Math 412. §3.3: More on Homomorphisms Professors Jack Jeffries and 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(1) = 1.

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 a ring isomorphism as a renaming of the elements of a ring, so that two isomorphic rings are “the same ring” if you just change the names of the elements.

A.WARM-UPON ARITHMETICIN RINGS: Let R be a ring, and a, b, c ∈ R. (1) Prove or disprove: a + c = b + c implies a = b. (2) Prove or disprove: ac = bc implies a = b. (3) Prove or disprove: every ﬁeld is an integral domain. (4) The deﬁnition of ring guarantees that additive inverses exist. Explain why additive inverses are unique in any ring. Hencefoth, we will write −a for the additive inverse of a.

(1) True. Assume a + c = b + c. Add the additive inverse of c to both sides: (a + c) + (−c) = (b + c) + (−c). Now use associativity: a + (c + −c) = b + (c + −c), and now the definition of additive inverse: a + 0 = b + 0. Finally, use the definition of the additive identity: a = b. QED. (2) False, even if c 6= 0. We have a counterexample in Z12. Let c = [4]12, a = [3]12 and b = [9]12. Then [3][4] = [9][4] = [0] in Z12. (3) True. Let F be a field. Take any x, y ∈ F such that xy = 0. We need to show that x or y is zero. If x = 0, we are done, so assume x 6= 0. Because F is a field, x has a multiplicative inverse, call it x−1. Multiply both sides by x−1 to get x−1(xy) = x−10. By Associativity of ×, we have (x−1x)y = x−10, and by definition of multiplicative inverses we have 1 · y = x−10. Then by definition of multiplicative identity, we have y = x−1 · 0. Since a field is a commutative ring, we have y = 0 · x−1, and last time we proved that 0 times anything is always 0, so we conclude that y = 0. So F is a domain. (4) Suppose that a and b are both additive inverses of c. That means 0 = a + c = b + c. From (1), we conclude that a = b. Note: In the beginning only including on Friday’s quiz will we write out every single use of the axioms as in these proofs. Soon, we will skip over this level of detail as we begin to fully understand it —-that is, soon, I will not demand that you point out every use of associativity, etc.

B.EXAMPLESOF HOMOMORPSHISMS: Which of the following mappings between rings is a homomorphism? Which are isomorphisms? In all cases, be sure you can prove it—note that you may have to name your map in order to do anything with it. n (1) The inclusion mapping Z ,→ Q sending each integer n to the rational number 1 . (2) Any inclusion of a subring S ⊂ R into a ring R. (3) The map Z → Zn sending each integer a to its congruence class [a]n. λ 0 (4) The map → M ( ) sending λ 7→ . R 2 R 0 λ 1 λ (5) The map → M ( ) sending λ 7→ . R 2 R 0 1 2

(6) The map M2(Z) → R sending each 2 × 2 matrix to its determinant.

The maps in (1)-(4) are homomorphisms, but not isomorphisms in general (for (2), we the inclusion map S,→ R is an isomorphism if and onlk if S = R; the others are never isomorphisms). The map in (5) is NOT a homomorphism, because the zero is not preserved. The map in (6) is NOT a homomorphism because the 1 0 0 0 1 0 0 0 addition is not preserved: det( + 6= det( + det . 0 0 0 1 0 0 0 1

C.BASIC PROOFSWITH HOMOMORPHISMS: 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 x ∈ S is nilpotent, then φ(x) is nilpotent.

(1) Note that 0S + 0S = 0S. Apply φ: φ(0S + 0S) = φ(0S) and use the fact that φ respects addition to write φ(0S) + φ(0S) = φ(0S). This is a statement in T . Now, add −φ(0S) to both sides as in Warm-up Problem A(1) to conclude: φ(0S) = 0T . (2) Given x ∈ S, we know that x + −x = 0S. Apply φ to get a statement in T : φ(x + −x) = φ(0S) = 0T . Because φ preserves addition: φ(x) + φ(−x) = 0T . This (together with the fact that + is commutative in any ring) exactly says that φ(−x) is the additive inverse of φ(x). So −φ(x) = φ(−x). n n (3) Suppose that x = 0S for some x ∈ S. Apply φ: φ(x ) = φ(0S) = 0T . Now because φ preserves n n multiplicative, φ(x ) = φ(x · x . . . x · x) = φ(x) · φ(x) . . . φ(x) · φ(x) = [φ(x)] = 0T . So φ(x) is also nilpotent.

D.KERNELOF RING HOMOMORPHISMS: By deﬁnition, the kernel of a ring homomorphism is the set of elements in the source that map to the ZERO of the target. That is, if φ : R → S is a ring homomorphism, ker φ = {r ∈ R | φ(r) = 0S}.

(1) Let φ : R → S be a ring homomorphism. Prove that 0R ∈ ker φ. (2) Compute the kernel of the canonical homomorphism: Z → Zn sending a 7→ [a]n. (3) Prove the following Theorem: A ring homomorphism φ : R → S is injective if and only if its kernel is {0R}.

(1) This follows since φ(0) = 0. (2) The kernel here is {nb | b ∈ Z}. (3) Suppose φ is injective. Take any x ∈ ker φ. Then φ(x) = φ(0) = 0. By injectivity, we have x = 0. So the kernel is {0R}. Conversely, assume that ker φ = {0R}. Suppose that φ(x) = φ(y) for x, y ∈ R. Then φ(x) + −φ(y) = 0S. Using C(2), we have φ(x) + φ(−y) = 0S, and since φ respects addition, φ(x + (−y)) = 0S. So x + −y is in the kernel. By assumption, x + −y = 0, so x = y.

E. Here are some operation tables for operations on the set S = {a, b, c, d}. 3 ♣ a b c d ♦ a b c d a a a a a a a b c d b a b c d b b c d a c a c a c c c d a b d a d c b d d a b c

♥ a b c d ♠ a b c d a a a a a a a b c d b a b c d b b a d c c a c c a c c d a b d a d a d d d c b a It turns out that S can be made into a ring (with identity) in two different ways. For both of these ring structures on S, the operations + and × are each one of {♣, ♦, ♥, ♠}. (1) Use a basic ring property1 to determine which of the above operations could be a multiplication and which addition on S. (2) Could S be a ring with ♦ for addition and ♥ for multiplication?2. (3) What are the two pairs of operations that make S a ring? (+, ×) = ( , ) and (+, ×) = ( , ). (4) Show that (S, ♦, ♣) is isomorphic to (Z4, +.×). Give an explicit isomorphism. (5) Show that (S, ♠, ♥) is isomorphic to Z2 × Z2. Give an explicit isomorphism. (6) Is Z2 × Z2 isomorphic to Z4?

(1) We know that 0 · x = 0 for all x in a ring. So look for some operation that has an element Q that will pair up to always give Q back with anything. We see that a plays this role in both ♥ and ♣. So a could be the zero of some ring, and ♥ and ♣ could be multiplication for some ring. Likewise ♦ and ♠ could be addition, since a really is an identity for both (by inspection of the tables), and every element has an inverse for both ♦ and ♠. Looking again, we see that b is an identity for both ♥ and ♣, so b could be the 1 in both rings. (2) Say S is a ring with addition ♦. How do we know whether ♥ or ♣ is its multiplication? The only ring axiom linking addition and multiplication is the DISTRIBUTIVE axiom, so we should try to use it to answer this question. We already know how 0 and 1 multiply, so we should try to distribute, say c. Compute c♥(b♦b) =? (c♥b)♦(c♥b). Using the tables, this amounts to c♥c =? c♦c which using the tables again we see is FALSE. So ♥ can not be paired with ♦ to get a ring structure on S. (3) So it must be that (S, ♦, ♣) is a ring. Also (S, ♠, ♥) is a ring. (4) (S, ♦, ♣) is isomorphic to Z4 with an explicit isomorphism given by φ([0]) = a, φ([1]) = b, φ([2]) = c, φ([3]) = d. (S, ♠, ♥) is isomorphic to Z2 × Z2 with explicit isomorphism given by ψ(0, 0) = a, ψ(1, 1) = b, ψ(1, 0) = c (or d), and ψ(0, 1) = d (or c). You can check this by writing out the charts for both operations on Z2 × Z2 and verify they are “the same” as ♦ and ♣.

1of 0, proved last time 2The only axiom relating addition and multiplication is the DISTRIBUTIVE law. Find a way to use it to answer this question 4

Actually, ﬁnding these isomorphisms to KNOWN RINGS is the easiest way to check that all the axioms hold for the exotic ring structures on S here. (5) The explicit isomorphism sends (0, 0) 7→ a, (1, 1) 7→ b, (1, 0) 7→ c (or d) and (0, 1) 7→ d (or c). (6) Z4 is NOT isomorphic to Z2 × Z2. The easiest way to see this is to note that under any isomorphism Z4 → Z2 × Z2, we must have [1]4 7→ ([1]2, [1]2) (by deﬁnition of homomorphism, since these are the respective multiplicative identities. But that mean also [1]4 + [1]4 7→ ([1]2, [1]2) + ([1]2, [1]2), so any such homomorphism would take [2]4 to ([0]2, [0]2). But alas, also [0]4 7→ ([0]2, [0]2), since this is the case in for any ring homomorphism (these being the additive identities). But then both [2] and [0] are taken to the same element of Z2 × Z2, so the map is not injective!

F. Let R be any ring (with identity). Prove that there exists a unique ring homomorphism Z → R.

Deﬁne φ : Z → R sending n 7→ (1R + ... 1R) where the sum is n-times. You need to check that this is a ring homomorphism: this is pretty easy since φ(n + m) will be the sum of 1R n + m times, and φ(nm) will be the sum of 1R nm, which is the product of (1R +···+1R)(1R +... 1R) (n and m times) by the distributive axiom in R. It is unique, because any ring homomorphism send 1 to 1. But that means that 2 = 1 + 1 must go to 1R + 1R and so on.

G.HOMOMORPHISMSBETWEENMODULARRINGS: Recall the canonical homomorphism

πn : Z → Zn, a 7→ [a]n.

(1) Prove that πn is a ring homomorphism. (2) Is πn surjective? Is it injective? What is the kernel? (3) Are there any ring homomorphisms Zn → Z? Explain. (4) Let n and d be integers such that d|n. Consider the mapping π : Zn → Zd sending [a]n → [a]d. Explain why π is well-deﬁned, meaning that it doesn’t depend on the choice of the representative a we take for each class. (5) Prove that π is a homomorphism. (6) Prove that π is surjective.

Don’t stress this one for now. We’ll return to it. But you do have the tools to understand and its good practice to try it if you have time.