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