9. VISITAS DESTACADAS Y ACTOS RELEVANTES
9.2. ACTOS RELEVANTES DEL CLAEX
Proof. The correspondence between subgroups described in (4.2.2) applies here. All
one has to do is verify that S is a subring (ideal) if and only if S/I is. Again we leave
it to the reader to fill in the details.
Exercises (6.2)
1. Classify each of the following subsets of a ring R as an additive subgroup, subring or ideal, as is most appropriate:
(a) {f ∈ R[t] | f (a) = 0} where R = R[t] and a ∈ R is fixed;
(b) the set of twice differentiable functions on[0, 1] which satisfy the differential equation f+ f= 0: here R is the ring of continuous functions on [0, 1];
(c) nZ where R = Z; (d) 12Z where R = Q.
(e) the set of n× n real matrices with zero first row where R = Mn(R). 2. The intersection of a set of subrings (ideals) is a subring (respectively ideal). 3. Use Exercise (6.2.2) to define the subring (respectively ideal) generated by a non- empty subset of a ring.
4. Let X be a non-empty subset of R, a commutative ring with identity. Describe the form of a typical element of: (a) the subring of R generated by X; (b) the ideal of R generated by X.
5. Let R be a ring with identity. If I is an ideal containing a unit, then I = R. 6. Let I and J be ideals of a ring R such that I ∩ J = 0. Prove that ab = 0 for all
a∈ I, b ∈ J .
7. Let a ∈ R and define θa : R[t] → R by θa(f ) = f (a). Prove that θa is a ring homomorphism. Identify Im(θa)and Ker(θa).
8. Let α : R → S be a surjective ring homomorphism and suppose that R has an identity element and S is not a zero ring. Prove that S has an identity element.
6.3
Integral domains, division rings and fields
The purpose of this section is to introduce some special types of ring with desirable properties. Specifically we are interested in rings having a satisfactory theory of division. For this reason we wish to exclude the phenomenon in which the product of two non-zero ring elements is zero.
If R is a ring, a left zero divisor is a non-zero element a such that ab= 0 for some
b = 0 in R. Of course b is called a right zero divisor. Clearly the presence of zero
Example (6.3.1) Let n be a positive integer. The zero divisors inZnare the congru- ence classes[m] where m and n are not relatively prime and 1 < m < n. Thus Znhas zero divisors if and only if n is not a prime.
For, if m and n are not relatively prime and d > 1 is a common divisor of m and
n, then[m][nd] = [md][n] = [0] since [n] = [0], while [nd] = [0]; thus [m] is a zero divisor. Conversely, suppose that[m][] = [0] where [] = [0]. Then n || m; thus, if
mand n are relatively prime, n|| and [] = [0] by Euclid’s Lemma. This contradiction shows that m and n cannot be relatively prime.
Next we introduce an important class of rings with no zero divisors. An integral
domain (or more briefly a domain) is a commutative ring with identity which has no
zero divisors. ThusZ is a domain, and by Example (6.3.1) Znis a domain if and only if n is a prime. Domains can be characterized by a cancellation property.
(6.3.1) Let R be a commutative ring with identity. Then R is a domain if and only if
the cancellation law is valid in R, i.e., ab= ac and a = 0 always imply that b = c. Proof. If ab = ac and b = c, a = 0, then a(b − c) = 0, so that a is a zero divisor
and R is not a domain. Conversely, if R is not a domain and ab = 0 with a, b = 0,
then ab= a0, so the cancellation law fails.
The next result tells us that when we are working with polynomials, life is much simpler if the coefficient ring is a domain.
(6.3.2) Let R be an integral domain and let f , g∈ R[t]. Then
deg(f g)= deg(f ) + deg(g).
Hence f g = 0 if f = 0 and g = 0, and thus R[t] is an integral domain.
Proof. If f = 0, then fg = 0 and deg(f ) = −∞ = deg(fg); hence the formula is
valid in this case. Assume that f = 0 and g = 0, and let atmand btn be the terms of highest degree in f and g respectively; here a = 0 and b = 0. Then fg = abtm+n + terms of lower degree, and ab = 0 since R is a domain. Therefore fg = 0 and
deg(f g)= m + n = deg(f ) + deg(g).
Recall that a unit in a ring with identity is an element with a multiplicative inverse. A ring with identity in which every non-zero element is a unit is called a division ring. Commutative division rings are called a fields. ClearlyQ, R and C are examples of fields, whileZ is not a field. Fields are one of the most frequently used types of ring since the ordinary operations of arithmetic can be performed in a field. Note that a
division ring cannot have zero divisors: for if ab = 0 and a = 0, then a−1ab = a−10= 0, so that b = 0. Thus two types of ring in which zero divisors cannot occur are integral domains and division rings.
6.3 Integral domains, division rings and fields 109
The ring of quaternions. The reader may have noticed that the examples of division rings given so far are commutative, i.e., they are fields. We will now describe a famous example of a non-commutative division ring, the ring of Hamilton’s1quaternions.
First of all consider the following three 2× 2 matrices over C:
I = i 0 0 −i , J = 0 1 −1 0 , K = 0 i i 0
where i = √−1. These are known in physics as the Pauli2Spin Matrices. Simple
matrix computations show that the following relations hold:
I2= J2 = K2= −1,
and
I J = K = −J I, J K = I = −KJ, KI = J = −IK.
Here 1 denotes the identity 2× 2 matrix and it should be distinguished from I. If a, b, c, d are rational numbers, we may form the matrix
a1+ bI + cJ + dK = a+ bi c + di −c + di a − bi .
This is called a rational quaternion. Let R be the set of all rational quaternions. Then
Ris a subring containing the identity of the matrix ring M2(C); for
(a1+ bI + cJ + dK) + (a1+ bI+ cJ + dK)
= (a + a)1+ (b + b)I + (c + c)J + (d + d)K,
while (a1+ bI + cJ + dK)(a1+ bI+ cJ + dK)equals
(aa− bb− cc− dd)1− (ab+ ab+ cd− cd)I
+ (ac+ ac+ bd− bd)J+ (ad+ ad+ bc− bc)K,
as may be seen by multiplying out and using the properties of I, J, K above.
The interesting fact about the ring R is that if 0= Q = a1 + bI + cJ + dK, then
Qis a non-singular matrix. For
det(Q)= a+ bi c + di −c + di a − bi
= a2+ b2+ c2+ d2= 0,
and so by elementary matrix algebra Q has as its inverse 1 det(Q) a− bi −c − di c− di a+ bi ,
which is also a quaternion. This allows us to state:
1William Rowan Hamilton (1805–1865) 2Wolfgang Ernst Pauli (1900–1958)
(6.3.3) The ring of rational quaternions is a non-commutative division ring.
Notice that the ring of quaternions is infinite. This is no accident since by a famous theorem of Wedderburn3a finite division ring is a field. This result will not be proved
here. However we will prove the corresponding statement for domains, which is much easier.
(6.3.4) A finite integral domain is a field.
Proof. Let R be a finite domain and let 0= r ∈ R; we need to show that r is a unit.
Consider the function α : R → R defined by α(x) = rx. Now α is injective since
rx = ry implies that x = y by (6.3.1). However R is a finite set, so it follows that α
must also be surjective. Hence 1= rx for some x ∈ R, and x = r−1. We shall now consider the role of ideals in commutative ring theory. First we show that the presence of proper non-zero ideals is counter-indicative for the existence of units.
(6.3.5) Let R be a commutative ring with identity. Then the set of non-units of R is
equal to the union of all the proper ideals of R.
Proof. Suppose that r is not a unit of R; then Rx = {rx | x ∈ R} is a proper ideal
containing r since 1 /∈ Rx. Conversely, if a unit r were to belong to a proper ideal I, then for any x in R we would have x = (xr−1)r ∈ I, showing that I = R. Thus no
unit can belong to a proper ideal.
Recalling that fields are exactly the commutative rings with identity in which each non-zero element is a unit, we deduce:
Corollary (6.3.6) A commutative ring with identity is a field if and only if it has no
proper non-zero ideals.
Maximal ideals and prime ideals. Let R be a commutative ring with identity. A
maximal ideal of R is a proper ideal I such that the only ideals containing I are I
itself and R. Thus maximal means maximal proper. For example, if p is a prime, pZ is a maximal ideal ofZ: for |Z/pZ| = p and the Correspondence Theorem (6.2.7) shows that no ideal can occur strictly between pZ and Z.
A related concept is that of a prime ideal. If R is a commutative ring with identity, a prime ideal of R is a proper ideal with the property: ab ∈ I implies that a ∈ I or
b∈ I, where a, b ∈ R.
There are enlightening characterizations of prime and maximal ideals in terms of quotient rings.
6.3 Integral domains, division rings and fields 111
(6.3.7) Let I be a proper ideal of a commutative ring with identity R.
(i) I is a prime ideal of R if and only if R/I is an integral domain; (ii) I is a maximal ideal of R if and only if R/I is a field.
Proof. (i) Let a, b∈ R; then ab ∈ I if and only if (a + I)(b + I) = I = 0R/I. Thus
I is prime exactly when R/I has no zero divisors, i.e., it is a domain.
(ii) By the Correspondence Theorem for ideals, I is maximal in R if and only if
R/Ihas no proper non-zero ideals; by (6.3.6) this property is equivalent to R/I being
a field.
Since every field is a domain, we deduce:
Corollary (6.3.8) Every maximal ideal of a commutative ring with identity is a prime
ideal.
On the other hand, prime ideals need not be maximal. Indeed, if R is any domain, the zero ideal is certainly prime, but it will not be maximal unless R is a field. More interesting examples of non-maximal prime ideals can be constructed in polynomial rings.
Example (6.3.2) Let R = Q[t1, t2], the ring of polynomials in t1, t2 with rational
coefficients. Let I be the subset of all polynomials in R which are multiples of t1.
Then I is a prime ideal of R, but it is not maximal.
For consider the function α : R → Q[t2] which carries a polynomial f (t1, t2)to
f (0, t2). Then α is a surjective ring homomorphism. Now if f (0, t2)= 0, then f is
a multiple of t1, and hence the kernel is I . By (6.2.4) R/I Q[t2]. Since Q[t2] is a
domain but not a field, it follows from (6.3.7) that I is a prime ideal of R which is not maximal.
The characteristic of an integral domain. Suppose that R is a domain; we wish to consider S= 1, the additive subgroup of R generated by 1. Suppose for the moment that S is finite, with order n say. Then we claim that n must be a prime. For suppose that n= n1n2where ni∈ Z and 1 < ni< n. Then 0= n1 = (n1n2)1= (n11)(n21)
by (6.1.3). However R is a domain, so n11 = 0 or n21 = 0, which shows that n
divides n1or n2, a contradiction. Therefore n is a prime.
This observation is the essence of:
(6.3.9) Let R be an integral domain and put S= 1. Then either S is infinite or else
it has prime order p. In the latter event pa= 0 for all a ∈ R.
To prove the last statement, simply note that pa= (p1R)a= 0a = 0.
Again let R be an integral domain. If1R has prime order p, then R is said to have characteristic p. The other possibility is that1R is infinite, in which case R is said to have characteristic 0. Thus the characteristic of R,
is either 0 or a prime. For example,Zp andZp[t] are domains with characteristic p, whileQ, R and R[t] all have characteristic 0.
The field of fractions of an integral domain. Suppose that F is a field and R is a subring of F containing 1F. Then R is a domain since there cannot be zero divisors in F . Conversely, one may ask if every domain arises in this way as a subring of a field. We shall answer the question positively by showing how to construct the field
of fractions of a domain. It will be helpful for the reader to keep in mind that the
procedure to be described is a generalization of the way in which the rational numbers are constructed from the integers.
Let R be any integral domain. We have first to decide how to define a fraction over R. Consider the set
S= {(a, b) | a, b ∈ R, b = 0}.
Here a will correspond to the numerator and b to the denominator of the fraction. We introduce a binary relation ∼ on S which will allow for cancellation between numerator and denominator:
(a1, b1)∼ (a2, b2)⇔ a1b2= a2b1.
Of course this is motivated by a familiar arithmetic rule: m1
n1 =
m2
n2 if and only if
m1n2= m2n1.
Next we verify that∼ is an equivalence relation on S. Only transitivity requires a comment: suppose that (a1, b1)∼ (a2, b2)and (a2, b2)∼ (a3, b3); then a1b2= a2b1
and a2b3 = a3b2. Multiply the first equation by b3 and use the second equation
to derive a1b3b2 = a2b3b1 = a3b2b1. Cancel b2 to obtain a1b3 = a3b1; thus
(a1, b1)∼ (a3, b3). Now define a fraction over R to be a∼-equivalence class
a
b = [(a, b)]
where a, b ∈ R, b = 0. Note that acbc = ab since (a, b)∼ (ac, bc); thus cancellation can be performed in a fraction.
Let F denote the set of all fractions over R. We would like to make F into a ring. To this end define addition and multiplication in R by the rules
a b+ a b = ab+ ab bb and a b a b = aa bb.
Here again we have been guided by the arithmetic rules for adding and multiplying fractions. It is necessary to show that these operations are well-defined, i.e., there is no dependence on the chosen representative (a, b) of the equivalent classab. For example, let (a, b)∼ (c, d) and (a, b)∼ (c, d). Then (ab+ ab, bb)∼ (cd+ cd, dd); for
6.3 Integral domains, division rings and fields 113 The next step is to verify the ring axioms; as an example let us check the validity of the distributive law
a b+ c d e f =a b e f +c d e f ,
leaving the reader to verify the other axioms. By definition a b e f +c d e f = ae bf + ce df = aedf + cebf bd f2 = ade+ bce bd f , which equals ad + bc bd e f =a b + c d e f , as claimed.
After all the axioms have been checked we will know that F is a ring; note that the zero element of F is 0F = 01RR. Clearly F is commutative and it has as its identity element 1F = 11RR. Furthermore, if a, b= 0, a b b a = ab ab = 1R 1R = 1F ,
so that, as expected, the inverse of ab is ab. Therefore F is a field, the field of fractions
of the domain R.
In order to relate F to R we introduce a natural function
θ : R → F
defined by θ (a)= a1. It is quite simple to check that θ is an injective ring homomor- phism. Therefore R Im(θ) and of course Im(θ) is a subring of F containing 1F. Thus the original domain is isomorphic with a subring of a field. Our conclusions are summed up in the following result.
(6.3.10) Let R be an integral domain and F the set of all fractions over R with the
addition and multiplication specified above. Then F is a field and the assignment a→ a1 determines is an injective ring homomorphism from R to F .
Example (6.3.3) (i) When R = Z, the field of fractions is, up to isomorphism, the
field of rational numbersQ. This example motivated the general construction. (ii) Let K be any field and put R = K[t]; this is a domain by (6.3.2). The field of fractions F of R is the field of rational functions in t over K; these are formally quotients of polynomials in t over K
f g
where f, g∈ R, g = 0. The notation
K{t}
Exercises (6.3)
1. Find all the zero divisors in the following rings: Z6,Z15,Z2[t], Z4[t], Mn(R). 2. Let R be a commutative ring with identity such that the degree formula deg(f g)= deg(f )+ deg(g) is valid in R[t]. Prove that R is a domain.
3. A left ideal of a ring R is an additive subgroup L such that ra∈ L whenever r ∈ R and a ∈ L. Right ideals are defined similarly. If R is a division ring, prove that the only left ideals and right ideals are 0 and R.
4. Let R be a ring with identity. If R has no left or right ideals except 0 and R, then
Ris a division ring.
5. Let θ : D → R be a non-zero ring homomorphism. If D is a division ring, show that it must be isomorphic with some subring of R.
6. Let I and J be non-zero ideals of a domain. Prove that I ∩ J = 0.
7. Let I be the principal ideal (Z[t])t of Z[t]. Prove that I is prime but not maximal. 8. The same problem for I = (Z[t])(t2− 2).
9. Let F be a field. If a, b∈ F and a = 0, define a function θa,b : F → F by the rule
θa,b(x)= ax + b. Prove that the set of all θa,b’s is a group with respect to functional composition.
10. Let F be the field of fractions of a domain R and let α: R → F be the canonical injective homomorphism r → 1r. Suppose that β : R → K is an injective ring ho- momorphism into some other field K. Prove that there is an injective homomorphism
θ : F → K such that θα = β. (So in a sense F is the smallest field containing an
Chapter 7
Division in rings
Our aim in this chapter is to construct a theory of division in rings that mirrors the familiar theory of division in the ring of integersZ. To simplify matters let us agree to restrict attention to commutative rings – in non-commutative rings questions of left and right divisibility arise. Also, remembering from 6.3 the unpleasant phenomenon of zero divisors, we shall further restrict ourselves to integral domains. In fact even this class of rings is too wide, although it provides a reasonable target for our theory. For this reason we will introduce some well-behaved types of domains.
7.1
Euclidean domains
Let R be a commutative ring with identity and let a, b∈ R. Then a is said to divide
b in R, in symbols
a|| b,
if ac= b for some c ∈ R. From the definition there follow very easily some elementary facts about division.
(7.1.1) Let R be a commutative ring with identity, and let a, b, c, x, y be elements of
R. Then
(i) a|| a and a || 0 for all a ∈ R; (ii) 0|| a if and only if a = 0;
(iii) if a|| b and b || c, then a || c, (so division is a transitive relation); (iv) if a|| b and a || c, then a || bx + cy for all x, y;
(v) if u is a unit, then u|| a for all a ∈ R;
(vi) if u is a unit, then a|| u if and only if a is a unit.
For example, taking the case of (iv), we have b = ad and c = ae for some
d, e∈ R. Then bx + cy = a(dx + ey), so that a divides bx + cy. The other proofs
are equally simple exercises that are left to the reader.
One situation we must expect to encounter is a pair of ring elements each of which divides the other: such elements are called associates.
(7.1.2) Let R be an integral domain and let a, b∈ R. Then a || b and b || a if and only
if b = au where u is a unit of R.
Proof. In the first place, a|| au and if u is a unit, then a = (au)u−1, so au|| a. Conversely, assume that a|| b and b || a. If a = 0, then b = 0, and the statement is certainly true. So let a = 0. Now a = bc and b = ad for some c, d ∈ R. Therefore
a= bc = adc, and by the cancellation law (6.3.1), dc = 1, so that d is a unit.
Thus two elements of a domain are associates precisely when one is a unit times the other. For example, two integers a and b are associates if and only if b= ±a.
Irreducible elements. Let R be a commutative ring with identity. An element a of R is called irreducible if it is neither 0 nor a unit and if its only divisors are units and associates of a, i.e., the elements that we know must divide a. Thus irreducible elements have as few divisors as possible.
Example (7.1.1) (i) The irreducible elements ofZ are the prime numbers and their
negatives.
(ii) A field has no irreducible elements since every non-zero element is a unit. (iii) If F is a field, the irreducible elements of the polynomial ring F[t] are the so-called irreducible polynomials, i.e., the non-constant polynomials which are not expressible as a product of polynomials of lower degree.
Almost every property of division in the integers depends ultimately on the Division Algorithm. Thus it will be difficult to develop a satisfactory theory of division in a ring unless some version of this algorithm is valid for the ring. This motivates us to introduce a special class of domains, the so-called Euclidean domains.
A domain R is called Euclidean if there is a function
δ: R − {0R} → N with the following properties:
(i) δ(a)≤ δ(ab) if 0 = a, b ∈ R;
(ii) if a, b∈ R and b = 0, there exist q, r ∈ R such that a = bq + r where either
r= 0 or δ(r) < δ(b).
The standard example of a Euclidean domain isZ where δ the absolute value function, i.e., δ(a)= |a|. Note that property (i) holds since |ab| = |a| · |b| ≥ |a| if b = 0. Of course (ii) is just the usual statement of the Division Algorithm forZ.
New and important examples of Euclidean domains are given by the next result.
(7.1.3) If F is a field, the polynomial ring F[t] is a Euclidean domain where the
7.1 Euclidean domains 117
Proof. We already know from (6.3.2) that R = F [t] is a domain. Also, by the same
result, if f, g = 0, then δ(fg) = deg(fg) = deg(f ) + deg(g) ≥ deg(f ) = δ(f ). Hence property (i) is valid. To prove the validity of (ii), put S = {f − gq | q ∈ R}. If 0∈ S, then f = gq for some q, and we may take r to be 0. Assuming that 0 /∈ S, we note that every element of S has degree≥ 0, so by the Well-Ordering Principle there exists r ∈ S with smallest degree, say r = f − gq where q ∈ R. Thus f = gq + r.
Now suppose that deg(r) ≥ deg(g). Write g = atm+ · · · and r = btn + · · · where m= deg(g), n = deg(r), 0 = a, b ∈ F and dots indicate terms of lower degree in t. Since m≤ n, we may form the polynomial
s = r − (a−1btn−m)g∈ R.
Now the term btncancels in s, so either s= 0 or deg(s) < n. But
s= f − (q + a−1btn−m)g,
so that s∈ S and hence s = 0, which contradicts the choice of r. Therefore deg(r) <
deg(g), as required.
A less obvious example of a Euclidean domain is the ring of Gaussian integers. A
Gaussian integer is a complex number of the form u+ iv
where u, v ∈ Z and of course i =√−1. It is easily seen that the Gaussian integers form a subring ofC containing 1, and hence constitute a domain.
Example (7.1.2) The ring R of Gaussian integers is a Euclidean domain.
Proof. In this case we define a function δ : R − {0} → N by the rule δ(u+ iv) = |u + iv|2= u2+ v2.
We argue that δ satisfies the two requirements for a Euclidean domain. In the first place, if 0= a, b ∈ R, then δ(ab) = |ab|2 = |a|2|b|2≥ |a|2since|b| ≥ 1.
Verifying the second property is a little harder. First write ab−1 = u+ ivwhere
u, v are rational numbers. Now choose integers u and v as close as possible to u and vrespectively; then|u − u| ≤ 12and|v − v| ≤ 12. Next
a= b(u+ iv)= b(u + iv) + b(u+ iv)
where u = u− u and v = v− v. Finally, let q = u + iv and r = b(u+ iv). Then a= bq + r; also q ∈ R and hence r = a − bq ∈ R. If r = 0, then
δ(r)= |b|2|u+ iv|2= |b|2(u2+ v2)≤ |b|2 1 4+ 1 4 = 1 2|b| 2<|b|2= δ(b)