## Ring Homomorphisms (Pt. I)

**Point of Post: **In this post we discuss the notion of ring homomorphisms, kernels, images, etc.

** Motivation** As is standard in math, especially in algebra after defining a structure and its subobjects we define the morphisms between the two objects. In particular, we would now like to define the morphisms between two rings and . What do we want these morphisms to do? Well, we clearly want the morphisms to preserve the ring structure. In other words, we’d like it to be additive (be a group homomorphism for the group structure) and multiplicative (be a semigroup homomorphism for the semigroup structure). After this we define the obvious notions of kernel and image and show that they are, in fact, subrings ,etc.

*Definitions and Basics*

Let and be rings. A map is a *ring homomorphism *(alternatively *homomorphism *or even just *morphism *when the context is clear) if for all one has that and . We denote the set of all ring homomorphisms by . Some thing An injective homomorphism is called a *monomorphism *and a surjective homomorphism is called an *epimorphism *and is often denoted . If is both a bijective homomorphism we call it an *isomorphism*. If there exists an isomorphism we say that and ‘ are *isomorphic *and write . Some things are obvious:

**Theorem: ***Let be rings and , . Then*

**Proof: **The first three of these follow immediately since is a group homomorphism from to . The fourth is immediate by induction. To see the fifth claim we know from first principles that is a group homomorphism and it’s clear it’s a semigroup homomorphism for the multiplicative structure since for all . The conclusion follows.

**Corollary: ***Let and be monomorphisms, epimorphisms, or isomorphism then is a monomorphism, epimorphism, and isomorphism respectively.*

* *

Moreover, we have that

**Theorem: ***Let is an isomorphism then is an isomorphism.*

**Proof: **We know that is a bijection so it suffices to prove its a homomorphism. The fact that is additive follows from basic ring theory and for any we have that we may conclude the theorem.

Since for any ring we evidently have that is an isomorphism we get the following (really quite expected) theorem:

**Theorem: ***The relation “” is an equivalence relation.*

So, what about elements of ? If this were a group homomorphism we would say something like “oh, well but is the same true here? Indeed it is:

*Image and Kernel*

Similar to the case for groups the notions of image and kernel of a ring homomorphism are important in the overall theory of rings. So, for a homomorphism we call (where is the zero element of ) the *kernel *of and denote it . We note that the kernel of is really the same as the kernel of the group homomorphism . We denote as and call it the *image *of . We first note that both of these respect the structure in an obvious way, namely:

**Theorem: ***Let be a homomorphism, then and are subrings of and respectively.*

**Proof: **Let then and and so is a subring as desired. Similarly, let then and and so and so the kernel is a subring as desired.

The fact that is also a group homomorphism gives us for free facts such as

**Theorem: ***Let be a homomorphism, then is injective if and only if is trivial.*

They also preserve other structure such as commutativity:

**Theorem: ***Let be a homomorphism with commutative, then is commutative.*

**Proof: **Let be arbitrary. Then, , and since were arbitrary the conclusion follows.

**References:**

1. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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