Dually, its limit can be constructed by taking a submodule of the direct product.

4. The tensor product was constructed by taking a quotient of a free (hence projective) module.

Exercise C (Coinduced Modules)

1. Let B be an A-algebra. Prove that for an A-module M, $M_B := \mathrm{Hom}_A(B, M)$, the set of all A-linear maps $B\to M$, has a natural structure of a B-module.

2. Prove that we get a functor $F :A\text{-}\mathbf{Mod} \longrightarrow B\text{-}\mathbf{Mod}, \quad M \mapsto M_B$

such that there is a natural bijection $\mathrm{Hom}_A(N, M) \cong \mathrm{Hom}_B(N, M_B).$

for any B-module N. Thus $F = \mathrm{Hom}_A(B, -)$ is right-adjoint to the forgetful functor $B\text{-}\mathbf{Mod} \to A\text{-}\mathbf{Mod}$.

We call $M_B$ the coinduced B-module from M. This entry was posted in Advanced Algebra and tagged , , , , , , , . Bookmark the permalink.