Commutative Algebra 9

Direct Sums and Direct Products

Recall that for a ring A, a sequence of A-modules M_1, \ldots, M_n gives the A-module M = M_1 \times \ldots \times M_n where the operations are defined component-wise. In this article, we will generalize the construction to an infinite collection of modules. Throughout this article, let (M_i) denote a collection of A-modules, indexed by i\in I.

Definition.

The direct product \prod_{i\in I} M_i is the set-theoretic product of the (M_i)_{i\in I}, with the structure of an A-module given by:

(m_i) + (m_i') := (m_i + m_i'), \quad a\cdot (m_i) := (am_i).

The direct sum \oplus_{i\in I} M_i is the submodule of \prod_{i\in I} M_i comprising of all (m_i)_{i\in I} such that m_i \ne 0 only for a finite number of i\in I.

Since there are only finitely many non-zero terms for an element (m_i) \in \oplus_i M_i, one often writes the element additively, i.e. \sum_i m_i.

For a simple example, note that if M_i = M for a fixed module M, then \prod_{i\in I} M is the set of all functions f : I\to M. On the other hand, \oplus_{i\in I} M is the set of all f such that f(i) \ne 0 for only finitely many i\in I. [We say f has finite support.]

Exercise

For an infinite collection (M_i) of A-modules and an ideal \mathfrak a\subseteq A, which of these claims is true?

\mathfrak a \left( \bigoplus_i M_i\right) = \left( \bigoplus_i \mathfrak a M_i\right), \quad \mathfrak a \left( \prod_i M_i\right) = \left( \prod_i \mathfrak a M_i\right).

[Note: if it seems too hard to find a counter-example, argue qualitatively why equality does not hold.]

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Universal Property

At first glance, it seems like the direct product is the right definition since it generalizes directly from direct product of groups, rings, topological spaces etc. However, generally the direct sum is better behaved for our needs. The above exercise should provide some evidence of that.

For now, we will show that the direct sum and direct product are actually dual of each other, by showing the corresponding universal properties.

Theorem (Universal Property of Direct Product).

Let M = \prod_{i\in I} M_i. For each j\in I, take the projection map

\pi_j : \prod_{i\in I} M_i \to M_j, \quad (m_i)_{i\in I} \mapsto m_j.

Now the collection of data (M, (\pi_i : M \to M_i)_{i\in I}) satisfies the following.

  • For any A-module N and collection of data (N, (\phi_i : N \to M_i)_{i\in I}) where each \phi_i is A-linear, there is a unique A-linear f : N\to M such that \pi_i\circ f = \phi_i for each i \in I.

Note

The idea is that homomorphisms N \to \prod_i M_i “classify” the collection of all I-indexed tuples (\phi_i : N\to M_i)_{i\in I}.

universal_property_direct_product

Proof.

For existence, define f : N\to \prod_i M_i to be n \mapsto (\phi_i(n))_{i\in I} \in \prod_i M_i. Now for any j\in I we have

\pi_j (f(n)) = \pi_j ((\phi_i(n)_i) = \phi_j(n).

For uniqueness, if j\in I, the j-th component of f(n) is \pi_j(f(n)) = \phi_j(n) so the tuple f(n) must be (\phi_i(n))_{i \in I}. ♦

The following result shows why universal properties are important.

Proposition.

Let M’ be an A-module and (M', (\pi'_i : M' \to M_i)) be a collection of data satisfying the above universal property. Then there is a unique isomorphism f : M\to M' such that \pi_i' \circ f = \pi_i for all i\in I.

Proof

1. We apply the universal property to N = \prod M_i = M and \phi_i = \pi_i for each i. There exists a unique f : M \to M such that \pi_i \circ f = \pi_i. We know one such f, namely 1_M. Hence this is the only possibility.

2. Next apply the universal property to NM’ and \phi_i = \pi_i'. There exists a unique g : M'\to M such that \pi_i\circ g = \pi_i' for each i.

3. But we know (M', (\pi_i')_i) also satisfies this same universal property. Swapping M and M’ there exists a unique f : M\to M' such that \pi_i' \circ f = \pi_i for each i.

4. Now \pi_i\circ (g\circ f) = \pi_i'\circ f = \pi_i for each i. By step 1, we have g\circ f = 1_M. By symmetry we get f\circ g = 1_{M'} too. ♦

Note

In fact, we have proven something stronger: that there is a unique homomorphism f : M\to M' such that \pi_i' \circ f = \pi_i for all i and this f must be an isomorphism.

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Direct Sum

Similarly, the direct sum satisfies the following.

Theorem (Universal Property of Direct Sum).

Let M = \oplus_{i\in I} M_i. For each j\in I, take the embedding

\epsilon_j : M_j \longrightarrow M, \quad m_j \mapsto (\ldots, 0, 0, m_j, 0, 0, \ldots).

Thus \epsilon_j(m_j)_i has component m_j at i = j and 0 if i\ne j. The collection of data (M, (\epsilon_i : M_i \to M)_{i\in I}) satisfies the following.

  • For any A-module N and collection of data (N, (\alpha_i : M_i \to N)_{i\in I}) where each \alpha_i is A-linear, there is a unique A-linear f : M\to N such that f\circ \epsilon_i = \alpha_i for each i\in I.

Proof.

Left as an exercise. ♦

We also have uniqueness for the above universal property.

Proposition.

Let M’ be an A-module and (M', (\epsilon'_i : M_i \to M)) be a collection of data satisfying the above universal property. Then there is a unique isomorphism f: M\to M' such that f\circ \epsilon_i = \epsilon'_i for each i\in I.

Proof.

Left as an exercise. ♦

The following exercise tests your conceptual understanding of universal properties. 

Important Exercise

Suppose we messed up and take the following maps instead:

  • embeddings \epsilon^0_j : M_j \to \prod_{i \in I} M_i which take m_j to (\ldots, 0, 0, m_j, 0, 0, \ldots);
  • projections \pi^0_j : \oplus_{i\in I} M_i which take (m_i)_{i\in I} to m_j.

Explain why (\prod_i M_i, (\epsilon^0_i)_{i\in I}) fails the universal property of direct sum, and (\oplus_i M_i, (\pi^0_i)_{i\in I}) fails the universal property of direct product.

[Hint (highlight to read): one of them fails the uniqueness property, the other fails the existence property.]

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Optional Example

Let I be the set of all primes. Consider the direct sum of \mathbb Z-modules A := \bigoplus_{p\in I} \mathbb Z / p\mathbb Z. Consider the \mathbb Z-linear homomorphism (i.e. additive homomorphism)

\bigoplus_{p\in I} (\mathbb Z / p\mathbb Z) \longrightarrow \mathbb Q/\mathbb Z, \quad (m_p)_{p\in I} \mapsto \sum_{p\in I} \frac{m_p}{p}.

Note that addition is well-defined because only finitely many m_p are non-zero. Also, \frac{m_p}p \in \mathbb Q/\mathbb Z is independent of our choice of integer \equiv m_p \pmod p.

The map is injective: suppose \frac{m_1}{p_1} + \ldots + \frac{m_k}{p_k} = 0 where p_i are distinct primes and each m_i \not\equiv 0 \pmod p_i.  Multiplying throughout by p_1 p_2 \ldots p_k gives

(p_1\ldots p_{k-1})m_k + (\text{some terms}) = 0

where each term in brackets is divisible by p_k. Since p_1\ldots p_{k-1} is coprime to q_k we have m_k \equiv 0 \pmod {p_k} which is a contradiction.

Claim: the image of the map is the set G of all rational \frac m n where n is square free. Indeed, suppose n in \frac m n is square free; write n = p_1 \ldots p_k as a product of distinct primes.

  • Now the subgroup H_n := \{\frac m n : 0 \le m < n\} of G has order n.
  • On the other hand, the submodule B_n := (\mathbb Z/p_1 \mathbb Z) \times \ldots \times (\mathbb Z/p_k \mathbb Z) \subset A maps into G injectively with image with H_n.
  • Since |B_n| = |H_n|, we see that H_n lies in the image.

Thus \bigoplus_p (\mathbb Z / p\mathbb Z) \cong G.

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2 Responses to Commutative Algebra 9

  1. Manas Jana says:

    Thank you for your beautiful notes. Can you give some references of your notes?

  2. limsup says:

    Hi. Some common references for Commutative Algebra include:

    – R. Y. Sharp, “Steps in Commutative Algebra” (possibly the easiest book to read among this list).
    – Atiyan & Macdonald, “Introduction to Commutative Algebra” (well-written but extremely terse).
    – David Eisenbud, “Commutative Algebra: With a View Toward Algebraic Geometry” (more advanced than the previous two books but has loads of nice examples and exercises, a bit verbose though).
    – Hideyuki Matsumura, “Commutative Ring Theory” (possibly the most advanced in this list, a good follow-up to A & M).

    But it’s best to supplement your reading of commutative algebra with some algebraic geometry. “Algebraic Curves” by Fulton is a nice introduction to the topic.

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