32 Divisibility Theory in Integral Domains

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1 3 Divisibility Theory in Integral Domains As we have already mentioned, the ring of integers is the prototype of integral domains. There is a divisibility relation on * : an integer b is said to be divisible by a nonzero integer a when there is an integer c such that ac = b. Integers with no nontrivial divisors are called prime, and every nonzero integer that is not a unit can be written as a product of prime numbers in a unique way. We want to investigate whether there are similar results in other integral domains. More generally, one can ask whether there are similar results in an arbitrary ring. However, in an arbitrary ring, one has to distinguish between left divisors and right divisors: if a,b,c are elements of a ring and ab = c, then a is called a left divisor, and b is called a right divisor of c. Here a may be a left divisor of c without being a right divisor of c, and vice versa. Furthermore, the existence of zero divisors in a ring complicates the theory. For these reasons, in this introductory book, we confine ourselves to integral domains. 3.1 Definition: Let D be an integral domain and let, D. If 0 and if there is a D such that =, then is called a divisor or a factor of and is said to be divisible by. We also say divides. We write when divides, and when 0 and does not divide. 3. Lemma: Let D be an integral domain and let,,,,, 1,,..., be elements of D. s (1) If, then,,. () If and, then. (3) If and 0, then. 1,,..., s, * More precisely, on \{0} 361

2 (4) If, then. (5) If and, then +. (6) If and, then. (7) If and, then +. (8) If 1,,..., s, then s s. (9) If 0, then 0. (10) 1 and 1. Proof: The claims are proved exactly as in the proof of Lemma 5.. We know that the units 1, 1 of any arbitrary integral domain (see Definition 9.14). divide every integer. This is true in 3.3 Lemma: Let D be an integral domain and D. Then is a unit of D (i.e., D ) if and only if for all D. Proof: If is a unit, then = 1 for some D; in particular, since D is an integral domain, = 1 0 and 0. For any D, we have ( ) = ( ) = 1 =, with D. Thus for any D. Conversely, if for all in D, then 1, so = 1 for some D. Thus has an inverse in D and is a unit of D. 3.4 Definition: Let D be an integral domain and, D. Then is said to be associate to if there is a unit D such that =. In this case, we write. 3.5 Lemma: Let D be an integral domain. Then is an equivalence relation on D. Proof: (i) For any D, we have = 1 and 1 is a unit. Hence and is reflexive. (ii) If, D and, then = for some D, then = 1 with 1 D (for D is a group by Theorem 9.15) and, so is symmetric. (iii) If,, D and,, then =, =, where 36

3 , are units in D. So =, with D (Theorem 9.15), thus and is transitive. Since is a symmetic relation, it is legitimate to say that and are associate when is associate to. The alert reader will have noticed that the group D acts on the set D in the sense of Definition 5.1, and the orbit of any D consists of the associates of (that is, elements of D which are associate to ). Lemma 3.5 is thus merely a special case of Lemma 5.5. For any D, the units and associates of are divisors of. A divisor of, which is neither a unit nor an associate of, is called a proper divisor of. An element need not have proper divisors; for instance, a unit has no proper divisors. The relation holds if and only if the relation 1 1 holds for any associate 1 of and for any associate of. In other words, as far as 1 divisibility is concerned, associate elements play the same role Examples: (a) The theory of divisibility in was discussed in 5. The units in are 1 and 1, and the associates of a are a and a. The terminology in this paragraph is consistent with that of 5. (b) Let D be a field. Then for any, D, 0 since 0 implies that there is an inverse 1 of in D and ( 1 ) = with 1 D. In particular, 1 for any D, 0. Hence any nonzero element in D is a unit and any two nonzero elements are associate. The divisibility theory is not very interesting in a field. (c) Let R = {a/b : (a,b) =1, 5 b} be the ring of Example 9.(e). It is easily seen that R is an integral domain. Let us find the units of R. The multiplicative inverse of a/b R ((a,b) = 1) is b/a, and b/a R if and only if 5 a. Thus R = {a/b R: (a,b) =1, 5 b, 5 a}. The associates of a/b R are the numbers x/y with (x,y) = 1, where a and x are exactly divisible by the same power of of

4 (d) We put [i] := {a + bi : a,b }. One easily checks that [i] is a subring of and that [i] is an integral domain. The elements of [i] are called gaussian integers (after C. F. Gauss ( ) who introduced them in his investigations about the so-called biquadratic reciprocity law). Since [i] is a subring of, each element = a + bi in [i] has a conjugate and a norm. The conjugate of = a + bi [i] is defined to be = a bi in [i] (a,b ). Notice that = for any, [i]. The norm N( ) of a + bi [i] is defined by N( ) = ; hence N(a + bi) = a + b (a,b ). Thus N( ) is a nonnegative integer for any [i], and equals 0 if and only if = 0 + 0i = 0. Moreover, N( ) =. =. = = N( )N( ) for any, [i]. Using this, it is easy to determine the units in [i]. We claim [i] is a unit in [i] if and only if N( ) = 1. Indeed, if is a unit in [i], then = 1, then N( )N( 1 ) = 1, where N( ),N( 1 ) are positive integers. This forces N( ) = 1, as claimed. Conversely, if N( ) = 1, then [i], and this yields 1, which means is a unit. = 1, where Thus a + bi is a unit if and only if N( ) = a + b = 1 (here a,b ) and a + b = 1 if and only if a = 1, b = 0 or a = 0, b = 1. Therefore is a unit if and only if of in [i] are the numbers,, i, i. (e) We put = theorem, 1 + 3i = 1, 1,i, i, so that [i] = {1, 1,i, i}. The associates. Thus = cos = cos 3 + isin 3 = e 4 3 = 3 + isin 3 1 3i = and. By de Moivre's 1 3 = cos isin 3 3 = 1. So 3 1 = 0, so ( 1)( + + 1) = 0. Since 1 0, we conclude From 3 = 1, we obtain = 0, which can also be verified directly. 4 =, whence ( ) = 0. We put [ ] := {a + b : a,b }. One easily checks that [ ] is a subring of and that [ ] is an integral domain. The closure of [ ] under multiplication follows from = 1 : (a + b )(c + d ) = ac + ad + bc + bd = ac + ad + bc + bd( 1 ) = (ac bd) + (ad + bc bd) 364

5 [ ] for all a + b, c + d [ ]. The ring [ ] was introduced independently by C. G. J. Jacobi ( ) and by G. Eisenstein ( ) in their investigations about the so-called cubic reciprocity law). Repeating the proof for [i], we see that a + b [ ] is a unit in [ ] if and only if N(a + b ) = 1. This is equivalent to a ab + b = 1, so equivalent to 4a 4ab + 4b = 4, so to (a b) + 3b = 4. The last equation holds if and only if a b =, b = 0 or a b = 1, b = 1 (a,b ). In this way, we get a + b = 1,, ( 1 ) =. The units in [ ] are 1,, ( 1 ) = ; the associates of [ ] are the numbers,, ( 1 ) =. (f) We put [ 5i] = {a + b 5i : a,b }. Again, it is easily verified that [ 5i] is an integral domain, and [ 5i] is a unit if and only if N( ) = 1. Now N(a + b 5i) = a + 5b = 1 if and only if a = 1, b = 0 (here a,b ). Thus 1 are the only units in [ 5i] and the associates of a number [ 5i] are the numbers. So far, the divisibility theory in an arbitrary integral domain has been completely analogous to the theory in, which culminates in the fundamental theorem of arithmetic asserting that every integer, not a zero or a unit, can be written as a product of prime numbers in a unique way. We proceed to investigate if a similar theorem is true in an arbitrary integral domain. First we introduce the counterparts of prime numbers. 3.7 Definition: Let D be an integral domain and D. Then is said to be irreducible if is neither zero nor a unit, and if, in any factorization = of, where, D, either or is a unit in D. When is neither zero nor a unit, and when is not irreducible in D, is said to be reducible. An irreducible element in D is therefore one which has no proper divisors. Clearly, when and are associates, is irreducible if and only if is irreducible. One might expect that such elements be called prime 365

6 rather than irreducible, but the term "prime" is reserved for another property (Definition 3.0). We now ask if every nonzero,. nonunit element in an integral domain D can be expressed as a product of finitely many irreducible elements. (cf. Theorem 5.13) Let us try to argue as in Theorem Given D ( 0, not a unit), is either irreducible or not.. In the former case, is a product of one irreducible element. In the latter case, = 1 for some suitable proper divisors of. Here 1 is either irreducible or not. In the former case, 1 is an irreducible divisor of. In the latter case, 1 = for some suitable proper divisors of 1. Here is either irreducible or not. Repeating this procedure, we get a sequence. = 0, 1,,... (s) of elements in D, where i+1 is a proper divisor of (i = 0,1,,... ). i When the sequence (s). stops after a finite number of steps, we obtain an irreducible divisor of.. However, we do not know that the sequence (s) ever terminates. In the case of, the absolute values of i, which are nonnegative integers, get smaller and smaller and,. since there are finitely many nonnegative integers less than,. the sequence (s) does come to an end.. But this argument cannot be extended to the general case, for there is no absolute value concept.. Let us suppose, however, that there is associated a nonnegative integer d( i ) to each i in such a way that d( i+1 ) sequence (s) does terminate.. d( i ). If this is possible, we can conclude that For example, when D is one of [i], [ ], [ 5i],. we may consider the norm N( i ) of i. The norm N( i ) is a nonnegative integer, and also N( i+1 ) N( i ) whenever i+1 is a proper divisor of. In fact, with the i norm function, there is a division algorithm in [i] and in [ ] Theorem: Let, be elements of [i] (resp. of [ ]), with 0. Then there are two elements and in [i] (resp. in [ ]) such that = + and N( ) N( ). Proof: (Cf. Theorem 5.3; note that and are not claimed to be unique.) The elements, of [i] (resp. of [ ]) are complex numbers, and 0. Thus /. Let us write 366

7 = x + yi (resp. = x + y ) with x,y. We want to be "approximately equal" to, with an "error" so small that N( ) 1. So we approximate x + yi (resp. x + y ) by an element we choose integers a,b in [i] (resp. in [ ]) as closely as we can. To this end, x such that 1 a and put = a + bi (resp. = a + b ). This is possible since the distance between x and the integer closest to x is less than or equal to 1. When x is half an odd integer, there are two choices for a, and therefore there can be no hope for uniqueness. In this case, we have in fact x a = 1 and the approximation above is the best possible one. The same remarks apply to y and b. We now put =. Then = +. It remains to show that N( ) 1. We have indeed N((x + yi) (a + bi)) N( ) = N( ) = N( ) = { N((x + y ) (a + b )) = { This completes the proof. y b N((x a) + (y b)i) N((x a) + (y b) ) = { (x a) + (y b) (x a) (x a)(y b) + (y b) ={ x a + y b x a + x a y b + y b { (1/) + (1/) (1/) + (1/)(1/) + (1/) = { /4 3/ What happens in [ 5i]? For a,b [ 5i], 0, we write = x + yi, with x,y. The best approximation to is given by = a + b 5i, where a,b are integers such that x conclude only a 1, y b 1. Putting =, we can 367

8 N( ) = N( ) = N( ) = N((x a) + (y b) 5i) = (x a) + 5(y b) (1/) + 5(1/) = 3/ ( ) instead of N( ) 1 as in [i], [ ]. 3.9 Theorem: In [ 5i], there are elements 0, 0 with 0 N( 0 0 ) N( 0 ) for all [ 5i]. 0 such that Proof: We choose 0, 0 in such a way that equality holds in ( ) above. This will be the case when x and y are half odd integers. So we set 1 + 5i, have 0 =. Then, for any = a + b 5i [ 5i] (with a,b ), we N( 0 0 ) = N( 0 )N( 0 0 ) = N( 0 0 )N( 0 0 b 5i)) 1+ 5i ) = N( 0 )N( 0 = (a + = N( 0 )[( 1 a) + 5( 1 as claimed. b) ] N( 0 )[(1/) + 5(1/) ] = N( 0 ) 3 N( 0 ), The integral domains on which there is a division algorithm are called Euclidean domains. The formal definition is as follows Definition: Let D be an integral domain. D is called a Euclidean domain if there is a function d: D\{0} {0} such that (i) d( ) d( ) for all, D\{0}, (ii) for any, D\{0}, there are, D satisfying = + and = 0 or d( ) d( ). The first condition (i) assures that the d-value of a divisor of D\{0} is less than or equal to the d-value of. It follows that d( ) = d( ) whenever and are associate. Using (ii) repeatedly, the analog of the Euclidean algorithm is seen to be valid, and the last nonzero remainder 368

9 is a greatest common divisor. It will be a good exercise for the reader to prove this result. is a Euclidean domain, with the absolute value function working as the function d of Definition This follows from Theorem 5.3, with b replaced by b. Also, [i] and [ ] are Euclidean domains, with the norm function working as the function d of Definition 3.10, as Theorem 3.8 shows. On the other hand, we do not yet know whether [ 5i] is a Euclidean domain. It does not follow from Theorem 3.9 that [ 5i] is not Euclidean. From Theorem 3.9, it follows only that either [ 5i] is not Euclidean, or [ 5i] is a Euclidean domain with a function d that is necessarily distinct from the norm function. In a Euclidean domain D, the sequence (s) terminates after a finite number of steps. We shall prove a more general statement (Theorem 3.14). Recall that { D: D} = D, where D, is the principal ideal generated by (Example 30.6(h)) Theorem: If D is a Euclidean domain, then every ideal of D is a principal ideal. Proof: Let D be a Euclidean domain, and let d be the function of Definition For any ideal A of D, we must find an such that A = D. We argue as in Theorem 5.4. When A = {0}, we clearly have A = D0, and the claim is true. Assume now A {0}. Then U = {d( ) {0}: A, 0} is a nonempty subset of the set of nonnegative integers. Let m be the smallest integer in U. Then m = d( ) for some A, 0; and d( ) d( ) for all A, 0. We show that A = D. First we have D A, because A and A has the "absorbing" property. To prove A D, take an arbitrary from A. There are, D such that = +, = 0 or d( ) d( ), provided 0 (Definition 3.10). Now A, so A, and since A as well, we see that A. Here d( ) d( ) is impossible, for then d( ) in U would be less than m, which is the smallest number in U. Hence 369

10 necessarily = 0 and = D. This shows D for all A, provided 0. Since 0 = 0 D as well, we get A D. Thus A = D. 3.1 Definition: An integral domain D is called a principal ideal domain if every ideal of D is a principal ideal. With this terminology, Theorem 3.11 can be reformulated as follows Theorem: Every Euclidean domain is a principal ideal domain. In any integral domain, if and only if D D, and if and only if D = D. Thus the sequence (s) gives rise to the chain D = D 0 D 1 D... of principal ideals in D. The sequence (s) breaks down if and only if this chain of ideals breaks down. For principal ideal domains, this is always true Definition: Let D be an integral domain. D is said to satisfy the ascending chain condition (ACC) if, for every chain A 0 A 1 A A 3... of ideals in D, there is an index k such that A m = A k for all m k; or, what is the same, every chain. B 0 B 1 B B 3... of ideals in D consists of finitely many terms. An integral domain satisfying the ascending chain condition is also called a noetherian domain (in honor of Emmy Noether ( )) Theorem: Every principal ideal domain satisfies the ascending chain condition (is noetherian). Proof: Let D be a principal ideal domain and let 370

11 A 0 A 1 A A 3... be a chain of ideals of D.. We must show there is an integer k such that A m = A k for all m k. To this and, we put B := A i=1 i. We claim B is an ideal of D. Indeed, if, B, then A j, A l for some indices j,l. Assuming j l without loss of generality, we have A j A l. Since A l is an ideal of D, we have + A l, so + B. Also A l, so B. This shows that B is a subgroup of D under addition.. Finally, if is an arbitrary element of D, then A l, since A l is an ideal, so B. Hence B is an ideal of D.. Since D is a principal ideal domain, B = D for some D.. As = 1 D = B = A i=1 i, we see A k for some k. We claim A m = A k for all m k. We know that A k A m for all m k because each ideal in the chain is contained in the next one (the chain is acsending).. On the other hand, for any m k, we have A A m i=1 i = B = D A k, because A k and A k is an ideal of D. Thus A m = A k for all m k and D satisfies the ascending chain condition.. Using Theorem 3.14, we shall prove the analog of Theorem 5.13 for any arbitrary principal ideal domain Theorem: Let D be a principal ideal domain. Then every element of D that is neither zero nor a unit can be exressed as a product of finitely many irreducible elements of D. Proof: First we prove that every element in D, which is neither zero nor a unit, has an irreducible divisor in D. Let D, 0, unit. Arguing as on page 366, we get a sequence = 0, 1,,... (s) of elements in D, where i+1 is a proper divisor of i (i = 0,1,,... ). In particular, none of the i is a unit. This sequence gives rise to the ascending chain D = D 0 D 1 D

12 of ideals of D (here D i D i+1 because i+1 is a proper divisor of i ). Since D is noetherian (Theorem 3.14), this chain breaks off: the chain consists only of the ideals D = D 0 D 1 D... D k say. Hence = 0, 1,,..., k are the only elements in the sequence (s). We claim that k is irreducible in D. Otherwise, there would be proper divisors k+1 and k+1 of k with k = k+1 k+1, and the sequence (s) would contain the term k+1 after k, and would not terminate with the term k, a contradiction. Hence k is an irreducible divisor of.. We proved that every element is neither zero nor a unit, has an irreducible divisor in D.. Let in D, which be an arbitrary nonzero, nonunit element in D.. We want to show that. can be written as a product of finitely many irreducible elements in D.. By what we proved above, we know that has an irreducible divisor, 1 say. We put = 1 1. Here 1 0. If 1 is not a unit, then 1 has an irreducible divisor, say. We put 1 =. Thus = 1. Here 0. If is not a unit, then has an irreducible divisor, 3 say. We put = 3 3. Thus = Continuing in this way, we get a sequence. = 0, 1,, 3,... of elements in D inducing a chain. D = D 0 D 1 D D 3... of ideals of D. By Theorem 3.14, this chain is finite, for example D = D 0 D 1 D D 3... D k We claim k is a unit. Otherwise k would have an irreducible divisor k+1 and, when we put k =, there would be, in the chain, an k+1 k+1 additional ideal D k+1 containing D k properly, a contradiction. Thus k is a unit.. Then = k 1 k k = k 1 ( k k ) is a product of the irreducible elements 1,, 3,..., k 1, k k. Having established the analog of Theorem 5.13, we proceed to work out the counterpart of Euclid's lemma (Lemma 5.15). For this we need the notion of greatest common divisor. 37

13 3.16 Definition: Let D be an arbitrary integral domain and let, D, not both zero. An element of D is called a greatest common divisor of and if (i) and, (ii) for all 1 in D, if 1 and 1, then 1. Notice that any associate of above satisfies the same conditions and hence any associate of a greatest common divisor of and is also a greatest common divisor of and. It is seen easily that any two greatest common divisor of and, if and have a greatest common divisor at all, are associates. So a greatest common divisor of and is not uniquely determined and we have to say a greatest common divisor, not the greatest common divisor. We let (, ) stand for any greatest common divisor of and. Thus (, ) is determined uniquely to within ambiguity among associate elements. Although we defined a greatest common divisor of two elements in an integral domain, this does not mean, of course, that any two elements (not both zero) in that domain do have a greatest common divisor. Introducing a definition does not create the definiendum. Given two elements (not both zero) in an integral domain, we cannot assert that they have a greatest common divisor. As a matter of fact, in an arbitrary integral domain, not every pair of elements (not both zero) has a greatest common divisor. For the special class of principal ideal domains, however, the following theorem holds Theorem: Let D be a principal ideal domain and let, be arbitrary elements in D, not both of them being zero. Then there is a greatest common divisor of and in D. Furthermore, if is a greatest common divisor of and, then there are, D such that = +. Proof: (cf. Theorem 5.4) As in the proof of Theorem 5.4, we consider the set A := { + :, D}. A is a nonempty subset of D. We claim that A is an ideal of D. To prove this, let 1, be arbitrary elements of A. Then 1 = 1 + 1, = + for some 1,, 1, D. Hence 373

14 1 + = ( ) + ( + ) = ( 1 + ) + ( 1 + ) D 1 = ( ) = ( 1 ) + ( 1 ) D and therefore A is a subgroup of D under addition. Also, for any D, 1 = ( ) = ( 1 ) + ( 1 ) D and thus A has the "absorbing" property as well. So A is an ideal of D. Since, are not both equal to zero, A {0}.. Now D is a principal ideal domain, so A = D for some D, and A {0} implies 0.. Also, since = 1 D = A, there are 0, 0 D with = is a greatest common divisor of and We prove now that 0 (i) = A = D, so = = for some D. Since we have 0, we can write. Likewise. (ii) If 1 D and 1, 1, then , hence 1. Thus is a greatest common divisor of and, and = 0, 0 D. The proof is complete for some In a principal ideal domain D, we see that D + D = D(, ) whenever, D are not both zero. Either from this remark, or better from Definition 3.16, it follows that (, ) = (, ). When (, ) 1, we say is relatively prime to, or and are relatively prime. In this case, there are, in D with + = Lemma: Let D be a principal ideal domain and,,, D. (1) If and (, ) 1, then. () If is irreducible and, then (, ) 1. Proof: (1) If (, ) 1, we have + = 1 for some, D. Hence + =. Now and, so and, so +, so. () Let be irreducible. Then 0. So (, ) exists by Theorem Let be a greatest common divisor of and. Then. Since is irreducible, either is associate to, or is a unit. In the first case, we 374

15 get (since ) against our hypothesis. Thus is a unit and 1, as claimed Lemma: Let D be a principal ideal domain and,, D. If is irreducible and, then or. Proof: If, the lemma is true. If, then (, ) 1 by Lemma 3.18() and so by Lemma 3.18(1), with in place of. 3.0 Definition: Let D be an arbitrary integral domain. If D is not zero or a unit, and if has the property that for all, D, or, then is called a prime element in D. In an arbitrary integral domain, all prime elements are irreducible. Indeed, let be prime. Then is not zero or a unit by definition. We show that has no proper divisors. Suppose =. Then and therefore or. Without restricting generality, let us assume. But as well, so and is a unit. Thus admits no proper factorization and is irreducible. The converse of this remark is not true. That is to say, in an arbitrary integral domain, there may be irreducible elements which are not prime (see Ex. 13). Lemma 3.19 asserts that irreducible and prime elements coincide in a principal ideal domain. This is the basic reason why there turns out to be a unique factorization theorem in principal ideal domains. 3.1 Lemma: Let D be an arbitrary ideal domain and 1, D. If is prime and 1... n, then 1 or or... or n.,..., n, Proof: Omitted. 375

16 3. Theorem: Let D be a principal ideal domain. Every element of D, which is not zero or a unit, can be expressed as a product of irreducible elements of D in a unique way, apart from the order of the factors and the ambiguity among associate elements. Proof: (cf. Theorem 5.17.) Let D, 0, unit. By Theorem 3.15, can be expressed as a product of irreducible elements of D. We must show uniqueness. Given two decompositions 1... r = = 1... s of into irreducible elements, we must show r = s and 1,,..., are, in r some order, associate to 1,,...,. This will be proved by induction on r. s First assume r = 1. Then 1 = = 1... s, so is irreducible. This forces s = 1 and 1 = = 1. This proves the theorem when r = 1. Now assume r and that the theorem is proved for r 1. This means, whenever we have an equation 1... r 1 = 1... t with irreducible,, there holds r 1 = t and 1,,..., some order, associates of 1,,..., t. r 1 are, in We have 1... r = = 1... s. So. So r r 1... s. Now r is prime by Lemma 3.19, and so r j for some j {1,,...,s}. (Here we use the fact that irreducible elements are prime in a principal ideal domain.. The conclusion r j is not valid in an arbitrary integral domain.) Reordering the 's if necessary, we may assume r s. Since s is irreducible, s has no proper divisors. Hence the divisor r of s is either a unit or an associate of s. But r is irreducible, so not a unit. Therefore r and s are associate. So r = s for some unit D. Then we obtain 1... r 1 ( s ) = = 1... s 1... ( r 1 ) = 1... s 1 and by induction, we get r 1 = s 1, 1,,..., ( r 1 ) are, in some order, associates of 1,,..., s 1. Hence and r = s 1,,..., r 1 are, in some order, associates of 1,,..., s 1 ; 376

17 and r is associate to s. This completes the proof. 3.3 Definition: Let D be an integral domain. If every element of D, which is not zero or a unit, can be expressed as a product of finitely many irreducible elements of D in a unique way, apart from the order of the factors and the ambiguity among associate elements, then D is called a unique factorization domain. With this definition, Theorem 3. reads as follows. 3. Theorem: Every principal ideal domain is a unique factorization domain. In particular, every Euclidean domain is a unique factorization domain. We generalize Lemma 3.19 to unique factorization domains. 3.4 Lemma: Let D be a unique factorization domain. Then every irre-ducible element of D is prime. Proof: Let be irreducible in D and, where, D. Thus there is a D with =. Then = 1... r, = 1... s, = 1... t, where,, are units and i, j, k are irreducible elements in D. From the uniqueness of the decomposition t = = = 1... r 1... s = 1... r 1... s we see that must be associate to one of the irreducible elements j or k. Thus divides or. So is prime. 377

18 There is the following generalization of Theorem 3.. If D is an integral domain in which every nonzero, nonunit element can be written as a product of finitely many irreducible elements, and if every irreducible element in D is prime, then D is a unique factorization domain. The proof of Theorem 3. is valid in this more general case. In a unique factorization domain D, any two elements, (not both zero) have a greatest common divisor. Clearly (, ) if = 0 and (, ) if = 0; and if 0, then = 1 m m 1m... r n r and = 1 n n 1... r r with suitable units,, irreducible elements 1,,..., r and nonnegative integers m i k, n i, and it is easily seen that = 1 k k 1... r r, where k i = min{m i, n i }, is a greatest common divisor of and. Thus (, ) exists in any unique factorization domain, provided only, are not both equal to zero. However, in an arbitrary unique factorization domain, (, ) cannot, in general, be expressed in the form +. There are unique factorization domains which are not principal ideal domains and there are principal ideal domains which are not Euclidean domains. 3.5 Theorem: Let D be a principal ideal domain and let D be a nonzero, nonunit element of D. Then is irreducible if and only if the factor ring D/D is a field. Proof: D/D is a commutative ring with identity 1 + D (Example 30.9(c)). Suppose is irreducible. We are to show that every nonzero element + D of D/D has an inverse in D/D. Let + D be distinct from the zero element 0 + D = D of D/D. This means D, so. Since is irreducible, we obtain (, ) 1 from Lemma 3.18(), and there are therefore, in D such that + = 1. So 1 D and ( + D )( + D ) = 1 + D. Thus + D is an inverse of + D. This proves that D/D is a field. We now prove that, if is not irreducible, then D/D is not a field. Indeed, if is not irreducible, then = for some, D, where neither nor is a unit. Here and, in view of this, would imply that ; then would be a unit, a contradiction. Hence and likewise. So D and D, so + D 0 + D + D, but 378

19 ( + D )( + D ) = + D = + D = 0 + D = zero element of D/D. Thus + D and + D are zero divisors in D/D and D/D cannot be a field. Exercises 1. Let D be an integral domain and D. Prove that is a prime element of D if and only if D is a prime ideal of D (see 30, Ex. 11).. Let D be a principal ideal domain and D. Prove that is an irreducible element of D if and only if D is a maximal ideal of D (see 30, Ex. 1). 3. Show that [ d] := {a + b d : a,b } is a Euclidean domain when d =,3,6. 4. Show that [ i] := {a + b i : a,b } is a Euclidean domain. 5. Let = domain. 1+ 7i. Show that [ ] := {a + b : a,b } is a Euclidean 6. Let D be a Euclidean domain, with the function d as in Definition Prove that D is a unit if and only if d( ) = d(1). 7. Find the decomposition into irreducible elements of in [i] and of 3 in [ ]. 8. Let p be an odd prime number. Prove that (i) p = p + 0i [i] is prime in [i] in case x 1 (mod p) has no solution and (ii) p [i] is not prime in [i], and in fact p = with a suitable prime element of [i], in case x 1 (mod p) has a solution. 9. Let [i]. Show that [i]/ [i] has exactly N( ) elements. 10. Using the Euclidean algorithm, find a greatest common divisor of 3 + 5i and + 3i; and of i and i in [i]. 379

20 11. Prove: an integral domain is a principal ideal domain if and only if there is a function d: D\{0} {0} satisfying and only if ; (i) d( ) d( ) for any, D\{0} with, and d( ) = d( ) if (ii) for all, D\{0} with and, there are,, D such that = + and d( ) min{d( ),d( )} i 1. Let =. Show that [ ] := {a + b : a,b } is a principal ideal domain, but not a Euclidean domain. 13. Prove that, 3, 1 + 5i, 1 5i are irreducible in [ 5i]. Show that,3 are not associate to 1 + 5i, 1 5i. Hence there are two essentially distinct decompositions. 3 = 6 = (1 + 5i)(1 5i) of 6 [ 5i] and therefore [ 5i] is not a unique factorization domain. 380

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