7. GRUPO DE ARMAMENTO
7.7. MANTENIMIENTO DE MISILES
by X+ Y = (X ∪ Y ) − (X ∩ Y ) and X · Y = X ∩ Y . Prove that (P (S), +, ·) is a commutative ring with identity. Show also that X2= X and 2X = 0P (S).
3. A ring R is called Boolean if r2 = r for all r ∈ R, (cf. Exercise 6.1.2). If R is a Boolean ring, prove that 2r = 0 and that R is necessarily commutative.
4. Let A be an arbitrary (additively written) abelian group. Prove that A is the underlying additive group of some commutative ring.
5. Find the unit groups of the following rings: (a) 2mn | m, n ∈ Z
; (b) Mn(R) with the standard matrix operations; (c) the ring of continuous functions on[0, 1].
6. Prove that the Binomial Theorem is valid in any commutative ring R, i.e., (a+b)n = n
i=0 n
i
aibn−iwhere a, b∈ R and n is a non-negative integer.
7. Let R be a ring with identity. Suppose that a is an element of R with a unique left
inverse b, i.e., b is the unique element in R such that ba = 1. Prove that ab = 1, so
that a is a unit. [Hint: consider the element ab− 1 + b.]
8. Let R be a ring with identity. Explain how to define a formal power series over R of the form∞n=0antnwith an ∈ R. Then verify that these form a ring with identity with respect to appropriate sum and product operations. (This is called the ring of
formal power series in t over R).
6.2
Subrings and ideals
In Chapter 3 the concept of a subgroup of a group was introduced, and already this has proved to be valuable in the study of groups and in applications. We aim to pursue a similar course for rings by introducing subrings.
Let (R,+, ×) be a ring and let S be a subset of the underlying set R. Then S is called a subring of R if (S,+S,×S)is a ring where+S and×S denote the binary operations+ and × when restricted to S. In particular S is a subgroup of the additive group (R,+).
From (3.3.3) we deduce a more practical description of a subring.
(6.2.1) Let S be a subset of a ring R. Then S is a subring of R precisely when S
contains 0R and is closed with respect to addition, multiplication and the formation of negatives, i.e., if a, b∈ S, then a + b ∈ S, ab ∈ S and −a ∈ S.
Example (6.2.1) (i)Z, Q, R are successively larger subrings of the ring of complex
numbersC.
(ii) The set of even integers 2Z is a subring of Z. Note that it does not contain the identity element; this is not a requirement for a subring.
(iv) Let S = 12Z, i.e., S = m2 | m ∈ Z. Then S is an additive subgroup of the ringQ, but it is not a subring since12 ×12 = 14 ∈ S. Thus the concept of a subring is/
more special than that of an additive subgroup.
Ideals. It is reasonable to expect there to be an analogy between groups and rings in which subgroups correspond to subrings. The question then arises: what is to correspond to normal subgroups? This is where ideals enter the picture.
An ideal of a ring R is an additive subgroup I such that ir∈ I and ri ∈ I whenever
i ∈ I and r ∈ R. So an ideal is an additive subgroup which is closed with respect to
multiplication of its elements by arbitrary ring elements on the left and the right. In particular every ideal is a subring.
Example (6.2.2) (i) Let R be a commutative ring and let x ∈ R. Define
Rx
to be the subset{rx | r ∈ R}. Then Rx is an ideal of R. For in the first place it is a subgroup since r1x+ r2x= (r1+ r2)xand−(rx) = (−r)x; also s(rx) = (sr)x for
all s∈ R, so Rx is an ideal. An ideal of this type is called a principal ideal.
(ii) Every subgroup ofZ has the form nZ where n ≥ 0 by (4.1.5). Hence every subgroup ofZ is an ideal.
(iii) On the other hand,Z is a subring but not an ideal of Q since 12(1) /∈ Z. So
subrings are not always ideals. Thus we have a hierarchy of distinct substructures of
rings:
ideal⇒ subring ⇒ subgroup.
Homomorphisms of rings. It is still not clear why ideals as defined above are the true analogs of normal subgroups. The decisive test of the appropriateness of our definition will come when homomorphisms of rings have been defined. If we are right, the kernel of a homomorphism will be an ideal.
It is fairly obvious how one should define a homomorphism from a ring R to a
ring S: this is a function θ : R → S which respects the ring operations in the sense
that:
θ (a+ b) = θ(a) + θ(b) and θ(ab) = θ(a)θ(b)
for all a, b∈ R. So in particular θ is a homomorphism of groups.
If in addition θ is bijective, θ is called an isomorphism of rings. If there is an isomorphism from ring R to ring S, then R and S are said to be isomorphic rings, in symbols
R S.
Example (6.2.3) (i) Let m be a positive integer. Then the function θm : Z → Zm defined by θm(x)= [x]mis a ring homomorphism. This is a consequence of the way sums and products of congruence classes were defined.
6.2 Subrings and ideals 105 (ii) The zero homomorphism 0: R → S sends every r ∈ R to 0S. Also the identity
isomorphism from R to R is just the identity function on R.
(iii) As a more interesting example, consider the set R of matrices of the form
a b
−b a
, (a, b∈ R).
These are easily seen to form a subring of the matrix ring M2(R). Now define a
function θ : R → C by the rule θ a b −b a
= a+ib where i =√−1. Then θ is a ring homomorphism:
indeed a1 b1 −b1 a1 a2 b2 −b2 a2 = a1a2− b1b2 a1b2+ a2b1 −a1b2− a2b1 a1a2− b1b2 ,
which is mapped by θ to (a1a2− b1b2)+ i(a1b2+ a2b1), i.e., to (a1+ ib1)(a2+ ib2).
An easier calculation shows that θ sends a1 b1 −b1 a1 + a2 b2 −b2 a2 to (a1+ ib1)+ (a2+ ib2).
Certainly θ is surjective; it is also injective since a+ib = 0 implies that a = 0 = b. Therefore θ is an isomorphism and we obtain the interesting fact that R C. Thus complex numbers can be represented by real 2× 2 matrices. In fact this is one way to define complex numbers.
Next let us consider the nature of the kernel and image of a ring homomorphism. The following result should be compared with (4.3.2).
(6.2.2) If θ : R → S is a homomorphism of rings, then Ker(θ) is an ideal of R and
Im(θ ) is a subring of S.
Proof. We know already that Ker(θ ) and Im(θ ) are subgroups by (4.3.2). Let k ∈
Ker(θ ) and r ∈ R. Then θ(kr) = θ(k)θ(r) = 0S and θ (rk) = θ(r)θ(k) = 0S since
θ (k)= 0S. Therefore Ker(θ ) is an ideal of R. Furthermore θ (r1)θ (r2)= θ(r1r2), so
that Im(θ ) is a subring of S.
(6.2.3) If θ : R → S is an isomorphism of rings, then so is θ−1: S → R.
Proof. We know from (4.3.3) that θ−1is an isomorphism of groups. It still needs to be shown that θ−1(s1s2)= θ−1(s1)θ−1(s2), (si∈ S). Observe that the image of each side under θ is s1s2. Since θ is injective, it follows that θ−1(s1s2)= θ−1(s1)θ−1(s2).
Quotient rings. Reassured that ideals are the natural ring theoretic analog of normal subgroups, we proceed to define the quotient of a ring by an ideal. Suppose that I is an ideal of a ring R. Then I is a subgroup of the additive abelian group R, so I R and we can form the quotient group R/I . This is an additive abelian group whose elements are the cosets of I . To make R/I into a ring, a rule for multiplying cosets must be specified: the obvious one to try is
(r1+ I)(r2+ I) = r1r2+ I, (ri ∈ R). To see that this is well-defined, let i1, i2∈ I and note that
(r1+ i1)(r2+ i2)= r1r2+ (r1i2+ i1r2+ i1i2)∈ r1r2+ I
since I is an ideal. Thus the rule is independent of the choice of representatives r1
and r2. A further straightforward check shows that the ring axioms hold; therefore
R/I is a ring, the quotient ring of I in R. Note also that the assignment r → r + I is a surjective ring homomorphism from R to R/I with kernel I ; this is the canonical
homomorphism, (cf. 4.3).
As one would expect, there are isomorphism theorems for rings similar to those for groups.
(6.2.4) (First Isomorphism Theorem) If θ: R → S is a homomorphism of rings, then
R/Ker(θ ) Im(θ).
(6.2.5) (Second Isomorphism Theorem) If I is an ideal and S a subring of a ring R,
then S+ I is a subring and S ∩ I is an ideal of S. Also S + I/I S/S ∩ I.
(6.2.6) (Third Isomorphism Theorem) Let I and J be ideals of a ring R with I ⊆ J .
Then J /I is an ideal of R/I and (R/I )/(J /I ) R/J .
Fortunately we can use the proofs of the isomorphism theorems for groups – see (4.3.4), (4.3.5), (4.3.6). The isomorphisms constructed in these proofs still stand if we allow for the additive notation. One has only to check that they are isomorphisms of rings.
For example, take the case of (6.2.4). From (4.3.4) we know that the assignment
r+ Ker(θ) → θ(r) yields a group isomorphism α : R/ Ker(θ) → Im(θ). Since α((r1+ Ker(θ))(r2+ Ker(θ))) = α(r1r2+ Ker(θ)) = θ(r1r2),
which equals
θ (r1)θ (r2)= α(r1+ Ker(θ))α(r2+ Ker(θ)),
we conclude that α is an isomorphism of rings: this proves (6.2.4). It is left to the reader to complete the proofs of the other two isomorphism theorems.
(6.2.7) (Correspondence Theorem for subrings and ideals) Let I be an ideal of a ring
R. Then the assignment S → S/I determines a bijection from the set of subrings of R that contain I to the set of subrings of R/I . Furthermore S/I is an ideal of R/I if and only if S is an ideal of R.