# Ring Homomorphisms

• If R and S are ring, a function is a ring homomorphism (or a ring map) if and for all . If R and S are rings with identity, it's customary to also require that .
• If is a ring map, then and .
• If is a ring map, then is an ideal in R and is a subring of S.
• If is a ring map and f is bijective, then f is a ring isomorphism. Rings R and S are isomorphic if there is a ring isomorphism .

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 , .

Most people also 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) Consider the function given by

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) The function given by

is not a ring homomorphism.

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) The function defined by

is not a ring map, since .

On the other hand, consider the function given by

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.

Lemma. 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.

Lemma. 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.

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

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

Example. ( Rings that aren't isomorphic) , because the two rings have different numbers of elements.

Note that it doesn't work the other way: If two rings have the same number of elements, they may or may not be isomorphic.

For example, 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 a 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 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) .

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 {\it 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) 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.

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Copyright 2008 by Bruce Ikenaga