RINGS AND FIELDS `07 RINGS AND FIELDS `07 De¯ne ring, subrings, modules, and submodules. Example 1. Rings: (a) We have the well-known rings; Z, Q, R, and C. (b) Inside C we ¯nd rings as follows; select an element ® that is n n¡1 n¡2 the root to some polynomial x +an¡1x +an¡2x +¢ ¢ ¢+a0 0 where the ajs are integers, and set R = Z[®] = ft 2 C j t = n¡1 n¡2 bn¡1® + bn¡2® + ¢ ¢ ¢ + b0; where each bi 2 Zg. (c) Given any ring R, we can construct a ring from the collection of n £ n matrices whose entries lie in R; S = Matn(R); where multiplication and addition are the standard matrix op- erations. (d) The upper, 2£2 matrices over a commutative ring R with equal diagonal entries forms a commutative subring of Mat2(R). Example 2. Modules Over a Ring R: (a) For an index set I, ¦I R and ©I R are modules over R. (b) An R-module F is called free on a subset X if for every map f : X ! A such that f(rx + sy) = rf(x) + sf(y) for x; y 2 X and r; s 2 R, where A is an R-module, there is a unique homomorphism h : M ! A such that hjX = f. (c) An R-module M is called projective if, for every epimorphism f : A ! B and homomorphism g : M ! B, there is a homo- morphism » Theorem 1. An R-module M is free if and only if M = ©I R. Theorem 2. An R-module M is projective if and only if M © N = F for some free module F . Let us use the projective and free modules as a foundation for our course. 1 2 RINGS AND FIELDS `07 Theorem 3. Every R-module is free if and only if R is a division ring. When is it true that every R-module is projective? Artin-Wedderburn Theorem . Every R-module is projective if and only if R is a ¯nite direct product of matrix rings over division rings; i.e., R = Matn1 (D1) £ Matn2 (D2) £ ¢ ¢ ¢ £ Matnk (Dk); for some natural numbers n1; n2; : : : ; nk and division rings D1;D2;:::;Dk. Example 3. Let H denote the (rational) Hamiltonian Quaternions; H = fa + bi + cj + dk j a; b; c; d 2 Qg; where addition is component-wise and multiplication extends linearly 2 2 2 from i = j = k = ¡1. Then, the left ideal of R = Mat2(H) consist- ing of the matrices with second column zero is a projective (left) ideal of R that is not free. p p pExample 4. Let R = Z[ ¡5] = fa + b ¡5 j a; b 2 Z and I = (3; 1 + ¡5). We claim that I is projective but not free. First we will argue that I 6= R. Notice 3Rp⊆ I. Modulop 3R, the general elementp of I can be represented as (1+ ¡5)¢p(a+b ¡5) = (a¡5b)+(a+b) ¡5, which is congruent to (a + b)[1 + ¡5] mod 3R. Therefore, I=3R »= Z=3Z, and since R=3R has order 9, I is properly contained in Rp. If I were free,p then I would be principal; say I = (a + b ¡5). Since I contains 3; 1+ ¡5, by computing norms, a2 +5b2 must divide 9 and 6. Hence a2 + 5b2 = 1 or 3. We rule out 1 since I 6= R. However, a2 + 5b2 = 3 is not possible. Therefore, I is not free. Set 2 p ® = (1 + ¡5); and ¯ = ¡1: 3 Note that ® ¢ I ⊆ R and ¯ ¢ I ⊆ R. De¯nep the homomorphism θ : F = R © R ! I by θ(r; s) = r3 + s(1 + ¡5). The epimorphism θ is split by the map ± : I ! F where ±(x) = (¯x; ®x). (Check: p 2 θ(±(x)) = θ((¯x; ®x)) = 3¯x + (1 + ¡5)®x = x(¡3 + 3 6) = x. Therefore, I is projective. p Example 5. The ring R = Z[ ¡5] of the last example has the property that every nonzero nonunit can be expressed as a product of irreducible elements (see homework), yet R is not a UFD (see the next example). RINGS AND FIELDS `07 3 Example 6. Let d be a square-free integer 6= 1, and de¯ne p R = Z[ d] provided d ´ 2; 3 (mod 4); or p 1 + d R = Z[ ] when d ´ 1(mod 4): 2 Then every ideal of R is projective. It is not known whether or not there are in¯nitely many d0s for which R is a PID. However, for such an R, R is a PID if and only if R is a UFD. 4 RINGS AND FIELDS `07 1. EXERCISE SET 1 In this homework set, R is an integral domain. A nonzero, nonunit r of R is said to be irreducible provided r = st for s; t 2 R can only happen when one of s or t is a unit. (Recall, u 2 R is a unit if uv = 1 for some v 2 R). A nonzero, nonunit r of R is said to be prime if r j ab for a; b 2 R, implies r j a or r j b. R is said to have the ascending chain condition on principal ideals, if any increasing chain of ideals (a1) ⊆ (a2) ⊆ (a3) ⊆ ¢ ¢ ¢ stablizes (i.e., there is an index n such that (am) = (an) for all m ¸ n). 1. Show that if R has the ascending chain condition on principal ideals, then every nonzero nonunit r of R is a product of ir- reducible elements. Hint: To argue that r has an irreducible factor, either r is irreducible, or r = r1s1 with r1; s1 nonzero nonunits. Note (r) is properly contained in (r1) in this case. Re- peat for r1 and use the hypothesis to eventually obtain r = s1t1 with s1 irreducible. Now write r = r1v1 = r1r2v2 = ::: with r1; r2;::: irreducible, and use the hypothesis to show that even- tually some vj is a unit. 2. Show that if R is a UFD then every irreducible element is prime and R has the ascending chain condition on principal ideals. 3. Show that if irreducible elements are prime in R and R has the ascending chain condition on principal ideals, then R is a UFD. 4. Show that R is a UFD if and only if every nonzero nonunit is a product of primes. 5. Show any PID is a UFD. RINGS AND FIELDS `07 5 2. Integral Domains Theorem 4. Let R be an integral domain. Every submodule of ©nR for any n is free if and only if R is a PID. Proof. An ideal of R is a submodule of R. Hence if the condition is in force, R is necessarily a PID. Conversely, assume that R is a PID. We will show, by induction on n, that every submodule of ©nR is free. When n = 1, the nonzero submodules of R are ideals and are principal (hence isomorphic to R). Inductively, let L be a nonzero submodule of F = ©nR. For ¼j : F ! th R equal to the projection map onto the j coordinate, I = ¼j(L) 6= 0 for some j. Since I = aR for some 0 6= a 2 R, we can de¯ne a splitting map f : aR ! L by f(ar) = rx where x 2 L is any element such that ¼j(x) = a. By the standard argument, » L = Im f © Ker ¼jjL = aR © Ker ¼jjL: th Note that Ker ¼jjL = fy 2 L j y has a zero j componentg · ©i6=jR; where the index i runs over 1; 2; : : : ; j ¡ 1; j + 1; : : : ; n. By induction, » » Ker ¼jjL = ©mR is free, and therefore, L = ©m+1R. ¤ Given an ideal I 6= 0 of R, de¯ne I¡1 = ft 2 Q j tIsubseteqRg: The ideal I is said to be invertible, if I ¢ I¡1 = R: Analogously, we have Theorem 5. Let R be an integral domain. Every submodule of ©nR for any n is projective if and only if every ideal of R is invertible (pro- jective). In order to see this we need a few observations. Bear Injective Test Lemma . The module U is injective if and only if U is injective relative to any sequence 0 ! I ! R where I is an ideal of R. Proof. Assume that U is injective with respect to any sequence 0 ! I ! R, and suppose that 0 ! A ! B is exact and f : A ! U. Consider the set C of all 2-tuples (g; C) where A · C · B and gjA = f: There is a maximal element (g0;C0). If C0 6= B let x 2 B n C0, and let I = fr 2 R j rx 2 C0g. Set 0 0 C = C0 + hxi. If I = 0, then hxi \ C0 = 0 and so C = C0 © hxi · B; 0 0 0 we then have the over-module C of C0 and can de¯ne g : C ! U by 6 RINGS AND FIELDS `07 0 g (c0; rx) = g(c0), contradicting the maximality of (g; C0). Therefore I 6= 0. By hypothesis, the map h : I ! U given by h(r) = g(rx) can be lifted to a map h0 : R ! U. We then de¯ne g0 : C0 ! U by 0 0 g (c0 + rx) = g(c0) + h (r). If c0 + rx = b0 + sx with c0; b0 2 C0 and r; s 2 R, then (r ¡ s)x = b0 ¡ c0 2 C0 and so r ¡ s 2 I. But then 0 0 0 g(b0 ¡ c0) = g((r ¡ s)x) = h(r ¡ s) = h (r ¡ s) = h (r) ¡ h (s) and 0 0 0 0 so g(b0) ¡ g(c0) = h (r) ¡ h (s) implying g(b0) + h (s) = g(c0) + h (r) making g0 well-de¯ned.