# Commutative Algebra/Kernels, cokernels, products, coproducts

## Kernels[edit]

**Definition 3.1**:

Let be a category with zero objects, and let be a morphism between two objects of . A **kernel** of is an arrow , where is what we shall call the object associated to the kernel , such that

- , and
- for each object of and each morphism such that , there exists a unique such that .

The second property is depicted in the following commutative diagram:

Note that here, we don't see kernels only as subsets, but rather as an object together with a morphism. This is because in the category of groups, for example, we can take the morphism just by inclusion. Let me explain.

**Example 3.2**:

In the category of groups, every morphism has a kernel.

**Proof**:

Let be groups and a morphism (that is, a group homomorphism). We set

and

- ,

the inclusion. This is indeed a kernel in the category of groups. For, if is a group homomorphism such that , then maps wholly to , and we may simply write . This is also clearly a unique factorisation.

For kernels the following theorem holds:

**Theorem 3.3**:

Let be a category with zero objects, let be a morphism and let be a kernel of . Then is a monic (that is, a monomorphism).

**Proof**:

Let . The situation is depicted in the following picture:

Here, the three lower arrows depict the general property of the kernel. Now the morphisms and are both factorisations of the morphism over . By uniqueness in factorisations, .

Kernels are essentially unique:

**Theorem 3.4**:

Let be a category with zero objects, let be a morphism and let , be two kernels of . Then

- ;

that is to say, and are isomorphic.

**Proof**:

From the first property of kernels, we obtain and . Hence, the second property of kernels imply the commutative diagrams

We claim that and are inverse to each other.

- and .

Since both and are monic by theorem 3.3, we may cancel them to obtain

- and ,

that is, we have inverse arrows and thus, by definition, isomorphisms.

## Cokernels[edit]

An analogous notion is that of a cokernel. This notion is actually common in mathematics, but not so much at the undergraduate level.

**Definition 3.5**:

Let be a category with zero objects, and let be a morphism between two objects of . A **cokernel** of is an arrow , where is an object of which we may call the object associated to the cokernel , such that

- , and
- for each object of and each morphism such that , there exists a unique factorisation for a suitable morphism .

The second property is depicted in the following picture:

Again, this notion is just a generalisation of facts observed in "everyday" categories. Our first example of cokernels shall be the existence of cokernels in Abelian groups. Now actually, cokernels exist even in the category of groups, but the construction is a bit tricky since in general, the image need not be a normal subgroup, which is why we may not be able to form the factor group by the image. In Abelian groups though, all subgroups are normal, and hence this is possible.

**Example 3.6**:

In the category of Abelian groups, every morphism has a cokernel.

**Proof**:

Let be any two Abelian groups, and let be a group homomorphism. We set

- ;

we may form this quotient group because within an Abelian group, all subgroups are normal. Further, we set

- ,

the projection (we adhere to the custom of writing Abelian groups in an additive fashion). Let now be a group homomorphism such that , where is another Abelian group. Then the function

is well-defined (because of the rules for group morphisms) and the desired unique factorisation of is given by .

**Theorem 3.7**:

Every cokernel is an epi.

**Proof**:

Let be a morphism and a corresponding cokernel. Assume that . The situation is depicted in the following picture:

Now again, , and and are by their equality both factorisations of . Hence, by the uniqueness of such factorisations required in the definition of cokernels, .

**Theorem 3.8**:

If a morphism has two cokernels and (let's call the associated objects and ), then ; that is, and are isomorphic.

**Proof**:

Once again, we have and , and hence we obtain commutative diagrams

We once again claim that and are inverse to each other. Indeed, we obtain the equations

- and

and by cancellation (both and are epis due to theorem 8.7) we obtain

- and

and hence the theorem.

## Interplay between kernels and cokernels[edit]

**Theorem 3.9**:

Let be a category with zero objects, and let be a morphism of such that is the kernel of some arbitrary morphism of . Then is also the kernel of any cokernel of itself.

**Proof**:

means

We set , that is,

In particular, since , there exists a unique such that . We now want that is a kernel of , that is,

Hence assume . Then . Hence, by the topmost diagram (in this proof), for a unique , which is exactly what we want. Further, follows from the second diagram of this proof.

**Theorem 3.10**:

Let be a category with zero objects, and let be a morphism of such that is the kernel of some arbitrary morphism of . Then is also the cokernel of any kernel of itself.

**Proof**:

The statement that is the cokernel of reads

We set , that is

In particular, since , for a suitable unique morphism . We now want to be a cokernel of , that is,

Let thus . Then also and hence has a unique factorisation by the topmost diagram.

**Corollary 3.11**:

Let be a category that has a zero object and where all morphisms have kernels and cokernels, and let be an arbitrary morphism of . Then

and

- .

The equation

is to be read "the kernel of is a kernel of any cokernel of itself", and the same for the other equation with kernels replaced by cokernels and vice versa.

**Proof**:

is a morphism which is some kernel. Hence, by theorem 3.9

(where the equation is to be read " is a kernel of any cokernel of "). Similarly, from theorem 3.10

- ,

where .

## Products[edit]

**Definition 3.12**:

Let be a category, and let be two objects of . A **product** of and , denoted , is an object of together with two morphisms

- and ,

called the **projections** of , such that for any morphisms and there exists a unique morphism such that the following diagram commutes:

- [[]]

**Example 3.13**:

**Theorem 3.14**:

If is a category, are objects of and are products of and , then

- ,

that is, and are isomorphic.

**Theorem 3.15**:

Let be a category, objects of and a product of and . Then the projection morphisms and are monics.

## Coproducts[edit]

**Definition 3.16**:

Let be a category, and let and be objects of . Then a **coproduct** of and is another object of , denoted , together with two morphisms and such that for any morphisms and , there exist morphisms such that and .

**Example 3.17**:

**Theorem 3.18**:

**Theorem 3.19**:

## Biproducts[edit]

**Definition 3.20**:

Let be a category that contains two objects and . Assume we are given an object of together with four morphisms that make it into a product, and simultaneously into a coproduct. Then we call a **biproduct** of the two objects and and denote it by

- .

**Example 3.21**:

Within the category of Abelian groups, a biproduct is given by the product group; if are Abelian groups, set the product group of and to be

- ,

the cartesian product, with component-wise group operation.

**Proof**: