Download Ring Homomorphisms and more Lecture notes Algebra in PDF only on Docsity!
Ring Homomorphisms
Definition. Let R and S be rings. A ring homomorphism (or a ring map for short) is a function f : R → S such that:
(a) For all x, y ∈ R, f (x + y) = f (x) + f (y).
(b) For all x, y ∈ R, f (xy) = f (x)f (y).
Usually, we require that if R and S are rings with 1, then
(c) f (1R) = 1S.
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 multipli- cation.
Example. (A ring map on the integers mod 2) Show that the following function f : Z 2 → Z 2 is a ring map: f (x) = x^2.
First, f (x + y) = (x + y)^2 = x^2 + 2xy + y^2 = x^2 + y^2 = f (x) + f (y).
2 xy = 0 because 2 times anything is 0 in Z 2. Next, f (xy) = (xy)^2 = x^2 y^2 = f (x)f (y).
The second equality follows from the fact that Z 2 is commutative. Note also that f (1) = 1^2 = 1. Thus, f is a ring homomorphism.
Example. (An additive function which is not a ring map) Show that the following function g : Z → Z is not a ring map: g(x) = 2x.
Note that g(x + y) = 2(x + y) = 2x + 2y = g(x) + g(y).
Therefore, g is additive — that is, g is a homomorphism of abelian groups. But g(1 · 3) = g(3) = 2 · 3 = 6, while g(1)g(3) = (2 · 1)(2 · 3) = 12.
Thus, g(1 · 3) 6 = g(1)g(3), so g is not a ring map.
Lemma. Let R and S be rings and let f : R → S be a ring map.
(a) f (0) = 0.
(b) f (−r) = −f (r) for all r ∈ R.
Proof. (a) f (0) = f (0 + 0) = f (0) + f (0), so f (0) = 0.
(b) By (a), 0 = f (0) = f (r + (−r)) = f (r) + f (−r).
But this says that f (−r) is the additive inverse of f (r), i.e. f (−r) = −f (r).
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 f (0) = 0 and f (−r) = −f (r), 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 f (0) = 0 and f (−r) = −f (r), then it isn’t a ring map.
Example. (Showing that a function is not a ring map) (a) Show that the following function f : Z → Z is not a ring map: f (x) = 2x + 5.
(b) Show that the following g : Z → Z is not a ring map:
g(x) = 3x.
(a) f (0) = 5 6 = 0.
(b) g(0) = 0 and g(−n) = −g(n) for all n ∈ Z. Nevertheless, g is not a ring map:
g(3 · 2) = g(6) = 3 · 6 = 18, but g(3) · g(2) = (3 · 3) · (3 · 2) = 54.
Thus, g(3 · 2) 6 = g(3) · g(2), so g does not preserve products.
Lemma. Let R, S, and T be rings, and let f : R → S and g : S → T be ring maps. Then the composite g · f : R → T is a ring map.
Proof. Let x, y ∈ R. Then
(g · f )(x + y) = g(f (x + y)) = g(f (x) + f (y)) = g(f (x)) + g(f (y)) = (g · f )(x) + (g · f )(y).
(g · f )(x · y) = g(f (x · y)) = g(f (x) · f (y)) = g(f (x)) · g(f (y)) = (g · f )(x) · (g · f )(y).
If, in addition, R, S, and T are rings with identity, then
(g · f )(1) = g(f (1)) = g(1) = 1.
Therefore, g · f 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 φ : R → S is
ker φ = {r ∈ R | φ(r) = 0}.
(a, x) + (b + c, y + z) = (a, x) + [(b, y) + (c, z)].
The third equality used associativity of addition in R and in S. The additive identity is (0, 0); the additive inverse −(r, s) of (r, s) is (−r, −s). 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 Z 6 ≈ Z 2 × Z 3.
Z 6 ≈ { 0 , 1 , 2 , 3 , 4 , 5 } with addition and multiplication mod 6. On the other hand,
Z 2 × Z 3 = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}.
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 Z 6 :
Here are the addition and multiplication tables for Z 2 × Z 3.
The two rings each have 6 elements, so it’s easy to define a bijection from one to the other — for example,
f (0) = (0, 0), f (1) = (0, 1), f (2) = (0, 2), f (3) = (1, 0), f (4) = (1, 1), f (5) = (1, 2).
However, this is not a ring isomorphism:
f (1 + 2) = f (3) = (1, 0), while f (1) + f (2) = (0, 1) + (0, 2) = (0, 0).
Thus, f (1 + 2) 6 = f (1) + f (2). It turns out, however, that the following map gives a ring isomorphism Z 6 → Z 2 × Z 3 :
f (0) = (0, 0), f (1) = (1, 1), f (2) = (0, 2), f (3) = (1, 0), f (4) = (0, 1), f (5) = (1, 2).
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 f (r + s) = f (r) + f (s) and another 36 cases for f (r · s) = f (r) · f (s).
Example. (Showing that a product of rings which is not isomorphic to another ring) Show that the rings Z 4 and Z 2 × Z 2 are not isomorphic.
Z 4 and Z 2 × Z 2 aren’t isomorphic as groups under addition. Since a ring isomorphism must give an isomorphism of the two rings considered as groups under addition, Z 4 and Z 2 × Z 2 can’t be isomorphic as rings. To see this directly, suppose f : Z 4 → Z 2 × Z 2 is an isomorphism. Then f (1) + f (1) = (0, 0), because everything in Z 2 × Z 2 gives 0 when added to itself. But since f is a ring map,
f (1) + f (1) = f (1 + 1) = f (2).
Therefore, f (2) = (0, 0). But I know that f (0) = (0, 0), because any ring map takes the additive identity to the additive identity. Now I have two elements 2 and 0 which both map to (0, 0), and this contradicts the fact that f is injective. Therefore, there is no such f , and the rings aren’t isomorphic.
