# Direct Sums and Direct Products

Recall that for a ring *A*, a sequence of *A*-modules gives the *A*-module 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 denote a collection of *A*-modules, indexed by .

Definition.The

direct productis the set-theoretic product of the , with the structure of an A-module given by:.

The

direct sumis the submodule of comprising of all such that only for a finite number of .

Since there are only finitely many non-zero terms for an element , one often writes the element additively, i.e. .

For a simple example, note that if for a fixed module *M*, then is the set of all functions . On the other hand, is the set of all *f* such that for only finitely many . [We say *f* has **finite support**.]

**Exercise A**

For an infinite collection of *A*-modules and an ideal , which of these claims is true?

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

# 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, g*enerally 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 . For each , take the

projectionmapNow the collection of data satisfies the following.

- For any A-module N and collection of data where each is A-linear, there is a unique A-linear such that for each .

**Note**

The idea is that homomorphisms “classify” the collection of all *I*-indexed tuples .

**Proof**.

For existence, define to be . Now for any we have

.

For uniqueness, if , the *j*-th component of is so the tuple must be . ♦

The following result shows why universal properties are important.

Proposition 1.Let M’ be an A-module and be a collection of data satisfying the above universal property. Then there is a unique isomorphism such that for all .

**Proof**

**1**. We apply the universal property to and for each *i*. There exists a unique such that . We know one such *f*, namely . Hence this is the only possibility.

**2**. Next apply the universal property to *N* = *M’* and . There exists a unique such that for each *i*.

**3**. But we know also satisfies this same universal property. Swapping *M* and *M’* there exists a unique such that for each *i*.

**4**. Now for each *i*. By step 1, we have . By symmetry we get too. ♦

**Note**

In fact, we have proven something stronger: that there is a unique *homomorphism* such that for all *i* and this *f* must be an isomorphism.

# Direct Sum

Similarly, the direct sum satisfies the following.

Theorem (Universal Property of Direct Sum).Let . For each , take the

embeddingThus has component at and 0 if . The collection of data satisfies the following.

- For any A-module N and collection of data where each is A-linear, there is a unique A-linear such that for each .

**Proof**.

Left as an exercise. ♦

We also have uniqueness for the above universal property.

Proposition 2.Let M’ be an A-module and be a collection of data satisfying the above universal property. Then there is a unique isomorphism such that for each .

**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 which take to ;
- projections which take to .

Explain why fails the universal property of direct sum, and fails the universal property of direct product.

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

# Optional Example

Let *I* be the set of all primes. Consider the direct sum of -modules . Consider the -linear homomorphism (i.e. additive homomorphism)

Note that addition is well-defined because only finitely many are non-zero. Also, is independent of our choice of integer .

The map is injective: suppose where are distinct primes and each . Multiplying throughout by gives

where each term in brackets is divisible by . Since is coprime to we have which is a contradiction.

Claim: the image of the map is the set *G* of all rational where *n* is square free. Indeed, suppose *n* in is square free; write as a product of distinct primes.

- Now the subgroup of
*G*has order*n*. - On the other hand, the submodule maps into
*G*injectively with image with . - Since , we see that lies in the image.

Thus .

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

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.

There is. map missing in in the important exercise, though it is obvious what you mean.

Also did you mean in “Since is coprime to we have which is a contradiction.

Thank you for all the bug fixes. 😀

I enjoy your exposition very much. Hope you don’t mind my comments.