Subsets of Euclidean Domains Possessing a Unique Division Algorithm Andrew D

Subsets of Euclidean domains possessing a unique division algorithm Andrew D. Lewis∗ 2009/03/16

Abstract Subsets of a Euclidean domain are characterised with the following objectives: (1) en- suring uniqueness of the quotient and remainder in the Division Algorithm; (2) permit- ting unique base expansion with respect to any nonzero nonunit in the ring; (3) allowing explicit solutions to B´ezout’sidentity with norm constraints.

The two most popular examples of Euclidean domains, the ring of integers Z and the ring F[ξ] of polynomials over a field F, possess slightly different properties. For example, in Z the quotient and remainder from the Division Algorithm are generally not unique (becoming so when one restricts to positive integers), while the quotient and remainder in F[ξ] are unique. Indeed, Jodeit Jr. [1967] shows that any Euclidean domain with a unique Division Algorithm is isomorphic to either a field or to the polynomial ring over a field. The differences in the two rings Z and F[ξ] also shows up in two other commonly presented results which derive from the Division Algorithm: (1) the expansion of elements of the ring in terms of a base (which is taken to be a nonzero nonunit); (2) the computation, using the Euclidean Algorithm, of solutions to Bézout’sidentity for coprime ring elements, and with constraints on the Euclidean norms of the solution. For the base expansion in Z, to ensure uniqueness one again needs to restrict to positive integers, whereas the base expansion is always unique in F[ξ]. Moreover, the proofs in the two cases are typically carried out separately, or the proof of one is (not entirely accurately) suggested to follow just like the proof of the other. This leads to the natural question, “Is there a property of subsets of Euclidean domains which ensures, in these subsets: (1) uniqueness of the quotient and remainder; (2) uniqueness of base expansion; (3) norm bounds in the Euclidean Algorithm.” We show that there is indeed such a property, and it is quite simple—we call this property “δ-positivity.” Let us review the basic features of Euclidean domains, and provide the new definitions that will be used to prove some useful results for Euclidean domains having these properties. If A is a subset of B we write A ⊆ B, using the notation A ⊂ B to denote proper inclusion. We denote by Z>0 the set of positive integers and by Z≥0 the set of nonnegative integers. For an integral domain R we let 1R denote the unit element and 0R denote the zero element. For a field F, F[ξ] denotes the polynomial ring with coefficients in F. By deg(A) we denote the degree of A ∈ F[ξ], with the convention that deg(0F[ξ]) = −∞. Since there is not perfect agreement on what properties should be assigned to a Euclidean norm, let us first say exactly what we mean in this paper by a Euclidean domain. A

∗Professor, Department of Mathematics and Statistics, Queen’s University, Kingston, ON K7L 3N6, Canada Email: [email protected], URL: http://www.mast.queensu.ca/~andrew/ Research supported in part by a grant from the Natural Sciences and Engineering Research Council of Canada

1 2 A. D. Lewis

Euclidean domain is a pair (R, δ), where R is an integral domain and where the map δ : R → Z≥0, called the norm, has the following properties:

1. if a, b ∈ R and if ab 6= 0R, then δ(ab) ≥ δ(a);

2. if a, b ∈ R with b 6= 0R, then there exists q, r ∈ R such that (a) a = qb + r and such that (b) δ(r) < δ(b).

We shall make use of the following facts about Euclidean domains, without explicit mention:

1. δ(0R) < δ(1R);

2. if a ∈ R \{0R} then δ(a) ≥ δ(1R);

3. δ(a) = δ(0R) if and only if a = 0R;

4. for a ∈ R \{0R}, δ(ab) = δ(a) if and only if b is a unit;

5. a ∈ R is a unit if and only if δ(a) = δ(1R). Now let us give a few new deﬁnitions. 1 Deﬁnition: Let (R, δ) be a Euclidean domain.

(i) A subset C ⊆ R is trivial if C = {0R}, and is nontrivial otherwise.

(ii) A nonempty subset C ⊆ R is δ-closed if, for each a, b ∈ C with b 6= 0R, there exists q, r ∈ C such that a = qb + r and such that δ(r) < δ(b). (iii) A subset C ⊆ R admits a unique Division Algorithm if, for each a, b ∈ C with b 6= 0R, there exists unique q, r ∈ C such that a = qb + r and such that δ(r) < δ(b). (iv) A nonempty subset P ⊆ R is δ-positive if, for each a, b ∈ P , we have δ(a − b) ≤ max{δ(a), δ(b)}. • In this paper we will be interested in nontrivial, δ-closed, and δ-positive subsemirings of Euclidean domains, recalling that a subsemiring S ⊆ R has the property that if a, b ∈ S then ab ∈ S and a + b ∈ S. Let us give the two primary examples which illustrate the preceding concepts. 2 Examples: 1. For the ring Z we take the usual Euclidean norm: δ(k) = |k|. One can easily verify using elementary properties of integers that the subset Z≥0 ⊆ Z is a nontrivial, δ-closed, and δ-positive subsemiring. Note, however, that Z is not a δ- positive subset of itself since, for example, δ(1 − (−2)) = 3 > max{δ(1), δ(2)}.

2. Let F be a ﬁeld and deﬁne δ : F[ξ] → Z≥0 by ( 0,A = 0 , δ(A) = F[ξ] deg(A) + 1,A 6= 0F[ξ]. Subsets of Euclidean domains possessing a unique division algorithm 3

The pair (F[ξ], δ) is then well known to be a Euclidean domain. We claim that F[ξ] is a δ-positive subset of itself. If either A or B is nonzero, then

δ(A − B) = deg(A − B) + 1 ≤ max{deg(A), deg(B)} + 1 = max{deg(A) + 1, deg(B) + 1} = max{δ(A), δ(B)},

and, if A = B = 0F[ξ], then δ(A−B) = max{δ(A), δ(B)}. This shows that F[ξ] is indeed δ-positive. • The following property of nontrivial δ-closed subsets is useful. 3 Lemma: If (R, δ) is a Euclidean domain and if S ⊆ R is a nontrivial δ-closed subsemiring, then 0R, 1R ∈ S.

Proof: Let b ∈ S − {0R}. Since S is δ-closed there exists q, r ∈ S such that b = qb + r with δ(r) < δ(b). We claim that this implies that q = 1R and r = 0R. Suppose that q 6= 1R. Then δ(b) ≤ δ((1R − q)b) = δ(r) < δ(b) which is a contradiction. Thus q = 1R, and it then follows that r = 0R. Let us ﬁrst explore the relationship between δ-positivity and uniqueness in the Division Algorithm. Note that, for the Euclidean domain (Z, δ), we do not generally have such uniqueness since, for example, we can write 6 = 1 · 5 + 1 = 2 · 5 − 4. 4 Proposition: If (R, δ) is a Euclidean domain and if S is a nontrivial, δ-closed, and δ- positive subsemiring of R, then S admits a unique Division Algorithm.

Proof: Suppose that a = q1b + r1 = q2b + r2 for q1, q2, r1, r2 ∈ S with δ(r1), δ(r2) < δ(b). Then (q1 − q2)b = r2 − r1, and so

δ((q1 − q2)b) = δ(r1 − r2) ≤ max{δ(r1), δ(r2)} < δ(b), using δ-closedness of S. This implies that (q1 − q2)b = 0R. Since b 6= 0R this implies that q1 − q2 = 0R and so q1 = q2. We then immediately have r1 = r2. The condition of δ-positivity is, in certain circumstances, also necessary for uniqueness in the Division Algorithm. 5 Proposition: Let (R, δ) be a Euclidean domain and let S ⊆ R be a nontrivial, δ-closed subsemiring with the following properties: (i) S generates R as a ring; (ii) S admits a unique Division Algorithm. Then S is δ-positive. Proof: Note that since S is a subsemiring, S generates R as a ring if and only if, for every r ∈ R, it holds that either r ∈ S or −r ∈ S. Suppose that S is not δ-positive so that δ(a − b) > max{δ(a), δ(b)} for some a, b ∈ S. Suppose that b − a ∈ S. Then

b = 0R · (b − a) + b, δ(b) < δ(b − a),

b = 1R · (b − a) + a, δ(a) < δ(b − a), which shows that S does not admit a unique Division Algorithm. An entirely similar argument gives the same conclusion when a − b ∈ S. 4 A. D. Lewis

Next we show that base expansion is valid in δ-positive subsets. Again, while base expansions exist for all integers, in order to ensure uniqueness of the coeﬃcients in the expansion, one needs to restrict to positive integers to obtain uniqueness. Much of the proof we give is to be found in standard texts, but we give all of the details in order to illustrate exactly where our additional hypothesis of δ-positivity is used. 6 Proposition: Let (R, δ) be a Euclidean domain, let S ⊆ R be a nontrivial, δ-closed, and δ-positive subsemiring, and let b ∈ S be a nonzero nonunit. Then, given a ∈ S \{0R}, there exists a unique k ∈ Z≥0 and unique r0, r1, . . . , rk ∈ S such that

(i) rk 6= 0R,

(ii) δ(r0), δ(r1), . . . , δ(rk) < δ(b) and 2 k (iii) a = r0 + r1b + r2b + ··· + rkb .

Proof: We prove the result by induction on δ(a). By Lemma3 we have

inf{δ(a) | a ∈ S} = 0.

Since we do not consider the case δ(a) = δ(0R), ﬁrst consider a ∈ R such that δ(a) = δ(1R). Then a is a unit. Thus, since b is a nonzero nonunit, we have δ(a) < δ(b), and the existence part of the result follows by taking k = 0 and r0 = a. Now suppose that the result holds for all a ∈ S such that δ(a) ∈ {δ(1R, . . . , m}. Let a be such that

δ(a) = inf{δ(r) | r ∈ S, δ(r) > m}.

If δ(a) < δ(b) then take k = 0 and r0 = a to give existence in this case. Otherwise, apply the Division Algorithm to give a = qb + r with δ(r) < δ(b). Since S is δ-closed, we can moreover suppose that q, r ∈ S. Now, since b is a nonzero nonunit, since we are supposing that δ(a) ≥ δ(b) > δ(r), and since S is δ-positive,

δ(q) < δ(qb) = δ(a − r) ≤ max{δ(a), δ(r)} = δ(a).

Therefore, we may apply the induction hypothesis to q to give

0 0 0 2 0 k q = r0 + r1b + r2b + ··· + rkb

0 0 0 for some k ∈ Z≥0 and for r0, r1, . . . , rk ∈ S. Then

0 0 0 2 0 k 0 0 2 0 k+1 a = (r0 + r1b + r2b + ··· + rkb )b + r = r + r0b + r1b + ··· + rkb , showing that the existence part of the result holds for δ(a) = inf{δ(r) | r ∈ S, δ(r) > m}. This proves the existence part of the result for all a ∈ S by induction. We also prove the uniqueness assertion by induction on δ(a). First we use a technical lemma concerning the general expansion of 0R in the base b. Subsets of Euclidean domains possessing a unique division algorithm 5

Lemma: Let (R, δ) be a Euclidean domain with b ∈ R a nonzero nonunit. If k ∈ Z≥0 and r0, r1, . . . , rk ∈ R satisfy 2 k (i) r0 + r1b + r2b + ··· + rkb = 0R and

(ii) δ(r0), δ(r1), . . . , δ(rk) < δ(b), then r0 = r1 = ··· = rk = 0R. Proof: We prove this by induction on k. For k = 0 the result is trivial. For k = 1 we have r0 + r1b = 0R, and we claim that r0 = r1 = 0R. Suppose that r1 6= 0R. Then

δ(b) ≤ δ(r1b) = δ(−r0) = δ(r0) < δ(b), which is a contradiction. Thus r1 = 0R, and then also r0 = 0R. Now suppose the result holds for k ∈ {0, 1, . . . , m} and consider the expression

2 m+1 m 0R = r0 + r1b + r2b + ··· + rm+1b = (r1 + r2b + ··· + rm+1b )b + r0.

Since the result holds for k = 1, it follows that

m r1 + r2b + ··· + rm+1b = 0R, r0 = 0R.

By the induction hypothesis, r1 = r2 = ··· = rm+1 = 0R, and so the result follows. H Now we carry on with the uniqueness part of the proof. First consider the case when δ(a) = δ(1R). Then, since b is a nonzero nonunit, δ(a) < δ(b). Suppose that

2 k k−1 a = r0 + r1b + r2b + ··· + rkb = (r1 + r2b + ··· + rkb )b + r0 (1) for r0, r1, . . . , rk ∈ S with δ(r0), δ(r1), . . . , δ(rk) < δ(b). By Proposition4 there is only one way to express a as qb + r with δ(r) < δ(b) and with q, r ∈ S, and from the existence part of the proof we know that this implies that

k−1 r1 + r2b + ··· + rkb = 0R, r0 = a.

By the lemma we can then assert that r1 = ··· = rk = 0R, and so we must have k = 0 and r0 = a as the unique solution to (1). Thus the result holds when δ(a) = δ(1R). Next suppose the result true for δ(a) ∈ {δ(1R, . . . , m}, and suppose that a ∈ S satisﬁes

δ(a) = inf{δ(r) | r ∈ S, δ(r) > m}.

Then suppose that

k 0 0 0 k0 a = r0 + r1b + ··· + rkb = r0 + r1b + ··· + rk0 b

0 0 0 0 0 for k, k ∈ Z≥0, r0, r1, . . . , rk ∈ S, and r0, r1, . . . , rk0 ∈ S satisfying δ(rj), δ(rj0 ) < δ(b) for 0 0 0 j ∈ {0, 1, . . . , k} and j ∈ {0, 1, . . . , k }. Also suppose that rk, rk0 6= 0R. Then

k−1 0 0 0 k0−1 0 (r1 + r2b + ··· + rkb ) b + r0 = (r1 + r2b + ··· + rk0 b ) b + r0. | {z } | {z } q q0 6 A. D. Lewis

0 0 By Proposition4 we have q = q and r0 = r0. First suppose that δ(a) < δ(b). Then, 0 0 by Proposition4, we have q = q = 0R and r0 = r0 = a. By the lemma it follows that 0 0 0 0 r1 = ··· = rk = 0R and r1 = ··· = rk0 = 0R, and so we have k = k = 0 and r0 = r0 = a. 0 Next suppose that δ(a) ≥ δ(b). Then it follows that q, q 6= 0R, since otherwise we have 0 0 a = r0 = r0, contradicting the fact that δ(r0), δ(r0) < δ(b). Then we have

δ(q) < δ(qb) = δ(a − r0) ≤ max{δ(a), δ(r0)} = δ(a)

0 since b is a nonzero nonunit and since δ(a) ≥ δ(b) > δ(r0). Similarly, δ(q ) < δ(a). There- fore, the induction hypothesis applies to q and q0 and we conclude that k − 1 = k0 − 1 and 0 rj = rj for j ∈ {1, . . . , k}, so proving the uniqueness part of the result by induction on δ(a). The preceding base expansion result has the following consequence which will be useful to us in our proof below of the norm bounds in the Euclidean Algorithm. 7 Proposition: Let (R, δ) be a Euclidean domain, let S ⊆ R be a nontrivial, δ-closed, and δ-positive subsemiring of R, and let

U = {r ∈ S | r is a unit} ∪ {0R}. If U ⊂ S and if x ∈ S satisﬁes

δ(x) = inf{δ(r) | r ∈ S, δ(r) > δ(1R)}, then, for a ∈ S \{0R}, there exists a unique k ∈ Z≥0 and c0, c1, . . . , ck ∈ U such that

(i) ck 6= 0R and k (ii) a = c0 + c1x + ··· + ckx .

Moreover, if U ⊂ S and if a, b ∈ S \{0R} are written as k l a = c0 + c1x + ··· + ckx , b = d0 + d1x + ··· + dlx for c0, c1, . . . , ck, d0, d1, . . . , dl ∈ U such that ck, dl 6= 0R, then δ(a) > δ(b) if and only if k > l. k Proof: Since x is a nonzero nonunit, from Proposition6 we can write a = c0 +c1x+···+ckx for unique c0, c1, . . . , ck ∈ S with ck 6= 0R and δ(c0), δ(c1), . . . , δ(ck) < δ(x). The hypotheses on x immediately give c0, c1, . . . , ck ∈ U. Now let a and b be as stated in the second assertion and write a = qb + r for q, r ∈ S with δ(r) < δ(b), this being possible by δ-closedness of S. Let us assume that δ(a) > δ(b). We will show by induction on δ(b) that k > l. First suppose that δ(b) = δ(1R) so that b ∈ U. Since δ(a) > δ(b) it follows that a is a nonzero nonunit and so, by the ﬁrst part of the result, k > 1, giving the result in this case. Assume the result holds for δ(b) ∈ {δ(1R), . . . , n} and suppose that δ(b) = inf{δ(r) | r ∈ c, δ(r) > n}.

We claim that the hypothesis that δ(a) > δ(b) implies that q is a nonzero nonunit. If q = 0R then a = r and so δ(b) > δ(r) = δ(a), in contradiction with our assumption. If q is a unit then δ(b) = δ(qb) = δ(a − r) = δ(a), Subsets of Euclidean domains possessing a unique division algorithm 7 the last equality holding since δ(r) < δ(b) < δ(a) and since δ(a − r) ≤ max{δ(a), δ(r)} by δ-positivity of S. Thus q being a unit leads to the contradiction δ(b) = δ(a). Since q is a m nonzero nonunit, by the ﬁrst conclusion of the proposition we have q = u0 +u1x+···+umx for m ∈ Z>0 with u0, u1, . . . , um ∈ U and um 6= 0R. Since δ(r) < δ(b) the induction p hypotheses imply that r = v0 + v1x + ··· + vpx for p < l with v0, v1, . . . , vp ∈ U and vp 6= 0R. Therefore,

k a = c0 + c1x + ··· + ckx m l p = (u0 + u1x + ··· + umx )(d0 + d1x + ··· + dlx ) + v0 + v1x + ··· + vpx , from which we deduce that k > l since R is a domain and since p < m + l. Now assume that k > l. Let us write

m p q = u0 + u1x + ··· + umx , r = v0 + v1x + ··· + vpx with u0, u1, . . . , um, v0, v1, . . . , vp ∈ U and um, vp 6= 0R. Since δ(r) < δ(b) the previous part of the proof gives p < l. By the uniqueness part of Proposition6 we must have m = k−l > 0. Therefore, again by the uniqueness part of Proposition6, we conclude that q is not a unit and so δ(q) > δ(1R). Therefore,

δ(b) < δ(qb) = δ(a − r) ≤ max{δ(a), δ(r)} = δ(a), the last equality holding since δ(r) < δ(b). This gives the result. Note that as a consequence of this, the characterisation of Jodeit Jr. [1967] of Euclidean rings admitting a unique division algorithm follows straightforwardly. 8 Corollary: If (R, δ) is a Euclidean domain that admits a unique Division Algorithm, then

(i) the set of units in R forms a ﬁeld which we denote by FR and

(ii) if FR ⊂ R then R is isomorphic to FR[ξ].

Proof: We claim that R admits a unique Division Algorithm if and only if δ(a + b) ≤ max{δ(a), δ(b)} for every a, b ∈ R. Certainly, if δ(a+b) ≤ max{δ(a), δ(b)} for every a, b ∈ R, then R is a δ-closed and δ-positive subsemiring of itself, and then uniqueness of quotient and remainder follows from Proposition4. Conversely, suppose that a, b ∈ R \{0R} satisfy δ(a + b) > max{δ(a), δ(b)}. Then we can write a = 0R · (a + b) + a with δ(a) < δ(a + b) and also a = 1R · (a + b) + (−b) with δ(−b) < δ(a + b). Thus R does not admit a unique Division Algorithm. That the units in R form a ﬁeld will follow if we can show that, if units a, b ∈ R satisfy a + b 6= 0R, then a + b is a unit. This, however, follows since

δ(1R) ≤ δ(a + b) ≤ max{δ(a), δ(b)} = δ(1R), and so δ(a + b) = δ(1R), implying that a + b is a unit. The ﬁnal assertion of the corollary follows from Proposition7 since every r ∈ R can be written as k r = a0 + a1x + ··· + akx 8 A. D. Lewis

for unique a0, a1, . . . , ak ∈ FR with ak 6= 0R and with x as deﬁned in the statement of Proposition7. We then easily see that the map

k k R 3 a0 + a1x + ··· + akx 7→ a0 + a1ξ + ··· + akξ ∈ FR[ξ] is the desired isomorphism.

The ﬁnal theorem we state concerns solutions to B´ezout’sidentity, which states that, if a, b ∈ R are elements of a principal ideal domain, then a and b are coprime if and only if there exists r, s ∈ R such that ra + bs = 1R. One way to compute r and s for Euclidean domains involves the Euclidean Algorithm. To establish notation, let us recall that the Euclidean Algorithm states that, if (R, δ) is a Euclidean domain and if a, b ∈ R with b 6= 0R, then there exists k ∈ Z≥0, q0, q1, . . . , qk ∈ R, and r0 = b, r1, . . . , rk ∈ R \{0R} such that

a = q0r0 + r1, δ(r1) < δ(r0),

r0 = q1r1 + r2, δ(r2) < δ(r1), . . (2)

rk−2 = qk−1rk−1 + rk, δ(rk) < δ(rk−1),

rk−1 = qkrk.

Moreover, it turns out that rk as it appears in the Euclidean Algorithm is a greatest common divisor for a and b. In particular, if a and b are coprime, then rk is a unit. Moreover, as we shall see in our next theorem, one can use the Euclidean Algorithm to ﬁnd r, s ∈ R such that ra + bs = 1R. In a Euclidean domain one can ask that r and s have norms satisfying some bound; the usual bounds are that δ(r) < δ(b) and δ(s) < δ(a). As we see in the following theorem, if one enforces δ-positivity, then the bounds are achieved by the (necessarily unique) solution obtained from the Euclidean Algorithm. Again, most of the steps in this theorem may be found in any textbook, but we give all of the details so as to reveal where the property of δ-positivity is used.

9 Theorem: If (R, δ) is a Euclidean domain and if a, b ∈ R\{0R} are coprime, let k ∈ Z≥0, q0, q1, . . . , qk ∈ R, and r0 = b, r1, . . . , rk−1 ∈ R \{0R} be such that

a = q0r0 + r1, δ(r1) < δ(r0),

r0 = q1r1 + r2, δ(r2) < δ(r1), . .

rk−2 = qk−1rk−1 + u, δ(u) < δ(rk−1),

rk−1 = qku, where u ∈ R is a unit (this being the case since a and b are coprime). Then let α0 = 1R and β0 = −qk−1, and recursively deﬁne α1, . . . , αk−1 ∈ R and β1, . . . , βk−1 ∈ R by

αj = βj−1, βj = αj−1 − qk−1−jβj−1, j ∈ {1, . . . , k − 1}. Subsets of Euclidean domains possessing a unique division algorithm 9

If we take ( ( 0 , δ(b) = δ(1 ), b−1, δ(b) = δ(1 ), r = R R s = R −1 −1 u αk−1, δ(b) > δ(1R), u βk−1, δ(b) > δ(1R), then ra + sb = 1R. Moreover, if S ⊆ R is a nontrivial, δ-closed, and δ-positive subsemiring, and if a and b additionally have the property that a, b ∈ S and that at least one of a and b is not a unit, then

(i) q0, q1 . . . , qk and r1, . . . , rk−1 may be chosen to lie in S and,

(ii) if q0, q1 . . . , qk and r1, . . . , rk−1 are so chosen, then r and s as deﬁned above additionally satisfy δ(r) < δ(b) and δ(s) < δ(a).

Proof: Let us ﬁrst reduce to the case when u = 1R. Multiply all equations in the Euclidean Algorithm for a and b by u−1:

−1 −1 −1 −1 −1 u a = q0u r0 + u r1, δ(u r1) < δ(u r0), −1 −1 −1 −1 −1 u r0 = q1u r1 + u r2, δ(u r2) < δ(u r1), . . −1 −1 −1 u rk−2 = qk−1u rk−1 + 1R, δ(1R) < δ(u rk−1), −1 u rk−1 = qk.

Note that the resulting equations hold if and only if the original equations hold, by virtue of R being an integral domain. The resulting equations are then the Euclidean Algorithm for −1 −1 −1 u a and u b, and at each step the remainders r0, r1, . . . , rk−1 are multiplied by u . The quotients q0, q1, . . . , qk remain the same, however. Thus the definitions of α0, α1, . . . , αk−1 and β0, β1, . . . , βk−1 are unchanged from the Euclidean Algorithm for a and b. Applying the conclusions of the theorem to the modified Euclidean Algorithm then gives r0, s0 ∈ R 0 −1 0 −1 such that r (u a) + s (u b) = 1R. Thus the conclusions of the first part of the theorem −1 0 −1 0 in the general case follow from those when u = 1R by taking r = u r and s = u s . Also −1 −1 note that the relation δ(u rj−1) < δ(u rj) is equivalent to the relation δ(rj−1) < δ(rj), j ∈ {0, 1, . . . , k − 1}. Therefore, the conclusions of the second part of the theorem in the general case also follow from those for the case when u = 1R. Thus, in the remainder of the proof we suppose that u = 1R. Let us also eliminate the case where δ(b) = δ(1R). If this is the case then we have −1 a = qb + r with δ(r) = δ(0R), and so r = 0R. Therefore, since b is a unit, q = ab . −1 Now, taking r = 0R and s = b , we have ra + sb = 1R. Moreover, for the second part of the theorem, δ(r) < δ(b) and δ(s) < δ(a) since s is a unit and a is not, the latter by the hypotheses of the theorem. Thus the conclusions of the theorem hold when δ(b) = δ(1R). Thus, in the remainder of the proof we suppose that b is a nonzero nonunit. We now prove the theorem by induction on k. If k = 1 then we have

a = q0 · r0 + 1R, δ(1R) < δ(r0),

r0 = q1. 10 A. D. Lewis

Thus 1R = 1R · a + (−q0) · b, and the theorem holds with r = α0 = 1R and s = β0 = −q0. Now suppose the theorem true for k ∈ {1, . . . , m − 1} and consider the Euclidean Algorithm for a and b = r0 of the form

a = q0r0 + r1, δ(r1) < δ(r0),

r0 = q1r1 + r2, δ(r2) < δ(r1), . .

rm−2 = qm−1rm−1 + 1R, δ(1R) < δ(rm−1),

rm−1 = qm. By the induction hypothesis, the conclusions of the theorem hold for the last m equations. But the last m equations are the result of applying the Euclidean Algorithm in the case where “a = r0” and “b = r1.” Thus, if we deﬁne α0 = 1R and β0 = −qk−1, and recursively deﬁne α1, . . . , αm−2 and β1, . . . , βm−2 by

αj = βj−1, βj = αj−1 − qm−1−jβj−1, j ∈ {1, . . . , m − 2}, 0 0 0 0 and if we take r = αm−2 and s = βm−2, then we have r r0 + s r1 = 1R. Since r0 = b we have 1R = αm−2r0 + βm−2(a − q0r0) = (αm−2 − q0βm−2)b + βm−2a, and so the theorem holds with r = αm−1 = βm−2 and s = βm−1 = αm−2 − q0βm−2, as desired. Now we proceed to the second part of the theorem, supposing that a, b ∈ S for a δ-closed and δ-positive subsemiring S ⊆ R. Since r0 = b, that q0 and r1 can be chosen to lie in S follows from the fact that S is δ-closed. This reasoning can then be applied to each line of the Euclidean Algorithm to ensure that all quotients and remainders can be chosen to lie in S. The following lemma records a useful property of these quotients and remainders. Lemma: Using the notation of the theorem statement, suppose that a, b ∈ S and that q0, q1, . . . , qk and r1, . . . , rk−1 are chosen to lie in S. Then, for j ∈ {0, 1, . . . , k − 1}, either (i) αj ∈ S and −βj ∈ S or

(ii) −αj ∈ S and βj ∈ S.

Proof: The lemma is proved by induction on j. For j = 0 we have α0 = 1R ∈ S and −β0 = qk−1 ∈ S. Suppose the lemma true for j ∈ {0, 1, . . . , m}. We have two cases. 1. αm ∈ S and −βm ∈ S: We immediately have −αm+1 = −βm ∈ S. Also, βm+1 = αm − qk−2−mβm ∈ S since αm ∈ S and qk−m−2(−βm) ∈ S, using the semiring property of S.

2. −αm ∈ S and βm ∈ S: This case follows, mutatis mutandis, in the manner of the previous case. H Now, the ﬁnal thing we need to show is that r and s constructed as above from a, b ∈ S satisfy δ(r) < δ(b) and δ(s) < δ(a). We prove this by induction on k. For k = 1 we have r = 1R and s = −q0. Therefore,

δ(r) = δ(1R) < δ(b) Subsets of Euclidean domains possessing a unique division algorithm 11 since we are assuming that b is a nonzero nonunit. Also, since b is a nonzero nonunit,

δ(s) = δ(−q0) < δ(−q0b) = δ(a − 1R) ≤ max{δ(a), δ(1R)} ≤ δ(a), using δ-positivity of S. So the ﬁnal assertion of the theorem holds for k = 1. Now suppose that this assertion holds for k ∈ {1, . . . , m − 1} and consider the Euclidean Algorithm for a and b of the form

a = q0r0 + r1, δ(r1) < δ(r0),

r0 = q1r1 + r2, δ(r2) < δ(r1), . .

rm−2 = qm−1rm−1 + 1R, δ(1R) < δ(rm−1),

rm−1 = qm.

Considering the last m equations, as in the first part of the proof we have the Euclidean 0 0 Algorithm for “a = r0” and “b = r1.” Therefore, considering r , s ∈ R as constructed in 0 0 the first part of the proof, we have δ(r ) < δ(r1) and δ(s ) < δ(r0). Again as in the first 0 0 0 part of the proof, we take r = s and s = r − q0s so that ra + sb = 1R. Then

0 δ(r) = δ(s ) < δ(r0) = δ(b).

It remains to show that δ(s) < δ(a). First suppose that δ(a) < δ(b). Then, by Proposition4 we have a = 0R · b + a as the unique output of the Division Algorithm in S. Thus we must have q0 = 0R and r1 = a. In this case,

0 δ(s) = δ(r ) < δ(r1) = δ(a), giving the norm bound for s if δ(a) < δ(b). Thus we consider the case when δ(b) ≤ δ(a). By 0 0 0 0 the lemma we have either (1) r ∈ S and −q0s ∈ S or (2) −r ∈ S and q0s ∈ S. Consider 0 0 the case r , −q0s ∈ S. We then have

0 0 a = q0b + r1, s = −q0s + r

0 0 0 0 with δ(r1) < δ(b), δ(s ) < δ(b), and δ(r ) < δ(r1). Since a, q0, b, r1, s, −s , r ∈ S we use Proposition7 to write these elements of S as uniquely deﬁned polynomials in x, where

δ(x) = inf{δ(r) | r ∈ S, δ(r) > δ(1R)}.

0 0 Let us denote these polynomials by Pa, Pq0 , Pb, Pr1 , Ps, P−s , and Pr . By Proposition7 we have

δ(r1) < δ(b) =⇒ deg(Pr1 ) < deg(Pb), 0 δ(s ) < δ(b) =⇒ deg(Ps0 ) < deg(Pb), 0 0 δ(r ) < δ(r1) =⇒ deg(Pr ) < deg(Pr1 ).

This immediately gives

0 0 deg(Pa) = deg(Pq0 ) + deg(Pb), deg(Ps) ≤ max{deg(Pq0 ) + deg(Ps ), deg(Pr )}. 12 A. D. Lewis

0 0 0 max{deg(Pq0 ) + deg(Ps ), deg(Pr )} = deg(Pq0 ) + deg(Ps ) then

0 deg(Pa) = deg(Pq0 ) + deg(Pb) > deg(Pq0 ) + deg(Ps ) ≥ deg(Ps) if

0 0 0 max{deg(Pq0 ) + deg(Ps ), deg(Pr )} = deg(Pr ) then

0 deg(Pa) = deg(Pq0 ) + deg(Pb) > deg(Pr1 ) > deg(Pr ) ≥ deg(Ps).

In either case we have deg(Ps) > deg(Pa), and then we apply Proposition7 again to give 0 0 0 0 δ(s) < δ(a) in the case when r , −q0s ∈ S. When −r , q0s ∈ S then −s ∈ S and we write

0 0 a = q0b + r1, −s = q0s + (−r ).

The steps above may now be repeated to give δ(s) = δ(−s) < δ(a) in this case.

References

Jodeit Jr., M. A. [1967] Uniqueness in the division algorithm, The American Mathematical Monthly, 74(1), pages 835–836, issn: 0002-9890, doi: 10.2307/2315810.