Math 371 Lecture #22 §6.2: Quotients and Homomorphisms, Part II

We are going to answer the question today of what is the relationship between rings R and S if there exists a surjective homomorphism f : R → S. If f is an isomorphism then R and S are the “same” rings in disguise. If f is not an isomorphism, then some “information” is lost by f in going from R to S. How much “information” is lost depends on a certain ideal of R. Definition. Let f : R → S be a homomorphism of rings. The kernel of f is the set

K = {r ∈ R : f(r) = 0S}.

Example. What is the kernel of the surjective homomorphism f : Z → Z15 defined by f(a) = [a]? Well, we have

K = {r ∈ Z : f(r) = [0]} = {r ∈ Z :[r] = [0]} = {r ∈ Z : r ≡ 0 (mod 15)} = {r ∈ Z : 15 | r} = {15k : k ∈ Z} = (15).

Is the kernel of a homomorphism f of rings always an ideal of the domain of f? Theorem 6.10. If f : R → S is a homomorphism of rings, then the kernel of f is ideal in R. You proved this in HW problem §6.1 #25 Ed. 2 (#27 Ed. 3). The largest the kernel of a homomorphism f : R → S of rings can be is all of R which means that f is the zero function f(r) = 0S for all r ∈ R; all of the “information” is lost by f.

The smallest the kernel of f can be is ideal (0R) in R. What does this say about f? Theorem 6.11. Let f : R → S be a homomorphism of rings and K the kernel of f. Then K = (0R) if and only if f is injective.

Proof. Suppose that K = (0R). To show that f is injective, suppose that f(a) = f(b) for some a, b ∈ R. Because f is a homomorphism we have

f(a − b) = f(a) − f(b) = 0S − 0S = 0S.

Hence a − b ∈ K, and with K = (0R) = {0R}, we obtain a − b = 0, or a = b. Conversely, suppose that f is injective.

For c ∈ K we have f(c) = 0S.

We know that 0S is also the image of 0R by f, i.e., f(0R) = 0S.

Since f is injective, we conclude that c = 0R, and so K = (0R).  Example. What is the kernel of the homomorphism f : Z15 → Z5 defined by f([a]15) = [a]5?

With K = {[a]15 ∈ Z15 : f([a]15) = [0]5}, we solve [a]5 = f([a]15) = [0]5 for [a]15 ∈ Z15.

The solutions are [0]15, [5]15, and [10]15, so K = {[0]15, [5]15, [10]15}. We know that every kernel of a homomorphism of rings is an ideal, but is every ideal the kernel of some homomorphism? Theorem 6.12. Let I be an ideal in a ring R. The map π : R → R/I given by π(r) = r + I is a surjective homomorphism with kernel K = I. Proof. The map π is surjective because any coset r + I in R/I is the image of r by π, i.e., π(r) = r + I. For r, s ∈ R we have π(r + s) = (r + s) + I = (r + I) + (s + I) = π(r) + π(s) and π(rs) = rs + I = (r + I)(s + I) = π(r)π(s). Thus π is a homomorphsm.

The kernel of homomorphism π of rings is the ideal K = {r ∈ R : π(r) = 0R + I} in R.

So r ∈ K if and only if r + I = 0R + I, which happens if and only if r ≡ 0 (mod I).

Therefore, r ∈ K if and only if r = r − 0R ∈ I.  The surjective homomorphism π : R → R/I is called the natural homomorphism from R to R/I. ∼ Example. An ideal of Z15 is K = {[0]15, [5]15, [10]15} = Z3 (where the isomorphism is given by [0]15 → [0]3, [10]15 → [1]3 and [5]15 → [2]3).

The natural surjective homomorphism π : Z15 → Z15/K is

π([a]15) = [a]15 + K, where the image of Z15 by π is

Z15/K = {[0]15 + K, [1]15 + K, [2]15 + K, [3]15 + K, [4]15 + K}.

Is Z15/K isomorphic to Z5? We will answer this shortly. This natural surjective homomorphism π : R → R/I of rings is a special case of a more general situation. Definition. Let R and S be rings. We say that S is a homomorphic image of R if there exists a surjective homomorphism f : R → S. Example. The ring Z5 is a homomorphic image of Z15. Theorem 6.13 (The First Isomorphism Theorem). Let f : R → S be a surjective homomorphism of rings with kernel K. The quotient ring R/K is isomorphic to S. Proof. We define a function ϕ : R/K → S using the given surjective homomorphism f : R → S. To each coset r + K in R/K we associate the element f(r) ∈ S, so that ϕ(r + K) = f(r).

This mapping may depend on the choice of representative r for the coset r + K. We show that ϕ is well-defined. Suppose that r + K = t + K. Then r ≡ t (mod K), meaning that r − t ∈ K. Since f is a homomorphism we obtain

0S = f(r − t) = f(r) − f(t).

Thus r +K = t+K implies ϕ(r +K) = f(r) = f(t) = ϕ(t+K), so that ϕ is well-defined. We next show that ϕ is surjective. For s ∈ S, there is r ∈ R for which f(r) = s because f is surjective. Thus s = f(r) = ϕ(r + K), so that ϕ is surjective. Next we show that ϕ is injective. Suppose that ϕ(r + K) = ϕ(c + K); then f(r) = ϕ(r + K) = ϕ(c + K) = f(c).

This implies that 0S = f(r) − f(c) = f(r − c) because f is a homomorphism. Then r − c ∈ K, meaning that r ≡ c (mod K), and so r + K = c + K. Finally we show that ϕ is a homomorphism: for c + K, d + K ∈ R/K, we have ϕ((c + K) + (d + K)) = ϕ((c + d) + K) = f(c + d) = f(c) + f(d) = ϕ(c + K) + ϕ(d + K), and ϕ((c + K)(d + K)) = ϕ(cd + K) = f(cd) = f(c)f(c) = ϕ(c + K)ϕ(d + K).

Therefore ϕ : R/K → S is an isomorphism.  Example. The kernel of the surjective homomorphism f : Z15 → Z5 defined by f[a]15) = [a]5 is ∼ K = {[0]15, [5]15, [10]15} = Z3. ∼ By the First Isomorphism Theorem, we know that Z15/K = Z5, so no need to write and compare the addition and multiplication tables for the two rings. ∼ Is it possible that Z15 = Z3 × Z5? (Yes, because 3 and 5 are relatively prime.)