* Definition.* Let R and S be rings. A * ring homomorphism* (or a * ring
map* for short) is a function such that:

(a) For all , .

(b) For all , .

Usually, we require that if R and S are rings with 1, then

(c) .

This is automatic in some cases; if there is any question, you should read carefully to find out what convention is being used.

The first two properties stipulate that f should "preserve" the ring structure --- addition and multiplication.

* Example.* (* A ring map on the
integers mod 2*) Show that the following function is a ring map:

First,

because 2 times anything is 0 in .

Next,

The second equality follows from the fact that is commutative.

Note also that .

Thus, f is a ring homomorphism.

* Example.* (* An additive
function which is not a ring map*) Show that the following
function is not a ring map:

Note that

Therefore, g is * additive* --- that is, g is a
homomorphism of abelian groups.

But

Thus, , so g is not a ring map.

* Lemma.* Let R and S be rings and let be a ring map.

(a) .

(b) for all .

* Proof.* (a)

(b) By (a),

But this says that is the additive inverse of , i.e. .

These properties are useful, and they also lend support to the idea that ring maps "preserve" the ring structure. Now I know that a ring map not only preserves addition and multiplication, but 0 and additive inverses as well.

*Warning!* A ring map f must satisfy and , but these are *not* part
of the *definition* of a ring map. To check that something is
a ring map, you check that it preserves sums and products.

On the other hand, if a function *does not* satisfy and , then it *isn't* a ring
map.

* Example.* (* Showing that a
function is not a ring map*) (a) Show that the following function
is not a ring map:

(b) Show that the following is not a ring map:

(a) .

(b) and for all . Nevertheless, g is not a ring map:

Thus, , so g does not preserve products.

* Lemma.* Let R, S, and T be rings, and let and be ring maps. Then the composite
is a ring map.

* Proof.* Let . Then

If, in addition, R, S, and T are rings with identity, then

Therefore, is a ring map.

There is an important relationship between ring maps and ideals. I'll consider half of the relationship now.

* Definition.* The * kernel*
of a ring map is

The * image* of a ring map is

The kernel of a ring map is like the null space of a linear transformation of vector spaces. The image of a ring map is like the column space of a linear transformation.

* Proposition.* The kernel of a ring map is a
two-sided ideal.

In fact, I'll show later that every two-sided ideal arises as the kernel of a ring map.

* Proof.* Let be a
ring map. Let , so and . Then

Hence, .

Since , .

Next, if , then . Hence, , so (why?), so .

Finally, let and let .

It follows that . Hence, is a two-sided ideal.

I'll omit the proof of the following result. Note that it says the
image of a ring map is a *subring*, not an *ideal*.

* Proposition.* Let be a
ring map. Then is a subring of S.

* Definition.* Let R and S be rings. A * ring isomorphism* from R to S is a bijective ring
homomorphism .

If there is a ring isomorphism , R and S are * isomorphic*. In this case, we write .

Heuristically, two rings are isomorphic if they are "the same" as rings.

An obvious example: If R is a ring, the identity map is an isomorphism of R with itself.

Since a ring isomorphism is a bijection, isomorphic rings must have the same cardinality. So, for example, , because the two rings have different numbers of elements.

However, and have the
"same number" of elements --- the same *
cardinality* --- but they are not isomorphic as rings. (Quick
reason: is a field, while is only an integral domain.)

I've been using this construction informally in some examples. Here's the precise definition.

* Definition.* Let R and S be rings. The * product ring* of R and S is the set
consisting of all ordered pairs , where and . Addition and multiplication are
defined component-wise: For and ,

I won't go through the verification of all the axioms; basically, everything works because everything works in each component separately. For example, here's the verification of the associative law for addition. Let , . Then

The third equality used associativity of addition in R and in S.

The additive identity is ; the additive inverse of is . And so on. Try out one or two of the other axioms for yourself just to get a feel for how things work.

* Example.* (* A ring isomorphic
to a product of rings*) Show that .

with addition and multiplication mod 6. On the other hand,

One ring consists of single elements, while the other consists of
pairs. Nevertheless, these rings are isomorphic --- *they are the
same as rings*.

Here are the addition and multiplication tables for :

Here are the addition and multiplication tables for .

The two rings each have 6 elements, so it's easy to define a
*bijection* from one to the other --- for example,

However, this is not a ring isomorphism:

Thus, .

It turns out, however, that the following map gives a ring isomorphism :

It's obvious that the map is a bijection. To *prove* that this
is a ring isomorphism, you'd have to check 36 cases for and another 36 cases for .

* Example.* (* Showing that a
product of rings which is not isomorphic to another ring*) Show
that the rings and are not isomorphic.

and aren't isomorphic as groups under addition. Since a ring isomorphism must give an isomorphism of the two rings considered as groups under addition, and can't be isomorphic as rings.

To see this directly, suppose is an isomorphism. Then , because everything in gives 0 when added to itself. But since f is a ring map,

Therefore, .

But I know that , because any ring map takes the additive identity to the additive identity. Now I have two elements 2 and 0 which both map to , and this contradicts the fact that f is injective.

Therefore, there is no such f, and the rings aren't isomorphic.

Copyright 2018 by Bruce Ikenaga