Problem 10: In a ring, we denote by (−x) the unique additive inverse of the element x. Prove that (−x)y = −(xy) = x(−y) and that (−x)(−y) = xy. You may only use ring axioms, including the uniqueness of 0 and of additive inverses, and 0x = 0, in this proof. Solution: The claim (−x)y = −(xy) means that (−x)y is the additive inverse of xy. To prove this claim, we have to show that (−x)y + xy = 0. By the distributive law, the left hand side can be rewritten as ((−x) + x)y, which is 0 · y = 0. The claim x(−y) = −(xy) is proved similarly, using the distributive law with the sum on the other side. Now (−x)(−y) = −(x(−y)) = −(−(xy)) from the previous two results. We will show that −(−z) = z for every z; then −(−(xy)) = xy follows in particular. But the claim −(−z) = z means that the additive inverse of (−z) is z, in other words, that (−z) + z = 0. But this is true because (−z) is the additive inverse of z. Problem 11: Given a set S, define the power set P(S) as the set that consists of all subsets of S. For A, B ∈ P(S) (i.e., for A, B ⊂ S), we define A + B := (A \ B) ∪ (B \ A) A · B := A ∩ B (These notations + and · are only for the purpose of the present problem and wouldn’t be understood automatically outside this context.) Show that (P(S), +, ·) is a commutative ring with one. Which sets are the 0 and the 1 in this ring? You may skip the trivial proofs of (C+), (C·) and (Ass·), but do show the proof of (Ass+), and the remaining axioms. Solution: Clearly + and · are binary operations on P(S), and are commutative. To show the additive group properties, we need to check (Ass+) and find the additive identity and for each A, its additive inverse. Notice that x ∈ A + B exactly if x is in A, but not in B, or in B, but not in A; in other words: x ∈ A + B if and only if x is in exactly one of the sets A, B, but not in both. To show (A + B) + C ⊂ A + (B + C), we assume x ∈ (A + B) + C. Then either (Case 1) x ∈ A + B but x ∈ / C, or (Case 2) x ∈ / A + B, but X ∈ C. Pursuing Case 1, suppose first (Case 1.1) that x ∈ A but x ∈ / B. And we still have x ∈ / C. Then, since x is neither in B nor in C, we know x ∈ / B +C. But since x ∈ A, we infer that x ∈ A+(B +C). — Alternatively, we have Case 1.2, where x ∈ / A, but x ∈ B. And still x ∈ / C. So x is in exactly one of B, C, i.e., x ∈ B + C. But since we also have x ∈ / A, we again infer that x ∈ A + (B + C). Pursuing Case 2 now, where x ∈ / A + B, we have two subcases: Case 2.1: x is in both A and B. Then, since also x ∈ C, we conclude that x ∈ / B + C. But we have x ∈ A, and therefore x ∈ A + (B + C). — Alternatively, we have Case 2.2: x is neither in A nor in B. But with x ∈ C, we now have x ∈ B + C. Since still x ∈ / A, we again conclude x ∈ A + (B + C). Gathering all the four subcases, we have concluded in each of them that x ∈ A + (B + C), provided x ∈ (A + B) + C. So we have proved (A + B) + C ⊂ A + (B + C). The same proof with the roles of the different sets exchanged gives (C + B) + A ⊂ C + (B + A). By commutativity, this means A + (B + C) ⊂ (A + B) + C. Altogether with the previously proved inclusion, this gives the associative law (A + B) + C = A + (B + C). We now claim that the empty set is the additive identity, i.e., that A + ∅ = A for all A. Indeed A + ∅ = (A \ ∅) ∪ (∅ \ A) = A ∪ ∅ = A . Finally we observe that each A is its own additive inverse, i.e., A + A = ∅. Indeed, A + A = (A \ A) ∪ (A \ A) = ∅ ∪ ∅ = ∅. With (C·) and (Ass·) clear, we still have to find a multiplicative identity and prove the distributive law. The multiplicative identity is S, since A ∩ S = A for every A ⊂ S. To show A · (B + C) = A · B + A · C, we assume x ∈ A · (B + C), i.e., x ∈ A and x ∈ B + C. So x is in exactly one of B, C; without loss of generality assume x ∈ B, but x ∈ / C. Then a fortiori, x∈ / A · C; but x ∈ A · B because x is both in A and in B. Therefore x ∈ A · B + A · C. Conversely, assume x ∈ A · B + A · C. So x is in exactly one of A · B and A · C. Without loss of generality, assume x ∈ A · B, but x ∈ / A · C. The first asserts that x ∈ A and x ∈ B. The second asserts that x is not in both of A, C, i.e., not in A or not in C. Since we know x ∈ A, we must have x∈ / C. From x ∈ B and x ∈ / C, we infer x ∈ B + C. Since also x ∈ A, we conclude x ∈ A · (B + C). So we have proved both inclusions of the distributive law A · (B + C) = A · B + A · C. Problem 12: Suppose (R, +, ·) is a ring. Consider the set R × R together with the following definition of + and ·: (a, b) + (c, d) := (a + c, b + d) and (a, b) · (c, d) = (ac − bd, ad + bc). Here, the minus sign stands for adding the additive inverse, as usual. (a) Show that R × R, together with these operations is again a ring. We will call this ring R[i]. (b) Show that R[i] has (1, 0) an identity element if R has 1 as identity element. Also show that R[i] is commutative, if R is. Solution: The abelian group properties for the addition are immediate, for instance (C+) is inherited from R as follows: (a, b) + (a! , b! ) = (a + a! , b + b! ) = (a! + a, b! + b) = (a! , b! ) + (a, b) Similarly for (Ass+). The additive identity is (0, 0), and the additive inverse of (a, b) is (−a, −b). I’ll tacitly use simple properties involving −, which can be concluded easily from the ring axioms, usually using the distributive law, next to other axioms. Associativity of · : ! " (s, t) · (x, y) · (u, v) = (sx − ty, sy + tx) · (u, v) = ((sx − ty)u − (sy + tx)v, (sx − ty)v + (sy + tx)u) (1) (2) = (sxu − tyu − syv − txv, sxv − tyv + syu + txu) (3) ! " (s, t) · (x, y) · (u, v) = (s, t) · (xu − yv, xv + yu) = (s(xu − yv) − t(xv + yu), s(xv + yu) + t(xu − yv)) (1) (2) = (sxu − syv − txv − tyu, sxv + syu + txu − tyv) (3) These equalities are justified as follows: (1),(2) definition of product; (3) distributive law in R; associativity of addition; associativity of multiplication. The results coincide, due to commutativity of addition. ! " Distributive Laws: I’ll show (s, t) + (x, y) · (u, v) = (s, t) · (u, v) + (x, y) · (u, v). The other distributive law is proved analogously. (Both are needed, logically.) ! " (s, t) + (x, y) · (u, v) = (s + x, t + y) · (u, v) = ((s + x)u − (t + y)v, (s + x)v + (t + y)u) (1) (2) = (su + xu − tv − yv, sv + xv + tu + yu) = ((su − tv) + (xu − yv), (sv + tu) + (xv + yu)) (3) (4) = (su − tv, sv + tu) + (xu − yv, xv + yu) = (s, t) · (u, v) + (x, y) · (u, v) (5) (6) These equalities are justified as follows: (1) definition of sum; (2) definition of product; (3) distributive law in R; associativity of addition; (4) commutativity of addition (and associativity again); (5) definition of sum; (6) definition of product; associativity of multiplication. Now if 1 is a unity in R, then (1, 0) · (a, b) = (1 · a − 0 · b, 1 · b + 0 · a) = (a − 0, b + 0) = (a, b) (1) (2) (3) where we have used: (1) definition of product; (2) property of 1, and the lemma 0 · x = 0; (3) property of 0. The opposite order (a, b) · (1, 0) = (a, b) is proved similarly. If R is commutative, then R[i] is, because (a, b) · (c, d) = (ac − bd, ad + bc) = (ca − db, da + cb) = (c, d) · (a, b) Problem 13: Show that R := {0, 2, 4, 6, 8} is a subring of Z10 with addition and multiplication modulo 10. The ring R does contain a multiplicative identity. Which is it? Solution: We have to show: If u, v ∈ R, then −u ∈ R, u + v ∈ R and uv ∈ R. Indeed, −0 = 0, −2 = 8, −4 = 6, . . . . We have the following addition and multiplication tables: + 0 2 4 6 8 0 0 2 4 6 8 2 2 4 6 8 0 4 4 6 8 0 2 6 6 8 0 2 4 8 8 0 2 4 6 · 0 2 4 6 8 0 0 0 0 0 0 2 0 4 8 2 6 4 0 8 6 4 2 6 0 2 4 6 8 8 0 6 2 8 4 This establishes closedness, and also 6 is the multiplicative identity in R. So we observe that a subring can have a different multiplicative identity than the bigger ring. For groups, this doesn’t happen, because the existence of an inverse allows to conclude: If ua = a for all a in the subgroup, then multiplying with a−1 (a calculation in the bigger group), we get u = 1. This argument is not available in a ring, since multiplicative inverses might not exist.