(c)Karen E. Smith 2017 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 412. Ring Homomorphisms Professor Karen E. Smith

φ DEFINITION:A ring homomorphism is a mapping R −→ S between two rings (with identity) which satisfies: (1) φ(x + y) = φ(x) + φ(y) for all x, y ∈ R. (2) φ(x · y) = φ(x) · φ(y) for all x, y ∈ R. (3) φ(1R) = 1S.

DEFINITION:A ring isomorphism is a bijective ring homomorphism. We say that two rings R and S are isomorphic if there is an isomorphism R → S between them. You should think of an isomorphism as a renaming: isomorphic rings are “the same ring” with the elements named differently.

φ DEFINITION: The kernel of a ring homomorphism R −→ S is the set of elements in the source that map to the ZERO of the target; that is,

ker φ = {r ∈ R | φ(r) = 0S}.

THEOREM: A ring homomorphism R → S is injective if and only if its kernel is zero.

A.EXAMPLESOF HOMOMORPHISMS: Which of the following mappings between rings is a homo- morphism? Which are isomorphisms? n (1) The inclusion mapping Z ,→ Q sending each integer n to the rational number 1 . (2) The doubling map Z → Z sending n 7→ 2n. (3) The residue map Z → Zn sending each integer a to its congruence class [a]n. (4) The “evaluation at 0” map R[x] → R sending f(x) 7→ f(0). df (5) The differentiation map R[x] → R[x] sending f 7→ dx . λ 0 (6) The map → M ( ) sending λ 7→ . R 2 R 0 λ 1 λ (7) The map → M ( ) sending λ 7→ . R 2 R 0 1 (8) The map M2(Z) → R sending each 2 × 2 matrix to its determinant.

For homomorphism: Yes, No (1 not preserved), yes, yes, no (does not preserve ×), yes, no (does not preserve 1), no (does not preserve +) NONE of these are isomorphisms.

B.BASIC PROOFS: Let φ : S → T be a homomorphism of rings.

(1) Show that φ(0S) = 0T . (2) Show that for all x ∈ S, −φ(x) = φ(−x). That is: a ring homomorphism “respects additive inverses”. (3) Show that if u ∈ S is a unit, then also φ(u) ∈ T is a unit.

(1) φ(0S) = φ(0S + 0S) = φ(0S) + φ(0S). Now, add −φ(0S) to both sides: get 0T = φ(0S). (2) 0T = φ(0S) = φ(x + −x) = φ(x) + φ(−x). So the additive inverse of φ(x) is φ(−x). That is, −φ(x) is φ(−x). 2

(3) We know uv = vu = 1S. Apply φ to get φ(uv) = φ(vu) = φ(1S) = 1T . So φ(u)φ(v) = φ(v)φ(u) = 1T and φ(u) is a unit.

C.KERNELOF RING HOMOMORPHISMS: Let φ : R → S be a ring homomorphism (1) Prove that ker φ is non-empty. (2) Compute the kernel of the canonical homomorphism: Z → Zn sending a 7→ [a]n. (3) Compute the kernel of the homomorphism Z → Z7 × Z11 sending n 7→ ([n]7, [n]11). (4) For arbitrary rings R,S, compute the kernel of the projection homomorphism R × S → R sending (r, s) 7→ r. Write your answer in set-builder notation.

(1) 0S ∈ ker φ. (2) Kernel is nZ, the set of multiples of n. (3) Kernel is multiples of 77, or 77Z. (4) Kernel is {(0, s) | s ∈ S}.

D. Prove the THEOREM: A ring homomorphism φ : R → S is injective if and only if its kernel is ZERO.

Assume φ is injective. Take arbitrary x ∈ ker φ. Then φ(x) = 0. But also φ(0) = 0. By injectivity, x = 0. So ker φ ⊂ {0}. Conversely, assume ker φ = 0. Take arbitrary x, y ∈ R such that φ(x) = φ(y). Then φ(x) − φ(y) = φ(x) + −φ(y) = φ(x) + φ(−y) = φ(x − y) = 0. So x − y ∈ ker φ, which means x − y = 0. So x = y and φ is injective.

φ E.ISOMORPHISM. Suppose that R −→ S is a ring isomorphism. True or False. (1) φ induces a bijection between units. (2) φ induces a bijection between zero-divisors. (3) R is a field if and only if S is a field. (4) R is an (integral) domain if and if S is an (integral) domain.

All true. Isomorphism just changes the name, but since it preserves 0 and 1, it preserves zero divisors and units. The last two statement then follow directly from the definitions (of domain, field, respectively) and (1) and (2) (respectively).

F. ISOMORPHISM. Consider the set S = {a, b, c, d} and the associative binary operations ♥ and ♠ listed below. You showed last time that (S, ♠, ♥) is a ring. ♥ a b c d ♠ a b c d ⊕ a b c d ⊗ a b c d a a a a a a a b c d a a b c d a a a a a b a b c d b b c d a b b a d c b a b c d c a c a c c c d a b c c d a b c a c c a d a d c b d d a b c d d c b a d a d a d (1) Discuss and recall some of the features of the ring (S, ♠, ♥). To what more familiar ring is it isomorphic? (2) The operations ⊕ and ⊗ (whose tables are listed below) define a different ring structure on S. What are the zero and one? Is the ring (S, ⊕, ⊗) isomorphic to (S, ♠, ♥)? Explain (3) Find an explicit isomorphism Z2 × Z2 → (S, ⊕, ⊗). (4) Find a different isomorphism Z2 × Z2 → (S, ⊕, ⊗)? (5) Can there be more than one isomorphism Z4 → (S, ♠, ♥)?

(1) Z4 (2) zero is a and one is b. (3) (0, 0) 7→ a, (0, 1) 7→ d, (1, 0) 7→ c, (1, 1) 7→ b, (4) (0, 0) 7→ a, (1, 0) 7→ d, (0, 1) 7→ c, (1, 1) 7→ b, 3

(5) No. Since [2] = [1] + [1], we know [2] must map to φ([1]) + φ([1]) = b + b = c and similarly φ([3]) = φ([1] + [1] + [1]) = b + b + b = b + c = d.

G.PRACTICEWITHWELL-DEFINEDNESS: (1) Show that Z4 → Z2 defined by [a]4 7→ [a]2 is a well-defined ring homomorphism. What does well-defined mean here? Remember the element of Z4 are sets. (2) Explain why the map Z5 → Z2 defined by [a]5 7→ [a]2 is not well defined. (3) Show that for any m, n ∈ Z, the map Zn → Zm is well-defined if and only if m|n. (4) We can define φ : Z5 → Z2 by [r]5 7→ [r]2 for 0 ≤ r < 5. Explain why this is well-defined. Explain why it is not a ring homomorphism, nor very ”natural”.

(1) The congruence class [a]4 is a set, and the notation in which we use “a” to stand for the whole set is biased. We could just as well use a + 4 or any number of the form a + 4k. So we need to check that the way we define the map does not depend on the choice of the representative a for the class [a]4. That is, we need to check that if b ∈ [a]4 (which is the same as saying [b]4 = [a]4), then also [b]2 = [a]2. To see this, note that [b]4 = [a]4 means that b = a + 4k for some k ∈ Z. But then also b = a + 2(2k), so b ∈ [a]2 as well. So if [b]4 = [a]4, then also [b]2 = [a]2. This tells us the map is well-defined: we get the same mapping no matter what representative of the set we take. (2) The “map” Z5 → Z2 defined by [a]5 7→ [a]2 is NOT well-defined. For example, [1]5 = [6]5 but [1]2 6= [6]2. (3) Suppose n = md for some d ∈ Z. We need to show that if [a]n = [b]n, then also [a]m = [b]m. Assume [a]n = [b]n. We know there is a k so that b = a + kn, but then also b = a + (kd)m, so [a]m = [b]m. Conversely, assume the map is well-defined. To show that m|n, we need to show that n = dm for some d. This is the same as showing that n ∈ [0]m. Since the map is well defined, we know the class [n]n = [0]n must go to the class [n]m = [0]m, so n ∈ [0]m, and m|n. (4) With this mapping, we are not making any choices in the notation: we list out each of the five elements of Z5 exactly once and say, unambiguously, where it goes under φ. However, note that φ([3]5 + [2]5) = φ([0]5) = [0]2, but φ([3]5) + φ([2]5) = [3]2 + [2]2 = [1]2 6= [0]2. So the map does not respect addition.

H.CAUTIONARY EXAMPLES: Let R and S be arbitrary rings. (1) Is the map R → R × S sending a 7→ (a, 0) a ring homomorphism? Explain. (2) Is the map R → R × S sending a 7→ (a, 1) a ring homomorphism? Explain.

Neither is. In the first, 1R 7→ (1, 0) which is not the multiplicative identity of R × S. In the second, 0 goes to (0, 1) which is not the additive identity.

I.CANONICAL RING HOMOMORPHISMS: Let R be any ring1. Prove that there exists a unique ring homomorphism Z → R. ∼ BONUS: Find a necessary and sufficient condition on integers n, m so that Zmn = Zn × Zm:

1with identity of course