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.