CHAPTER 1

Revision of theory

1.1. Basic definitions and examples In this chapter we will revise and extend some of the results on rings that you have studied on previous courses. A ring is an algebraic object in which we can operate in a similar way as we do with integers. Inside the Z of integers we can perform operations of , substraction and multiplication, but in general Z is not closed under division; for instance the quotient 3/2 is not an integer. Addition and multiplication have a series of well known properties; the abstraction of those properties is what constitutes the formal notion of ring.

DEFINITION 1.1.1. A ring is a nonempty set R together with two operations, a sum + and a product satisfying the following properties: · Sum S1 Associativity: (a + b)+c = a +(b + c) for all a, b, c R, ∈ S2 Commutativity: a + b = b + a for all a, b R, ∈ S3 Zero: There is an element 0 R such that a +0=a =0+a for each a R, ∈ ∈ S4 Inverses: For each a R there is an element a R such that a +( a)=0. ∈ − ∈ − In what follows we will always write a b to denote a +( b). − − REMARK. These properties simply mean that (R, +) is an abelian group. Product P1 Associativity: a(bc)=(ab)c for all a, b, c R, ∈ P2 Unit: There is an element 1 R such that 1a = a1=a for all a R, ∈ ∈ REMARK. This properties mean that (R, ) is a multiplicative monoid. · P3 Distributivity: For all a, b, c R, ∈ a(b + c)=ab + ac, (a + b)c = ac + bc. Some rings satisfy an additional property for the product: P4 Commutativity: ab = ba for all a, b R. ∈ When this extra property is satisfied, we will say that R is a commutative ring. In this course, we will be mostly interested on commutative rings, although we will occasionally deal with some noncommutative examples.

EXAMPLE 1.1.2. R = 0 with the trivial operations. This is called the trivial ring, { } and is the only ring for which one has 1=0. In all what follows, we will assume our rings to be nontrivial, i.e. 0 =1. ￿ EXAMPLE 1.1.3. The integers Z with the usual addition and multiplication. EXAMPLE 1.1.4. The fields of rational numbers Q, real numbers R or complex num- bers C, or in general any field F.

1 2 1. REVISION OF RING THEORY

EXAMPLE 1.1.5. The rings of integers modulo n, Z/nZ (sometimes also denoted by Zn) consisting of the set a a Z = 0, 1,...,n 1 , where a is the residue class { | ∈ } { − } of a modulo n, so a = a = rn r Z . Addition and multiplication are defined as { | ∈ } a + b := a + b, ab := ab EXAMPLE 1.1.6. Let R be any ring, and define R[x] to be the set of all polynomials a + a x + + a xn where the coefficients a are elements of the ring R. Then R[x] 0 1 ··· n i is a ring with addition and multiplication defined in the usual way (check as an exercise!). The ring R[x] is called the polynomial ring in one variable with coefficients in R. EXAMPLE 1.1.7. The polynomial ring in n variables with coefficients in R, denoted by R[x1,...,xn] and defined inductively by R[x1,...,xn]:=R[x1,...,xn 1][xn]. − EXAMPLE 1.1.8. The ring M (R) of n n matrices (a ) with coefficients n × i,j i,j=1,...,n ai,j in R, and the usual matrix addition and multiplication. EXAMPLE 1.1.9. The power set ring. Let X be a set, and let (X)= Y Y X P { | ⊆ } the set of all subsets of X. On (X) consider the operations: P Y + Z := Y Z =(Y Z) (Y Z) the symmetric difference as addition, • ￿ ∪ \ ∩ YZ:= Y Z the intersection as a product. • ∩ With this operations (X) becomes a ring with zero element 0=∅ and unit element P 1=X. EXAMPLE 1.1.10. Let V be a vector space, consider the set End(V ):= f : V { → V f is a linear map , then End(V ) is a ring with pointwise addition (f +g)(v):=f(v)+ | } g(v) and multiplication given by composition fg(v):=f(g(v)), where the zero element is the constant map 0(v)=0and the unit element is the identity map Id(v)=v. REMARK. The ring of endomorphisms of a vector space is nothing but a matrix ring in disguise. We will state more precisely what we mean by this once we talk about ring isomorphisms. EXAMPLE 1.1.11. Consider the set C(R):= f : R R f continuous of { → | } real-valued continuous functions. The C(R) is a ring with pointwise addition (f +g)(x):= f(x)+g(x) and multiplication (fg)(x):=f(x)+g(x). REMARK. One might wonder whether one could define a different ring structure on C(R) by replacing pointwise multiplication by composition as a product. Unlike it hap- pened in the case of linear map, this operation does not turn C(R) into a ring. EXAMPLE 1.1.12. Let X be a set and R be a ring. In a similar fashion to the previous example, the set XR of all R-valued maps f : X R on X becomes naturally a ring with → pointwise addition and multiplication. EXAMPLE 1.1.13. Quaternion algebras. Let F be a field (of characteristic different from 2), and let α,β F. The quaternion algebra αFβ is defined as the set a + bi + ∈ { cj + dk a, b, c, d F with standard sum and product defined by the rules ij = k = ji, | ∈ } − i2 = α, j2 = β. If F is a subfield of the real numbers, αFβ can also be described as the subring a + b√αc√β + d√αβ of M ( ) consisting of matrices of the form , where 2 C c√β d√αβ a b√α ￿ − − ￿ a, b, c, d F. ∈ EXAMPLE 1.1.14. The ring of power series with real coefficients

n R[[x]] := anx an R n .  | ∈ ∀  n 0 ￿≥  In general one can construct the ring of formal power series R[[x]] with coefficients in any   ring R. 1.2. SUBRINGS, IDEALS AND QUOTIENT RINGS 3

EXAMPLE 1.1.15. Let G be a group, R a ring, the group ring R[G] is defined as R[G]:= a x a R x, a =0for all x except a finite number x G x | x ∈ ∀ x = f : G∈ R f has a finite support . {￿￿ → | } ￿ The addition and product are defined (in the functional notation) as follows: (f + g)(x):=f(x)+g(x), • 1 (fg)(x):=(f g)(x)= y G f(y)g(y− x). • ∗ ∈ The product in R[G] receives the name￿ of product of functions. REMARK. Note that in this case the convolution product actually defines a different ring structure in the set of functions f : G R than the pointwise product, so this example → is actually different from Example 1.1.12; for instance, if R is a commutative ring then GR (with the pointwise product) is also commutative, whereas the group ring R[G] will be noncommutative whenever G is.

1.2. Subrings, ideals and quotient rings Let R be a ring, we look at subsets of R which are in fact themselves rings in their own right when we restrict to them the sum and product of R. More precisely, we say that S is a subring of R if: (i) 1 S, R ∈ (ii) S is an additive subgroup of R. In other words, whenever a, b S, one has a b S. ∈ − ∈ (iii) S is closed under the product of R, in other words, S is a multiplicative submonoid of R, i.e. for all a, b S one has ab S. ∈ ∈ We will write S R to denote that S is a subring of R. ≤ EXAMPLES 1.2.1. (1) Z Q R C. ≤ ≤ ≤ (2) The trivial ring 0 is NOT a subring of Z, as 1 / 0 . { } ∈{ } (3) For any ring R, one has R R[x], where R consists of all constant polynomials ≤ in R[x]. (4) The rings of matrices Mn(R) contain several interesting subrings. Some exam- ples are The ring D (R) of diagonal matrices. • n The ring U (R) of upper-triangular matrices. • n The ring L (R) of lower-triangular matrices. • n (5) If Si i I is a family of rings such that Si R for all i I, then Si R. { } ∈ ≤ ∈ i I ≤ (6) Let X R be a subset of a ring R, define [X]:= S R X∈ S the ⊆ { ≤ | ⊆ } intersection of all subrings of R containing X. Then [X] is a subring￿ of R, ￿ called the subring generated by X.

EXERCISE 1.2.2. Show that [X] can be identified with the set of all sums of the form x x where x X 1 . ± 1 ··· n i ∈ ∪{ } ￿ We move now to the key notion of ideal. Ideals are certain subsets of rings that play a similar role to that of normal subgroups in group theory, in the sense they allow us to build quotients of rings. Also, knowing the ideals of a ring in full detail often lead to a complete description of all the modules, so understanding ideals is a fundamental topic of this course.

DEFINITION 1.2.3. Let R be a commutative ring. A subset I R of R is said to be ⊆ an ideal of R if it has the following properties: I1 Additive closure: I (R, +) is an additive subgroup of R, i.e. I = ∅ is ≤ ￿ nonempty and for all a, b I one has a b I. ∈ − ∈ I2 Absorbency: For all r R and for all a I one has ra = ar I. ∈ ∈ ∈ 4 1. REVISION OF RING THEORY

When I is an ideal of R we will write I ￿ R. REMARK 1.2.4. In the previous definition we are assuming that R is commutative. For noncommutative rings one has to be more careful and distinguish among the notions of left, right and two-sided ideals.

EXAMPLES 1.2.5. (1) 0:= 0 R the zero ideal, which is an ideal for every { } ￿ ring R. (2) R￿R the total ideal, also an ideal for every ring. The ideals I which are different from the total ideal are called proper ideals. (3) Let R be a ring, I,J R ideals, then I J R and I+J := i + j i I, j J ￿ ∩ ￿ { | ∈ ∈ }￿ R are ideals of R, respectively called the intersection and the sum of I and J. I + J

I J

I J ∩ I J is the greatest ideal (with respect to the inclusion ordering) contained in ∩ both I and J, whilst I + J is the least ideal containing both I and J. (4) If Iα α A is a family of ideals of R, then α A Iα ￿ R is also an ideal of R. { } ∈ ∈ (5) If I1,...,In are ideals of R, define ￿ I + + I := i + + i i I , 1 ··· n { 1 ··· n| j ∈ j} then I + + I R is also an ideal of R. 1 ··· n ￿ (6) Let R be a ring, and a R an element of R, and define (a)=Ra := ra r R . ∈ { | ∈ } Then (a) ￿ R is an ideal, called the principal ideal generated by a. More gener- ally, if a ,...,a R are elements of R, then the set Ra + + Ra = 1 n ∈ 1 ··· n r a + + r a r R is an ideal of R, called the ideal generated by { 1 1 ··· n n| i ∈ } a1,...,an. (7) Particular examples of principal ideals are the trivial ideal 0 = (0) and the total ideal R = (1). If R = Z, the principal ideal generated by 2 is the set (2) = 2n n Z of even numbers. { | ∈ } Let R be a ring and I R an ideal. For each a R we define the coset of a with ￿ ∈ respect to I as the set a = a + I := a + i i I . { | ∈ } Since I is an additive subgroup of R, and (R, +) is a commutative subgroup, I is normal in R, and thus the set R/I = a a R is an additive group with addition defined by { | ∈ } a + b := a + b, i.e. (a + I)+(b + I):=(a + b)+I. The absorbency property of an ideal also ensures that the product of cosets ab = ab is well defined, endowing the set R/I with a ring structure. In particular, the zero and unit elements of R/I are 0 and 1, respectively.

EXAMPLES 1.2.6. (1) R/0=R, (2) R/R =0, (3) Z/(n)=Z/nZ = Zn.

1.3. Ring homomorphisms DEFINITION 1.3.1. Let R and S be rings; a map f : R S is said to be a ring → homomorphism (or ring morphism for short) if (1) f(1) = 1, 1.3. RING HOMOMORPHISMS 5

(2) f(a + b)=f(a)+f(b) for all a, b R, ∈ (3) f(ab)=f(a)f(b) for all a, b R. ∈ If f is injective we say it is a monomorphism, if it is surjective it is called an epimorphism, and if it is a bijection it is called an isomorphism. In this case we say that R is isomorphic to S and write R ∼= S. DEFINITION 1.3.2. Let f : R S be a ring morphism. We define the of f as → the set Im(f):=f(R)= f(r) r R S, { | ∈ }⊆ and the kernel of f as the set 1 Ker(f):=f − (0) = r R f(r)=0 R. { ∈ | }⊆ LEMMA 1.3.3. Let R and S be rings, and f : R S be a ring homomorphism, then → the following properties hold: (1) Im(f) S is a subring of S, ≤ (2) Ker(f) ￿ R is an ideal of R. PROOF. (1). 1 = f(1 ), and hence 1 Im(f). If s ,s Im(f) then there are r ,r R such S R S ∈ 1 2 ∈ 1 2 ∈ that f(r1)=s1,f(r2)=s2, but then using that f is a ring morphism one has s s = f(r ) f(r )=f(r r ) Im(f), 1 − 2 1 − 2 1 − 2 ∈ s s = f(r )f(r )=f(r r ) Im(f), 1 2 1 2 1 2 ∈ thus Im(f) S subring. ≤ (2). f(0R)=0S, so 0R Ker(f), and thus Ker(f) = ∅. Now, if a, b Ker(f), one has ∈ ￿ ∈ f(a b)=f(a) f(b)=0 0=0, − − − hence a b Ker(f), so Ker(f) is an additive subgroup of R. Now, for any a Ker(f), − ∈ ∈ and for any r R one has ∈ f(ra)=f(r)f(a)=f(r)0 = 0, thus ra Ker(f), and consequently Ker(f) R is an ideal of R. ∈ ￿ ￿ THEOREM 1.3.4 (First isomorphism theorem). Let R and S be rings, and f : R → S be a ring homomorphism, then the mapping f(r) r = r +Ker(f) provides an ↔ isomorphism R/ Ker(f) ∼= Im(f). PROOF. Let a = a +Ker(f) R/ Ker(f) for a R. One has ∈ ∈ a = b a b Ker(f) f(a b)=0 f(a)=f(b), ⇔ − ∈ ⇔ − ⇔ so the application π : R/ Ker(f) Im(f) given by π(a):=f(a) is well defined and → injective. Now, if b Im(f) then there is some a R such that b = f(a), and thus ∈ ∈ b = π(a); henceforth, π is surjective, and so it is a bijection. Now, π((1)) = f(1) = 1, and for any a, b R one has ∈ π(a b)=π(a b)=f(a b)=f(a) f(b)=π(a) π(b), − − − − − π(ab)=π(ab)=f(ab)=f(a)f(b)=π(a)π(b), hence, π is a ring homomorphism, and since it is bijective we have a ring isomorphism R/ Ker(f) ∼= Im(f). ￿ EXAMPLES 1.3.5. (1) The identity map Id : R R is a ring isomorphism. If S R is a subring, the → ≤ inclusion map i : S R is a ring monomorphism. S → 6 1. REVISION OF RING THEORY

(2) The complex conjugation map σ : C C defined by σ(z)=z is a ring isomor- → phism. (3) Let I R be an ideal of R, The canonical projection π : R R/I is defined ￿ I → by π(r):=r = r + I. It is easy to see that πI is a ring homomorphism, π (a)=0 a =0 a + I = I a I, so Ker(π )=I. For any a R/I I ⇔ ⇔ ⇔ ∈ I ∈ we can write a = πI (a), thus πI is an epimorphism, and hence Im(πI )=R/I. The statement of the first isomorphism theorem in this particular case is just a tautology, saying that R/I is isomorphic to itself. (4) For any n 2, the ring Zn = Z/nZ is the image of the canonical projection ≥ πn : Z Z/(n)=Zn. → (5) Let R S subring, and a S. We define the evaluation map e : R[x] S ≤ ∈ a → by p(x) p(a), i.e. if p(x)=r + r x + , +r xn then e (p(x)) = ￿→ 0 1 ··· n a r + r a + , +r an S. The evaluation map is a ring homomorphism for 0 1 ··· n ∈ which we have Im(e )= p(a) p R[x] = r ai r R = R[a] S, a { | ∈ } i | i ∈ ≤ Ker(e )= p(x) ￿￿R[x] p(a)=0￿. a { ∈ | } Note that, if we assume that a R, one has ∈ p(x) Ker(e ) p(a)=0 (x a) p(x) p(x)=(x a)r(x) ∈ a ⇔ ⇔ − | ⇔ − p(x) ((x a)). ⇔ ∈ − In this case one gets Im(e ) = R[x]/ Ker(e )=R/((x a)) = R. a ∼ a − ∼ LEMMA 1.3.6. Let f : R S and g : S T be ring homomorphisms. Then the → → composition g f : R T is also a ring homomorphism. ◦ → PROOF. The proof is immediate and follows from applying repeatedly that f and g are homomorphisms: (g f)(0) = g(f(0)) = g(0) = 0. ◦ (g f)(a + b)=g(f(a + b)) = g(f(a)+f(b)) = g(f(a)) + g(f(b)) ◦ =(g f)(a)+(g f)(b). ◦ ◦ (g f)(1) = g(f(1)) = g(1) = 1. ◦ (g f)(ab)=g(f(ab)) = g(f(a)f(b)) = g(f(a))g(f(b)) ◦ =(g f)(a) (g f)(b). ◦ · ◦ ￿ THEOREM 1.3.7 (Second isomorphism theorem). Let R be a ring, I ￿ R an ideal, S R a subring; then the following properties hold: ≤ (1) S + I R is a subring of R, ≤ (2) I ￿ S + I is an ideal of S + I, (3) S I S is an ideal of S, ∩ ￿ (4) There is a ring isomorphism S + I S = I ∼ S I ∩ PROOF. (1). Let s ,s S, i ,i I, then 1 2 ∈ 1 2 ∈ (s + i ) (s + i )=(s s )+(i i ) S + I, 1 1 − 2 2 1 − 2 1 − 2 ∈ and thus S + I is an additive subgroup of R. Similarly, (s + i )(s + i )=s s +(s i + i s + i i ) S + I, 1 1 2 2 1 2 1 2 1 2 1 2 ∈ 1.3. RING HOMOMORPHISMS 7 and 1 S S + I, so S + I is a submonoid of S, and hence S + I R is a subring. ∈ ⊆ ≤ (2). The I is additively closed because it is a subgroup of R; moreover, since S + I R ⊆ the absorbency property for I with respect to S + I is immediate. Hence, I ￿ S + I is an ideal of S + I. (3). As both I and S are additive subgroups of R, S I is also additively closed, and since ∩ it S I S, is is an additive subgroup of S. For the absorbency, let x S I, s S; ∩ ⊆ ∈ ∩ ∈ since x I and s S R, absorbency property for I tells us that xs I. Also, as ∈ ∈ ⊆ ∈ x S, s S and S is a subring, we get xs S, and thus sx S I. This proves that ∈ ∈ ∈ ∈ ∩ S I is an ideal of S. ∩ (4). Consider the map ϕ : S (S + I)/I given by ϕ(s)=s = s + I. Since ϕ can → be seen as the composition of the inclusion S S + I with the canonical projection → S + I (S + I)/I, and the composition of ring morphisms is a ring morphism by the → previous lemma, we get the ϕ is a ring morphism. Now, for any s + i (S + I)/I, we have ϕ(s)=s, but since (s + i) s = i I ∈ − ∈ we have s = s + i, and thus ϕ(s)=s + i. Hence ϕ is surjective, i.e. Im ϕ =(S + I)/I. Moreover, one has Ker ϕ = s S ϕ(s)=0 { ∈ | } = s S s =0 { ∈ | } = s S s I { ∈ | ∈ } = S I, ∩ so applying the first isomorphism theorem to the morphism ϕ we obtain the result. ￿ THEOREM 1.3.8 (Third isomorphism theorem). Let R be a ring, I J R ideals of ⊆ ￿ R, then J/I is an ideal of R/I and moreover R/I = R/J. J/I ∼

PROOF. It is immediate to check that J/I is an ideal of R/I. consider the map ϕ : R/I R/J given by ϕ(x + I):=x + J. Assume that x + I = x￿ + I, i.e. x x￿ I; as → − ∈ I J, one gets x x￿ J, so x + J = x￿ + J; and thus ϕ is well defined. The mapping ⊆ − ∈ ϕ is also obviously surjective. Now, for the kernel of ϕ we have Ker ϕ = x + I R/I ϕ(x + I)=0 { ∈ | } = x + I R/I x + J { ∈ | } = x + I R/I x J { ∈ | ∈ } = J/I, and the desired result follows from the first isomorphism theorem. ￿ COROLLARY 1.3.9 (Correspondence theorem). Let R be a ring and I ￿ R an ideal of R; the map S S/I defines a correspondence between the set of subrings of R containing ￿→ I and the set of subrings of R/I. Similarly, the map J J/I gives a correspondence ￿→ between the set of ideals of R containing I and the set of ideals of R/I.

REMARK 1.3.10. Note that in principle we cannot state that different ideals of R containing I will give rise to different ideals in the quotient ring. The only bit we can be sure of is that all ideals of R/I will be of the form J/I for some J.