# REMARKS ABOUT EUCLIDEAN DOMAINS. 1. Introduction. Definition 1.2. An integral domain R is called Euclidean if there is a function d: R {0}

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1 REMARKS ABOUT EUCLIDEAN DOMAINS KEITH CONRAD 1. Introduction The following definition of a Euclidean (not Euclidian!) textbooks. We write N for {0, 1, 2,... }. domain is very common in Definition 1.1. An integral domain R is called Euclidean if there is a function d: R {0} N with the following two properties: (1) d(a) d(ab) for all nonzero a and b in R, (2) for all a and b in R with b 0 we can find q and r in R such that a = bq + r, r = 0 or d(r) < d(b). We will call (1) the d-inequality. Sometimes it is expressed in a different way: for nonzero a and b, if a b then d(a) d(b). This is equivalent to the d-inequality, taking into account of course the different roles of a and b in the two descriptions. Examples of Euclidean domains are Z (with d(n) = n ), F [T ] for any field F (with d(f) = deg f; this example is the reason that one doesn t assume d(0) is defined), and Z[i] (d(α) = N(α)). A possibly silly kind of example is any field F with d(a) = 1 for all a 0 (here all remainders are 0). Definition 1.1 is in [1, 2.1], [3, 37], [4, 7.2], [5, 6.5], [7, 3.7], [8, Chap. III 3], [10, 12.4], [13, 3.6], and [14, 18]. However, in [2, 8.1] we find a different definition, where the d-inequality is missing: Definition 1.2. An integral domain R is called Euclidean if there is a function d: R {0} N such that for all a and b in R with b 0 we can find q and r in R such that (1.1) a = bq + r, r = 0 or d(r) < d(b). Any function d: R {0} N that satisfies (1.1) will be called a Euclidean function on R. Thus a Euclidean domain in Definition 1.2 is an integral domain that admits a Euclidean function, while a Euclidean domain in Definition 1.1 is an integral domain that admits a Euclidean function satisfying the d-inequality. Does Definition 1.2 describe a larger class of rings than Definition 1.1? No. We will show in Section 2 that any Euclidean domain (R, d) in the sense of Definition 1.2 can be equipped with a different Euclidean function d such that d(a) d(ab) for all a and b in R, so (R, d) is Euclidean in the sense of Definition 1.1. The main reason that the d-inequality is not included in the definition of a Euclidean domain in [2] is that it is irrelevant to prove the two main theorems about Euclidean domains: that every Euclidean domain is a PID 1 and that the Euclidean algorithm in a Euclidean domain terminates after finitely many steps and produces a greatest common divisor. The 1 Not every PID is a Euclidean domain, as we ll see in Section 5. 1

2 2 KEITH CONRAD Euclidean algorithm provides a practical method of computing greatest common divisors when it is available. Why is the d-inequality nearly always mentioned in the (textbook) literature if it s actually not needed? Well, it is not needed for the two specific results cited in the previous paragraph, but it is convenient to use the d-inequality if we want to prove factorization into irreducibles in a Euclidean domain without proving the result more generally in a PID. There is factorization into irreducibles in any PID, which subsumes the same result for Euclidean domains since Euclidean domains are PIDs, but the proof of the existence of irreducible factorizations in a PID is less concrete than a proof available in Euclidean domains. We will see why in Sections 3 and 4. In Section 5 we discuss Euclidean domains among quadratic rings. 2. Refining the Euclidean function Suppose (R, d) is a Euclidean domain in the sense of Definition 1.2. We will introduce a new Euclidean function d: R {0} N, built out of d, which satisfies d(a) d(ab). Then (R, d) is Euclidean in the sense of Definition 1.1, so the rings that admit Euclidean functions in either sense are the same. Here s the definition (trick?): for nonzero a in R, set d(a) = min b 0 d(ab). That is, d(a) is the smallest d-value on the nonzero multiples of a. (We have ab 0 when b 0 since R is an integral domain.) So d(a) = d(ab 0 ) for some nonzero b 0 and d(ab 0 ) d(ab) for all nonzero b. For example, d(1) = min b 0 d(b) is the smallest d-value on R {0}. An additional property is (2.1) d(a) d(a) for all nonzero a in R, since a is a multiple of a. Theorem 2.1. Let (R, d) be a Euclidean domain in the sense of Definition 1.2. Then (R, d) is Euclidean in the sense of Definition 1.1. Proof. For any nonzero a and b in R we have d(a) d(ab). Indeed, write d(ab) = d(abc) for some nonzero c in R. Since abc is a nonzero multiple of a, d(a) d(abc) = d(ab). We now show R admits division with respect to d. Pick a and b in R with b 0. Set d(b) = d(bc) for some nonzero c R. Using division of a by bc (which is nonzero) in (R, d) there are q 0 and r 0 in R such that a = (bc)q 0 + r 0, r 0 = 0 or d(r 0 ) < d(bc). Set q = cq 0 and r = r 0. Since d(bc) = d(b) and d(r) d(r) (by (2.1)), the inequality d(r) < d(bc) implies d(r) < d(b). Thus a = bq + r, r = 0 or d(r) < d(b).

3 REMARKS ABOUT EUCLIDEAN DOMAINS 3 Hence (R, d) is a Euclidean domain in the sense of Definition 1.2. We end this section with a brief discussion of two other possible refinements one might want in a Euclidean function (but which we will not need later): uniqueness of the quotient and remainder it produces and multiplicativity. In Z we write a = bq + r with 0 r < b and q and r are uniquely determined by a and b. There is also uniqueness of the quotient and remainder when we do division in F [T ] (relative to the degree function) and in any field (the remainder is always 0). Are there any other Euclidean domains where the quotient and remainder are unique? Division in Z[i] does not have a unique quotient and remainder relative to the norm on Z[i]. For instance, dividing 1 + 8i by 2 4i gives 1 + 8i = (2 4i)( 1 + i) 1 + 2i and 1 + 8i = (2 4i)( 2 + i) + 1 2i, where both remainders have norm 5, which is less than N(2 4i) = 20. Theorem 2.2. If R is a Euclidean domain where the quotient and remainder are unique then R is a field or R = F [T ] for a field F. Proof. See [9] or [12]. This might be a surprise: Z isn t in the theorem! Aren t the quotient and remainder unique there? Yes if we take the remainder r so that 0 r < b, but not if we try to fit it into the setting of Euclidean domains using r < b. For instance, 51 = and 51 = The point is that using the Euclidan function on the remainder too, in the case of Z, permits negative remainders so in fact Z does not have a unique quotient and remainder when we measure the remainder by its absolute value. The Euclidean function in many basic examples satisfies a stronger property than d(a) d(ab), namely d(ab) = d(a)d(b) with d(a) 1 when a 0. For instance, this holds in Z (d(n) = n ) and Z[i] (d(α) = N(α)). The degree on F [T ] is not multiplicative, but d(f) = 2 deg f is multiplicative. There is also a multiplicative Euclidean function on any field F : d(a) = 1 for all a 0. It is an open question to decide whether every Euclidean domain has a multiplicative Euclidean function [11, p. 5]. 3. Features of the d-inequality Let R be a Euclidean domain. By Theorem 2.1 we may assume our Euclidean function d satisfies the d-inequality: d(a) d(ab) for all nonzero a and b in R. Using this inequality we will prove a few properties of d: Theorem 3.1. Let (R, d) be a Euclidean domain where d satisfies the d-inequality. Then (1) d(a) d(1) for all nonzero a R, (2) if b R then d(ab) = d(a) for all nonzero a, (3) if b R and b 0 then d(ab) > d(a) for all nonzero a. In particular, for nonzero a and b, d(ab) = d(a) if and only if b R. Proof. (1): By the d-inequality, d(1) d(1 a) = d(a). (2): By the d-inequality, d(a) d(ab). To get the reverse inequality, let c be the inverse of b, so the d-inequality implies d(ab) d((ab)c) = d(a).

4 4 KEITH CONRAD (3): We want to show the inequality d(a) d(ab) is strict when b is not a unit and not 0. The proof is by contradiction. Assume d(a) = d(ab). Now use division of a by ab: We rewrite this as a = (ab)q + r, r = 0 or d(r) < d(ab). a(1 bq) = r, r = 0 or d(r) < d(a). Since b is not a unit, 1 bq is nonzero, so a(1 bq) is nonzero (because R is an integral domain). Thus r 0, so the inequality d(r) < d(a) becomes d(a(1 bq)) < d(a). But this contradicts the d-inequality, which says d(a) d(a(1 bq)). Thus it is impossible for d(a) to equal d(ab) when b is not a unit. Note Theorem 3.1 is not saying two elements having the same d-value are unit multiples, but rather that a multiple has the same d-value if and only if it is a unit multiple. For example, in Z[i] we have N(1 + 2i) = N(1 2i) but 1 + 2i and 1 2i are not unit multiples. Remember this! Corollary 3.2. Let (R, d) be a Euclidean domain where d satisfies the d-inequality. We have d(a) = d(1) if and only if a R. That is, the elements of least d-value in R are precisely the units. Proof. Take a = 1 in parts 2 and 3 of Theorem 3.1. We can see Corollary 3.2 working in Z and F [T ]: the integers satisfying n = 1 are ±1, which are the units of Z. The polynomials f in F [T ] satisfying deg f = deg 1 = 0 are the nonzero constants, which are the units of F [T ]. Corollary 3.3. Let (R, d) be a Euclidean domain where d satisfies the d-inequality. If a and b are nonunits, then d(a) and d(b) are both less than d(ab). Proof. This is immediate from part (3) of Theorem 3.1, where we switch the roles of a and b to get the inequality on both d(a) and d(b). 4. Irreducible factorization To see the simplicity introduced by a Euclidean function, we will prove irreducible factorization in both Euclidean domains and in PIDs. Definition 4.1. Let R be an integral domain. A nonzero element a of R is called irreducible if it is not a unit and in any factorization a = bc, one of the factors b or c is a unit. A nonzero nonunit that is not irreducible is called reducible. There are three types of nonzero elements in an integral domain: units (the invertible elements, whose factors are always units too), irreducibles (nonunits whose factorizations into two parts always involve one unit factor), and reducibles (nonunits that admit some factorization into a product of two nonunits). Notice that in a field there are no reducible or irreducible elements: everything is zero or a unit. So if we want to prove a theorem about irreducible factorization, we avoid fields. Theorem 4.2. In any Euclidean domain that is not a field, every nonzero nonunit is a product of irreducibles.

5 REMARKS ABOUT EUCLIDEAN DOMAINS 5 Proof. Let (R, d) be a Euclidean domain that is not a field. By Theorem 2.1 we may assume d(a) d(ab) for all nonzero a and b in R. Therefore Corollary 3.3 applies. We will prove the existence of irreducible factorizations by induction on the d-value. From Corollary 3.2, the units of R have the smallest d-value. Any a R with second smallest d-value must be irreducible: if we write a = bc and b and c are both nonunits, then d(b) and d(c) are both less than d(a) by Corollary 3.3. Therefore b and c are units, so a is a unit. This is a contradiction. Assume now that a R is a nonunit and all nonunits with smaller d-value admit an irreducible factorization. To prove a admits an irreducible factorization too, we may suppose a is not irreducible itself. Therefore there is some factorization a = bc with b and c both nonunits. Then d(b) < d(a) and d(c) < d(a) by Corollary 3.3, so b and c both have irreducible factorizations by induction. Thus their product a has an irreducible factorization. The conclusion of Theorem 4.2 is true for PIDs, but the proof will require more abstract methods than induction. Theorem 4.3. In a PID that is not a field, any nonzero nonunit is a product of irreducibles. The proof of Theorem 4.3 relies on the following lemma, which has a recursive flavor. Lemma 4.4. If R is an integral domain and a R is a nonzero nonunit that does not admit a factorization into irreducibles then there is a strict inclusion of principal ideals (a) (b) where b is some other nonzero nonunit that does not admit a factorization into irreducibles. Proof. By hypothesis a is not irreducible, so (since it is neither 0 nor a unit either) there is some factorization a = bc where b and c are nonunits (and obviously are not 0 either). If both b and c admitted irreducible factorizations then so does a, so at least one of b or c has no irreducible factorization. Without loss of generality it is b that has no irreducible factorization. Since c is not a unit, the inclusion (a) (b) is strict. Now we can prove Theorem 4.3. Proof. Suppose there is an element a in the PID that is not 0 or a unit and has no irreducible factorization. Then by Lemma 4.4 there is a strict inclusion (a) (a 1 ) where a 1 has no irreducible factorization. Then using a 1 in the role of a (and Lemma 4.4 again) there is a strict inclusion (a 1 ) (a 2 ) where a 2 has no irreducible factorization. This argument (repeatedly applying Lemma 4.4 to the generator of the next larger principal ideal) leads to an infinite increasing chain of principal ideals (4.1) (a) (a 1 ) (a 2 ) (a 3 ) where all inclusions are strict. This turns out to be impossible in a PID. Indeed, suppose a PID contains an infinite strictly increasing chain of ideals: and set I 0 I 1 I 2 I 3 I = n 0 I n.

6 6 KEITH CONRAD This union I is an ideal. The reason is that the I n s are strictly increasing, so any finite set of elements from I lies in a common I n. (This is the key idea.) So I is closed under addition and arbitrary multiplications from the ring since each I n has these properties. (Make sure you understand that step.) Because we are in a PID, I is principal: I = (r) for some r in the ring. But because I is the union of the I n s, r is in some I N. Then (r) I N since I N is an ideal, so I = (r) I N I, which means I N = I. But this is impossible because the inclusion I N+1 I becomes I N+1 I N and we were assuming I N was a proper subset of I N+1. Because of this contradiction, nonzero nonunits in a PID without an irreducible factorization do not exist. The proof that a PID does not contain an infinite strictly increasing chain of ideals holds for a broader class of rings than PIDs. Theorem 4.5. A commutative ring in which every ideal is finitely generated does not contain an infinite strictly increasing chain of ideals. Proof. The second half of the proof of Theorem 4.3 works in this more general context. All we have to do is show the logic works when I is finitely generated rather than principal. The point is that if I = (x 1,..., x m ) then the finitely many x i s all lie in some common I N (because the I n s are an increasing chain). And now the contradiction is obtained just as before: I N = I but then I N+1 I N, contradiction. Corollary 4.6. In an integral domain where every ideal is finitely generated, every nonzero nonunit has an irreducible factorization. Proof. If there were an element a that is not 0 or a unit and that did not admit an irreducible factorization then, as in the proof of Theorem 4.3, we could produce an infinite strictly increasing chain of (principal) ideals. But there are no infinite strictly increasing chains of ideals in the ring, by Theorem 4.5. Definition 4.7. A commutative ring where every ideal is finitely generated is called a Noetherian ring. These rings are named after Emmy Noether, who was one of the pioneers of abstract algebra in the first half of the 20th century. Their importance, as a class of rings, stems from the stability of the Noetherian property under many basic constructions. If R is a Noetherian ring, so is any quotient ring R/I (which may not be an integral domain even if R is), any polynomial ring R[X] (and thus R[X 1,..., X n ] by induction on n, viewing this as R[X 1,..., X n 1 ][X n ]), and any formal power series ring R[[X]] (and thus R[[X 1,..., X n ]]). The PID property behaves quite badly, e.g., if R is a PID other than a field then R[X] is not a PID. For instance, R[X, Y ] = R[Y ][X] is never a PID for any integral domain R. But if R is Noetherian then R[X, Y ] is Noetherian. Briefly, the property ideals are finitely generated of Noetherian rings is much more flexible than the property ideals are singly generated of PIDs. Using this terminology, Corollary 4.6 says in any Noetherian integral domain any element other than 0 or a unit has an irreducible factorization. It is worth comparing the proof of this general result (Corollary 4.6) to the special proof we gave in the case of Euclidean domains,

9 REMARKS ABOUT EUCLIDEAN DOMAINS 9 Proof. Pick any x R. By division in R we can write x = aq+r where r = 0 or d(r) < d(a). If r 0, then the inequality d(r) < d(a) forces r to be a unit. Since x r mod a, we conclude that R/(a) is represented by 0 and by units. Example When R = Z we can use a = 2. Then Z/2Z is represented by 0 and 1. When R = Z[i] we can use a = 1 + i. Then Z[i]/(1 + i) is represented by 0 and 1. These examples show some units could be congruent modulo a, but at least every element of the ring is congruent to 0 or some (perhaps more than one) unit. We have shown that if R is a Euclidean domain, there are elements of R (namely nonunits with least d-value) modulo which everything is congruent to 0 or to a unit from R. A ring where no element has this property therefore can t be a Euclidean domain. Remark In a domain R, an element a for which the ring R/(a) is represented by 0 and units in R is called a universal side divisor in the literature. This terminology seems stupid. Side divisor? What s that? Remember the property, but forget the label (and don t ever use it, because nobody will know what you re talking about). Example The quadratic ring R = Z[(1 + 19)/2] is a PID that is not Euclidean. This is proved in [2, pp. 277,282]. We will include here the proof that R is not Euclidean by showing R contains no element a for which R/(a) is represented by 0 and units. First we compute the norm of a typical element α = x + y(1 + 19)/2: ( (5.1) N(α) = x 2 + xy + 5y 2 = x + y ) 2 19y This norm always takes values 0 (this is clearer from the second expression for it than the first) and once y 0 we have N(α) 19y 2 /4 19/4 > 4. In particular, the units are solutions to N(α) = 1, which are ±1: R = {±1}. The first few norm values are 0, 1, 4, 5, 7, and 9. In particular, there is no element of R with norm 2 or 3. This and the fact that R {0} has size 3 are the key facts. If R were Euclidean then there would be a nonunit a in R such that R/(a) is represented by 0 and units, so by 0, 1, and 1. Perhaps 1 1 mod a, but we definitely have ±1 0 mod a (since a is not a unit, (a) is a proper ideal so R/(a) is not the zero ring). Thus R/(a) has size 2 (if 1 1 mod a) or 3 (if 1 1 mod a). We show this can t happen. If R/(a) has size 2 then 2 0 mod a (think about R/(a) as an additive group of size 2), so a 2 in R. Therefore N(a) 4 in Z. There are no elements of norm 2, so the only nonunits with norm dividing 4 are elements with norm 4. A check using (5.1) shows the only such numbers are ±2. However, R/(2) = R/( 2) does not have size 2. For instance, 0, 1, and (1 + 19)/2 are incongruent modulo ±2: the difference of any two of these, divided by two, is not of the form x + y(1 + 19)/2 for x and y in Z. Similarly, if R/(a) has size 3 then a 3 in R, so N(a) 9 in Z. Since there is no element with norm 3, a must have norm 9 (it doesn t have norm 1 since it is not a unit). The only elements of R with norm 9 are ±3, so a = ±3. The ring R/(3) = R/( 3) does not have size 3: 0, 1, 2, and (1 + 19)/2 are incongruent modulo ±3 since their differences divided by 3 are not in Z[(1 + 19)/2]. Since #(R {0}) = 3 and R has no element a such that R/(a) has size 2 or 3, R can t be a Euclidean domain.

10 10 KEITH CONRAD References [1] S. Alaca and K. S. Williams, Introductory Algebraic Number Theory, Cambridge Univ. Press, [2] D. Dummit and R. Foote, Abstract Algebra, 3rd ed., Wiley, [3] J. Durbin, Modern Algebra: An Introduction, 5th ed., J. Wiley, [4] J. B. Fraleigh, A First Course in Abstract Algebra, 6th ed., Addison-Wesley, [5] F. Goodman, Algebra: Abstract and Concrete (Stressing Symmetry), 2nd ed., Prentice-Hall, [6] M. Harper, Z[ 14] is Euclidean, Canad. J. Math. 56 (2004), [7] I. Herstein, Topics in Algebra, 2nd ed., Wiley, [8] T. Hungerford, Algebra, Springer-Verlag, [9] M. A. Jodeit, Uniqueness in the division algorithm, Amer. Math. Monthly 74 (1967), [10] C. Lanski, Concepts in Abstract Algebra, Brooks/Cole, [11] F. Lemmermeyer, The Euclidean Algorithm in Algebraic Number Fields, uni-heidelberg.de/ hb3/publ/survey.pdf. [12] T. S. Rhai, A characterization of polynomial domains over a field, Amer. Math. Monthly 69 (1962), [13] J. Rotman, Advanced Modern Algebra, Prentice-Hall, [14] B. L. van der Waerden, Modern Algebra, Ungar, New York, 1953.

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