Integral Domains: ED, PID and UFDs

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1 CHAPTER 2 Integral Domains: ED, PID and UFDs 2.1. Domains In this whole chapter, R will denote a commutative ring. We will write R := R \{0} to refer to the set of all nonzero elements of R. DEFINITION Let R be a ring, a R. We say that a is a unit if there is some b R such that ab =1=ba. When such an element b exists, it is called the inverse of a, and denoted by b = a 1. The set of all units of the ring R is denoted by U(R) EXERCISE Show that U(R) is a multiplicative group. DEFINITION Let R be a nontrivial ring. We say that R is a field if every nonzero element of R is a unit, i.e. if for each a R there is some b R such that ab = ba =1. REMARK For historical reasons, normally one only considers commutative fields. There are examples of noncommutative rings such that every nonzero element is invertible, for example this happens for the quaternion ring. Some books call those rings noncommutative fields, but it is much more common to find them under the denomination of division rings. EXAMPLES i) Q, R and C are all fields. ii) Z p is a field whenever p is a prime number. iii) R(x) = f(x) g(x), f,g R[x],g = 0 the field of rational functions. DEFINITION Let R be a nontrivial ring. We say that R is an integral domain (ID for short) if whenever a, b R are such that ab =0, then necessarily a =0or b =0. In other words, if whenever a, b = 0then one has ab = 0. EXAMPLES i) The ring of integers Z. ii) Every field F is an integral domain. iii) If R is an ID, and S R is a subring of R, then S is also an ID. In particular, any subring of a field is an integral domain. iv) If R is an integral domain, then the polynomial ring R[x] is also an integral domain. Integral domains have very nice arithmetical properties similar to some well known ones that we find in the ring of integers. An example is the following: PROPOSITION (Cancellation law). Let R be an ID, and let a, b, c R such that ab = ac. If a = 0, one has b = c. PROOF. From ab = ac one gets 0=ab ac = a(b c); as a = 0it must be b c =0, and thus b = c. We know that every subring of a field is an integral domain. Actually, the converse is also true, i.e. for any integral domain R we can find a field Q such that R is a subring of Q. THEOREM (Field of fractions). Let R be an integral domain, then there is a field Q satisfying the following properties: 9

2 10 2. INTEGRAL DOMAINS: ED, PID AND UFDS (1) R Q subring, (2) Every q Q can be written as q = ab 1 for some a, b R, b = 0. The field Q is unique (up to isomorphism) and receives the name of field of fractions (or field of quotients) of R. PROOF. The proof is constructive, giving an explicit description of the field Q. Actually, the construction mimics the procedure of constructing the rational numbers from integers. More precisely, we define Q as where we define the equivalence relation Q := {(a, b) a R, b R } /, (a, b) (c, d) ad = bc. If the equivalence class of a pair (a, b) under the above equivalence relation is denoted by (a, b), we define the operations in Q as With these operations, Q is a field. 1 (a, b)+(c, d) :=(ad + bc, bd) (a, b)(c, d) :=(ac, bd). DEFINITION Let R be a nonzero ring. We say that R is simple if its only ideals are the zero ideal {0} and the total ideal R. PROPOSITION A (commutative) ring R is simple if and only if it is a field. PROOF. Assume that R is simple, and let a R. Since R is simple, one gets (a) =R, thus 1 (a), and hence there must exist some b R such that 1=ab, and hence a U(R); thus, every nonzero element of R is a unit. Conversely, if R is a field, and 0 = I R, let a I a nonzero element of I. Since R is a field, a must be a unit, so there is b R such that ab =1, but since a I, the absorbency property ensures that 1=ab I, and thus I = R, so R is simple. DEFINITION Let R be a ring and I R an ideal. We say that I is a maximal ideal if I = R is a proper ideal and whenever I J R one has J = I or J = R. In other words, I is maximal, with respect to inclusion, among proper ideals, that is, there is no proper ideal J such that I J. PROPOSITION Let R be a ring and I R an ideal. The quotient ring R/I is a field if and only if I is maximal. PROOF. R/I is a field if and only if it is simple, if and only if R/I = 0and for all K R/I, either K =0or K = R/I. By the correspondence theorem, each ideal K of R/I must be of the form J/I for I J R, so R/I is a field if and only if J/I =0or J/I = R/I for all J I, or equivalently if and only if for all J containing I, either J = I or J = R, i.e. if and only if I is maximal. DEFINITION An ideal I R is called a prime ideal if it is a proper ideal of R such that whenever ab I then either a I or b I. PROPOSITION Let I R for a ring R, then R/I is an integral domain if and only if I is a prime ideal. PROOF. R/I ID if and only if R/I = 0and for all a, b R/I such that ab =0then either a =0or b =0. This latest conditions are equivalent to I = R and whenever ab I, then either a I or b I, i.e. if and only if I is a prime ideal. 1 I will include a fully detailed proof of this theorem in the way it was stated on the exercise sheet. Lecture notes will be upgraded after Reading Week to include that detailed proof.

3 2.3. PRIMES AND IRREDUCIBLES 11 COROLLARY Every maximal ideal is a prime ideal. PROOF. If I maximal, then R/I is a field, in particular R/I is an ID, and thus I is prime Ideals and divisibility DEFINITION Let R a ring, a, b R. We say that a divides b, or that b is a multpiple of a, or that b is divisible by a, and write a b if there is some r R such that b = ar. Divisibility can be nicely described in terms of ideals, since a b if and only if b = ar for some r R, if and only if b (a), if and only if (b) (a). DEFINITION Let a, b R for some ring R, we say that b is associated to a, or that a and b are associates if there is some unit u U(R) such that b = ua. If a and b are associates, we will write a b. PROPOSITION Let R be an integral domain, a, b R. The following properties hold: (1) a b if and only if a b and b a, if and only if (a) =(b). (2) a 1 if and only if a U(R), if and only if (a) = (1) = R. (3) a 0 if and only if a =0, if and only if (a) = (0) = 0. (4) Being associates is an equivalence relation on R. PROOF. (1). If a b then b = au for some u U(R), so a b, but also a = bu 1, so b a. If a b, b a there are some c, d R such that b = ac, a = bd, and thus one gets b = ac = bdc. By the cancellative law it follows that cd =1, so both c, d U(R) and we get a b. Also, a b and b a if and only if (a) (b) and (b) (a), if and only if (a) =(b). (2). If a 1 then a = u1 =u U(R). Conversely, if a U(r), as a = a1 we get a 1. (3). a 0 if and only if (a) =(0), if and only if a =0. (4). By (1), a b if and only if (a) =(b), so being associates is an equivalence relation. EXAMPLES (1) U(Q) =Q, or more generally for any field F one has U(F) =F, and the only classes of associates are {0} and F. (2) Let R be an ID, then U(R[x]) = U(R). So for instance (x + 1) 2(x + 1) in Q[x], but not in Z[x], as U(Z[x]) = U(Z) ={ 1, 1}. (3) U(Z n )={a {1,...,n 1} gcd(a, n) =1} Primes and irreducibles DEFINITION Let R be a (commutative) ID. An element a R is said to be a prime, or a prime element of R if: (1) a = 0, a/ U(R), i.e. a is neither 0 nor a unit of R. (2) Whenever b, c R are such that a bc, then either a b or a c. The above conditions can be rewritten in terms of the principal ideal generated by a as follows: (1 ) (0) (a) R (2 ) If bc (a), then either b (a) or c (a). In other words, for a R, we have that a is prime if and only if the principal ideal (a) is a prime ideal. DEFINITION Let R be an ID, a R. We say that b R is a proper divisor of a if b a and b is neither a unit or an associate of a, i.e. if (a) (b) R.

4 12 2. INTEGRAL DOMAINS: ED, PID AND UFDS DEFINITION Let R be an ID. We say that an element a R is an irreducible element (also called an atom) if (1) a = 0, a/ U(R), (2) a has no proper divisors, i.e. if b a then either b U(R) or b a. In terms of ideals, a is irreducible if whenever (a) (b) then either (a) =(b) or (b) =R. In other words, a is irreducible if the principal ideal (a) is maximal among proper principal ideals. EXAMPLES (1) 2 and 3 are irreducible in Z (actually, in Z irreducible elements an primes are the same thing). (2) x 2 +1 is irreducible in R[x], but not in C[x], as one has (x 2 +1) = (x i)(x+i). (3) 6 is not irreducible in Z, as 6=2 3. The notion of irreducibility admits several equivalent descriptions: PROPOSITION Let a R a nonzero, nonunit element of an ID R. The following are equivalent: (1) a is irreducible. (2) If a = bc for some b, c R, then either b U(R) or c U(R). (3) If a = bc for some b, c R, then either b a or c a. The proof is completely straightforward. PROPOSITION Let R be an integral domain. Then every prime element a R is also irreducible. PROOF. Let R be a prime, and suppose a = bc, then one has b a, c a. As a bc and a is prime, then either a b or a c, yielding either a b or a c Principal ideal domains DEFINITION Let R be an ID. One says that R is a principal ideal domain (PID for short) if every ideal is a principal ideal, i.e. for any I R there is some a R such that I =(a). REMARK. If R is a PID and I =(a), then the element a is unique up to associates, as for any integral domain one has (a) =(b) if and only if a b. EXAMPLES (1) Let F be a field. If I R ideal, then either I = 0 = (0) or I = R = (1), so F is a principal ideal domain. (2) Let I Z, as Z is a cyclic group, every additive subgroup is also cyclic, and thus I =(n) for some n; hence, Z is a PID. PROPOSITION Let R be a PID, then every irreducible a R is also prime. PROOF. Let a R be irreducible, we will show that a is prime by showing the ideal (a) is a maximal ideal, and thus a prime ideal. Let I be an ideal such that (a) I. As R is a PID, it must be I =(b) for some b R, thus we have a (b) and hence a = bc for some c R. As a is irreducible, either b U(R), and hence (b) =R, or c U(R), in which case a b and thus (a) =(b) =I. So, (a) is a maximal ideal and by Corollary it follows that (a) is a prime ideal, and hence a is prime. COROLLARY In a PID the notions of prime and irreducible are equivalent.

5 2.4. PRINCIPAL IDEAL DOMAINS 13 Note that from the proof of Proposition it follows that if R is a PID and a R is irreducible, then (a) is a maximal ideal and thus R/(a) is a field. We know already that Z is a PID, and we will show later in this chapter that for any field F the polynomial ring F[x] is also a PID. As we will see, the technique that proves F[x] is PID is very similar to the one that allow us to prove that Z is a PID: There is some kind of norm or measure of how big nonzero elements are. There is a division with remainder with the remainder being strictly smaller than the divisor. These properties can be axiomatized, leading to a family of particularly nice rings. DEFINITION An Euclidean domain (ED for short) is an ID R endowed with a map N : R N such that ED1 For a, b R, if a b then N(a) N(b). ED2 For any a, b R, with b = 0, there exist q, r R such that a = bq + r and either r =0or N(r) <N(a). REMARK If (R, N) is a ED, since for all a R one has 1 a, one must have N(1) N(a). In other words, the norm of the unit element is the smallest amont the norms of all the elements in the ring. EXAMPLES (1) Z, with the map N defined as N(a) := a for all a = 0. The division is the usual one. (2) For any field F, the polynomial ring F[x] is and ED, with norm map N(f) := deg(f). Here we use the usual division of polynomials. (3) The ring of Gaussian integers Z[i] ={a + bi a, b Z}, with norm N(z) = zz = a 2 + b 2 for any z = a + bi Z[i], is an ED. PROOF. The ring Z[i] is an integral domain because we have Z[i] C, and C is a field. ED1. If z w, then w = zt, and then N(w) =N(zt) =ztzt = zztt = N(z)N(t), thus N(z) N(w). ED2. Let z,w Z[i], with w = 0. As Z[i] Q(i) and Q(i) is a field we can take zw 1 = a+bi in Q(i). Pick now integers u, v Z such that a u 1/2, b v 1/2 (u and v are the rounding of a and b, respectively); the element q = u + vi belongs to Z[i]. Now let s =(a u)+(b v)i Q(i), and define r := sw. Note that qw + r = qw + sw =(q + s)w =(a + bi)w = zw 1 w = z Z[i], and thus r = z qw Z[i]. Finally observe that N(r) = N(s)N(w) =ssn(w) =((a u) 2 +(b v) 2 )N(w) N(w) <N(w), 4 so q and r satisfy the required properties. PROPOSITION If R is an ID with a map N : R N satisfying ED2, then R is a PID. In particular, every Euclidean Domain is a PID. PROOF. Let I R be a (nonzero) ideal. Then there is some a I, a = 0, such that N(a) is minimal. Let b I; by ED2, there are some q, r R such that b = aq + r and wither r =0or N(r) <N(a), but r = b qa I (as both a, b I) and since N(a) is minimal, we must have r =0, and hence b = qa is a multiple of a, henceforth I =(a).

6 14 2. INTEGRAL DOMAINS: ED, PID AND UFDS COROLLARY The rings Z, F[x], Z[i] are all PDIs. PROPOSITION Let R be an ED, a R, then a U(R) if and only if N(a) = N(1). PROOF. Let a U(R), then there is some b R such that ab =1, thus a 1 and by ED1, N(a) N(1). But one always has 1 a and thus N(1) N(a). Henceforth N(a) =N(1). Using the division algorithm, write 1=aq + r. If it were r = 0one would have N(r) <N(a) =N(1) but that is not possible, since 1 r and thus N(1) N(r); thus, it must be r =0, and hence 1=aq, implying a U(R). EXAMPLES (1) U(Z) ={u Z u =1} = {1, 1}, (2) U(F[x]) = {f F[x] deg f =deg1=0} = F, (3) U(Z[i]) = a + bi Z[i] a 2 + b 2 =1 = {1, 1,i, i} Unique Factorization Domains DEFINITION An ID R is said to be a unique factorization domain (UFD for short) if every nonzero, nonunit element a R \ U(R) can be written as a product of irreducible elements a = p 1 p s, and such a expression is unique up to reordering of the factors and associates. PROPOSITION Let R be an ID. The following are equivalent: (1) R is a UFD. (2) Every a R \ U(R) can be written as a product of primes. (3) Every irreducible in R is prime, and every a R \ U(R) can be written as a product of irreducibles, PROOF. 1 3 By definition of UFD, every a R \ U(R) can be written as a product of irreducibles. Let a R be an irreducible element, and suppose that a bc, then there is some d R such that ad = bc. If bc =0then either b =0or c =0, and trivially a 0, so either a b or a c. So let us assume bc = 0. If either b or c are units, we immediately get that a divides the other one, so we can also assume b, c nonunits. Write b = b i, c = c j, d = d k where b i,c j,d k irreducibles. Then we get ad 1 d r = b 1 b s c 1 c t. Since R is a UFD, it must be a b i or a c j for some i or j; in the first case, a divides b, in the second a divides c, thus a is prime. 3 2 Obvious. 2 1 Let a R \ U(R). By (2), a = p 1 p r for p i primes. Every prime is also irreducible, so a is a product of irreducibles. Suppose p 1 p r = q i q s for some irreducibles q 1,...,q s. We will show by induction that r = s, and after some relabelling we get p i q i. For r =1, one has a = p 1 irreducible; if p 1 = q 1 q s, as p 1 is irreducible we must have s =1and q 1 = p 1. Assume now (inductive hypothesis) the property holds for r 1, with r>1, and any s, and let a = p 1 p r = q 1 q s. One has p r q 1 q s, and as p r is prime then p r divides some q j. After relabelling we can assume p r q s ; then, as q s is irreducible, it must be q s = up r for some u U(R), i.e. p r q s, and hence p 1 p r = q 1 q s = q 1 q s 1 p r u.

7 2.6. CHAIN CONDITIONS 15 Applying the cancellation law, one gets p 1 p r 1 = q 1 q s 1 u, and the result follows by the inductive hypothesis Chain conditions Our goal now is to show that every PID is also a UFD. We have already shown that in a PID every ireducible is prime, so we only have to check that every element a R \ U(R) in a PID can be written as a product of irreducibles. DEFINITION A ring R is said to satisfy the ascending chain condition on principal ideals (ACC on principal ideals) if whenever we have an ascending chain of principal ideals I 1 I 2 I n I n+1 then there is some N N such that I n = I N for all n N. REMARK For some rings, the ascending chain condition can be satisfied for all ideals, not just principal ones. When this happens, we say that the ring is noetherian. Noetherian rings are an important topic in other areas of mathematics such as Number Theory and Algebraic Geometry. PROPOSITION Let R be a ring satisfying the ACC on (principal) ideals, and let S be a nonempty family of (principal) ideals of R, then S has a maximal element, i.e. there exists some J Ssuch that for all I Ssatisfying J I one has I = J. PROOF. By contradiction, assume that S does not have such a maximal element. As S is not empty, let I 1 S. By our assumption, I 1 is not maximal in S, so there is some I 2 Ssuch that I 1 I 2. Again, I 2 cannot be maximal in S so there must exist I 3 S such that I 1 I 2 I 3. By proceeding along the same lines, we end up providing an infinite chain I 1 I 2 I n of (principal) ideals of R, contradicting the ACC for (principal) ideals. Thus, if R satisfies ACC, then S must have a maximal element. REMARK The converse of the result is also true (the proof is an easy exercise). We have only stated the result for principal ideals, but the same proof works verbatim for any family of ideals if R is noetherian. EXAMPLE Let R be a UFD, a R \ U(R), and write a = a 1 a s for some irreducible elements a i R. Suppose that (a) (b), then b a = a 1 a s. As R is a UFD, it must be b a i1 a it for some 1 i 1 < <i t s, thus there are only a finite number of principal ideals containing (a), namely the ones generated by some product of the irreducibles appearing in the factorization of a. Thus, every UFD satisfies the ACC on principal ideals. PROPOSITION Let R be an ID. If R satisfies the ACC on principal ideals, then every nonzero, nonunit element of R can be written as a product of irreducibles. PROOF. Suppose there is an element a R \ U(R) which is not a product of irreducibles, then the family S := {(a) a is not a product of irreducibles} is non-empty and since R satisfies ACC for principal ideals, by Proposition we can pick a such that (a) is maximal in S. In particular, a cannot be irreducible, so there are some b, c R \ U(R) such that a = bc, where b, c are proper divisors of a, and thus one gets (a) (b) R and (a) (c) R. Since (a) is maximal in S, bith b and c must be written as a product of irreducibles, b = b 1 b s, c = c 1 c s, but then we get a = b 1 b s c 1 c t product of irreducibles, which is a contradiction.

8 16 2. INTEGRAL DOMAINS: ED, PID AND UFDS PROPOSITION Any PID satisfies the ACC on (principal) ideals. PROOF. Let I 1 I 2 an ascending chain of ideals of R. Consider I = n 0 I n. We claim that I R is an ideal of R: 0 I 1 I, hence 0 I and I1 is satisfied. Let a, b I, then a I r, b I s for some r, s, and then a, b I max{r,s}, hence a + b I max{r,s} I, thus I is an additive subgroup of R. Let a I, r R, then a I n for some n, and since I n is an ideal, ar I n I, hence the absorbency property is also satisfied, and thus I is an ideal of R. Now, since R is a PID, the ideal I must be principal, i.e. there must exist some a R such that I =(a). But then a I = I n, so there must be some N N such that a I N, and hence I N (a) =I, and it follows I N = I. Consequently, for all n N one has (a) I N I n I =(a), and thus I n = I N for all n N, so R satisfies ACC on (principal) ideals. COROLLARY ED PID UFD GCD, LCM and factorization DEFINITION Let R be a UFD, and let a, b R. An element d R is said to be a greatest common divisor (gcd for short) of a and b if the following conditions are satisfied: (1) d a, d b, (2) For any e R such that e a and e b one has e d. These conditions can be restated in terms of ideals as follows: (1 ) (a) (d), (b) (d), (2 ) For any e R such that (a) (e) and (b) (e) one has (d) (e). Note that if both d and d are gcd s of a and b, by the second property one gets (d) (d ) and (d ) (d), thus (d) =(d ) and thus d d. Henceforth, the greatest common divisor (if it exists at all!) is unique up to associates. REMARK. In some particular UFDs we can make canonical choices for gcd s. For instance in the ring of integers Z we can always pick the positive gcd, or in a polynomial ring F[x] with coefficients in a field we can choose the monic one. For more general UFDs, there is no way of making a canonical choice. EXAMPLES (1) If a =0, then for any b R one has gcd(a, b) =b. (2) If a U(R), for any b R one has gcd(a, b) =1 a. (3) If R is a UFD, and a, b R \ U(R), then we can write a = up α1 1 pαr r and b = vp β1 1 pβr r where u, v U(R) are units, p i R are distinct primes, and α i,β i 0. In this case, the element d = p γ1 1 pγr r is a greatest common divisor of a and b. In particular, gcd s always exist in a UFD. (4) Let R be a PID (and hence a UFD), a, b R. The ideal (a) +(b) ={ra + sb r, s R} must be principal, so there must exist d R such that (a) + (b) =(d). Then d is a greatest common divisor of a and b, and moreover, since d (d) =(a)+(b) there must exist h, k R such that d = ha + kb. In other words, we have a Bézout s identity for gcd s in a PID.

9 2.8. PROPERTIES OF POLYNOMIAL RINGS OVER DOMAINS 17 (5) Let R be an ED with norm map N (in particular, R is a PID); then we know that gcd(a, b) exists for all a and b and can be written as d = ah + bk for some h, k R. In this case, both d, h and k can be explicitly computed by repeated use of the division algorithm. The resulting algorithm is exactly the same as the well-known Euclidean algorithm for computing the gcd of integers. DEFINITION Let R be an ID, a, b R. An element e R is a least common multiple (lcm) of a and b if the following properties are satisfied: (1) a e, b e, (2) For any f R such that a f and b f one has e f. Or in terms of ideals: (1 ) (e) (a), (e) (b), (2 ) For any f R such that (f) (a) and (f) (b) one has (f) (e). PROPOSITION (1) If a =0, then for any b R one has lcm(a, b) =0. (2) If a U(R), then for any b R one has lcm(a, b) =b. (3) If R is a UFD, a, b R \U(R) elements that can be written as a = up α1 1 pαt t and b = vp β1 1 pβt t, where u, v U(R) are units, p i R are distinct primes, and α i,β i 0, and denote δ i := max{α i,β i }. Then lcm(a, b) =p γ1 1 pγt t. (4) If R is a UFD, a, b R, then (a) (b) =(lcm(a, b)). DEFINITION Let R be a UFD, we say that elements a, b R are coprime if gcd(a, b) =1, i.e. if (a)+(b) =R. PROPOSITION Let R be a UFD, r R. The one has gcd(ra 1,...,ra k )= r gcd(a 1,...,a k ). In particular, if d = gcd(a 1,...,a k ) then one has gcd( a1 d,..., a k d )= Properties of polynomial rings over domains We already know some properties of the ring of polynomials for certain types of rings of coeeficients. For instance: (1) If F is a field, then F[x] is NOT a field. In fact, the polynomial ring R[x] is never a field. (2) If R is an ID, then R[x] is also an ID. (3) If F is a field, then F[x] is a ED, and hece a PID. Our goal in this section is to show that R is a UFD if and only if R[x] is also a UFD. The only if part is fairly trivial, but the if part, i.e. showing that if R is a UFD then R[x] is also a UFD, is trickier. The main idea that we will use for this proof will be to go from the ring R[x] to the ring Q[x], where Q is the field of fractions of R, and then take advantage of the fact that Q[x] is a PID and in particular a UFD. Note that a polynomial f R[x] might be irreducible in Q[x] but not in R[x], for instance f(x) =2x 2 +4is irreducible in Q[x], but in Z[x] we have f(x) = 2(x 2 + 2). This happens because 2 is a unit in Q but not in Z. To look at irreducibles f R[x] we will need to chack that f does not have a non-unit constant factor in R. DEFINITION Let R be a UFD, 0 = f = a 0 + a 1 x + + a n x n R[x]. The polynomial f is said to be primitive if gcd(a 0,...,a n )=1, i.e. if there isn t any prime p R such that p a i for all i =0,...,n. EXAMPLES (1) If f = x n + a n 1 x n a 0 is monic, then f is also primitive. (2) The polynomial 3+4x+2x 2 is primitive in Z[x], but 4+4x 2x 2 +10x 3 +20x 4 is not, since gcd(4, 4, 2, 10, 20) = 2.

10 18 2. INTEGRAL DOMAINS: ED, PID AND UFDS (3) If f R[x] is irreducible, then f is primitive. LEMMA Let R be a UFD with field of fractions Q = a b a, b R, b = 0, and let f Q[x] a non-zero polynomial. The f can be written as f = λ f where λ Q and f R[x] is primitive. Moreover λ and f are unique up to multiplication by units in R. PROOF. Let f = a0 b an b n x n Q[x]. Take r = b 0 b n R, and a i = a ir/b i. Let d = gcd(a 0,...,a n), and denote c i := a i /d R. We then have f = d r (c c n x n ) Obviously d/r Q. By Proposition 2.7.6, gcd(c 0,...,c n )=1, thus f := c 0 + +c n x n is primitive. Assume now that f = λ f = µ g where f and g are primitive. Write λ = a/b, µ = c/d, where a, b, c, d R, and let f = a 0 + +a n x n, g = b 0 + +b n x n. From the equality λ f = µ g we get ad(a a n x n )=bc(b b n x n ), thus ada i = bcb i for all i =0,...,n. One then has ad ad gcd(a 0,...,a n ) = gcd(ada 0,...,ada n ) = gcd(bcb 0,...,bcb n ) = bc gcd(b 0,...,b n ) bc, where we are using the fact f and g are primitive. Since the gcd is only defined up to a unit in R, we get that there must exist u U(R) such that bc = uad, and therefore µ = uλ. Now, since we have ada i = bcb i = uadb i, and a, d = 0, we obtain b i = u 1 a i, and thus g = u 1 f, so λ and f are unique up to multiplication by a unit of R. DEFINITION If f Q[x] is written as f = λ f as in the previous lemma, the element λ Q is called the content of f, and denoted by λ = c(f), and the polynomial f is called the primitive part of f. EXAMPLE Let f = x +2x2 Q[x], then c(f) = 2 21 and f = x + 21x 2. PROPOSITION Let R be a UFD with field of fractions Q, f Q[x], f = 0. The following properties hold: i) If λ Q, then c(λf) =λc(f) and λf = f. ii) f R[x] if and only if c(f) R. iii) f R[x] is primitive if and only if c(f) =1. iv) If f,g R[x] are primitive, and f g in Q[x], then f g R[x]. PROOF. i) f = c(f) f, where f primitive. Thus λf = λc(f) f. By uniqueness of the content and the primitive part, c(λf) =λc(f) and λf = f. ii) If f = a a n x n, and d = gcd(a 0,,a n ), then f = d( a0 d + + an d xn ), and gcd(a 0 /d,..., a n /d) =1, thus c(f) =d R. If c(f) R, as f R[x], then obviously f = c(f) f R[x]. iii) If f is primitive, then f = f = c(f) f, thus c(f) =1. Conversely, if c(f) =1, then f = c(f) f = f, and thus f is primitive. iv) Let f,g be primitive, then c(f) =c(g) =1. If f g in Q[x] there is some λ U(Q[x]) = Q such that g = λf, but then 1=c(g) =c(λf) =λc(f) =λ,

11 2.8. PROPERTIES OF POLYNOMIAL RINGS OVER DOMAINS 19 so λ =1(up to units in R), and thus f g in R[x]. THEOREM (Gauss lemma). Let R be a UFD, if f,g R[x] are primitive, then fg is also primitive. PROOF. Write f = a a n x n, g = b b m x m ; then fg = c c m+n + x m+n, where c i = j+k=i a jb k = a 0 b i + a i b 0. Suppose that fg is not primitive, then there must exist p R prime such that p c i for all i. Now, since f and g are primitive, they must have a first coefficient which is not divisible by p, i.e. there are some i, j such that p a 0,...,p a i 1,p a i, and p b o,...,p b j 1,p b j. But then one has c i+j = a 0 b i+j + a i i b j+1 +a i b j + a i+1 b j a i+j b 0. A B Now, p divides A because it divides a 0,...,a i 1, and p divides B because it divides b 0,...,b j 1. By assumption p divides c i+j, and as a i b j = c i+j A B, one gets that p divides a i b j, since p is prime, it must divide either a i or b j which is a contradiction. Henceforth, fg must be primitive. PROPOSITION Let R be a UFD with field of fractions Q, and let f,g Q[x], then one has fg = f g and c(fg)=c(f)c(g). PROOF. As f = c(f) f and g = c(g) g, one gets fg = c(f) fc(g) g = c(f)c(g) f g. By Gauss Lemma, f g is primitive, so it is the primitive part of fg, i.e. fg = f g, and by the uniqueness of the content-primitive part decomposition we get c(fg)=c(f)c(g). By using the previous results, we can describe the irreducibles in R[x] in terms of the irreducibles in Q[x]: PROPOSITION Let R be a UFD with field of fractions Q, and let f R[x], f = 0. The following properties hold: i) If deg f =0(so f R ) then f is irreducible in R[x] if and only if it is irreducible in R. ii) If deg f 1, then f is irreducible in R[x] if and only if f is primitive and is irreducible in Q[x]. PROOF. i) If f is not irreducible in R[x], then f = gh for some g, h R[x]. As deg f =0we must have deg g =degh =0as well, but then g and h are also elemets of R, and thus f is not irreducible in R. The converse statement is immediate. ii) If f is irreducible in R[x], then f is primitive. Assume f is not irreducible in Q[x], then we can write f = gh for some g, h Q[x], and then we have f = f = c(g)c(h) g h. As by Gauss Lemma f g is primitive, we get c(g)c(h) = 1, and hence f = g h with g, h R[x], contradicting the irreducibility of f in R[x]. Hence, f must be irreducible in Q[x]. The converse statement is obvious. THEOREM If R is a UFD, then R[x] is also a UFD. PROOF. Let f R[x] a nonzero, nonunit element of R[x]. If deg f =0, then f R, and since R is a UFD we can write f = p 1 p s where p i are irreducible in R. By Proposition 2.8.9, p i are also irreducible in R[x]. If deg f 1, as R[x] Q[x] we can look at f as an element in Q[x]. SInce Q is a field, Q[x] is a PID (in particular a UFD), and hence we can write f = f 1 f k where f i are irreducible in Q[x]. By taking content-primitive part we get f = c(f) f = c(f 1 ) c(f k ) f 1 f k,

12 20 2. INTEGRAL DOMAINS: ED, PID AND UFDS where now each f i is primitive and belongs to R[x]. As f i are irreducible in Q[x] and f i f i in Q[x], we have f i irreducible in Q[x]; by proposition 2.8.9, we obtain that f i are irreducible in R[x]. Now, as f R[x] then c(f) R, so we can write c(f) = p 1 p s,with p i irreducible in R (and thus in R[x]). Putting everything together, we can write f = p 1 p s f1 f k, where each term is irreducible in R[x]. Thus, every nonzero, nonunit element of R[x] can be written as a product of irreducibles. Now we need to show uniqueness of the factorization. Suppose that f = p 1 p s f 1 f k = q 1 q s g 1 g k where p i,q j are irreducibles in R and f i,g j are irreducibles in R[x] of degree bigger than 0. As every irreducible is primitive, taking content and primitive part we get p 1 p s = q 1 q s, and f 1 f k = g 1 g k. Now, using the fact that R is a UFD we obtain that s = s and (after possibly some reordering) p i q i for all i =1,...,s. Now since all f i and g j are irreducible in R[x], by they are also irreducible in Q[x], so we have two equal decompositions as products of irreducibles f 1 f k = g 1 g k in Q[x]. Since Q is a field, Q[x] is a UFD, so we have k = k and, up to reordering, f i g i in Q[x]. As f i and g i are primitive, by proposition f i g i in R[x]. REMARK By repeated use of the previous theorem, we obtain that for any UFD R the multivariate polynomial ring R[x 1,...,x n ] s also a UFD.