HIGHER EUCLIDEAN DOMAINS

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1 HIGHER EUCLIDEAN DOMAINS CHRIS J. CONIDIS Abstract. Samuel and others asked for a Euclidean domain with Euclidean rank strictly greater than ω, the smallest infinite ordinal. Via a limited technique Hiblot produced such a ring of Euclidean rank at most ω 2. Using a novel and more powerful technique, we are able to construct Euclidean domains of arbitrarily large Euclidean rank. This is a new and interesting solution to Samuel s question that is likely to have applications in Mathematical Logic. 1. Introduction Throughout this article all rings will be commutative with identity 1. The usual definition of a Euclidean domain R says that it is an integral domain with a function called a Euclidean norm ϕ : R N satisfying: (1) for any given a, b R, b 0, there exist elements q, r R so that a = bq + r and ϕ(r) < ϕ(b). One of the main reasons to consider Euclidean rings is that they form a subclass of the principal ideal domains. In other words, in a Euclidean ring R every ideal I R is generated by a single element x I. Indeed, it is well-known that x can be taken to be any element of I that minimizes ϕ(x). Thus, the guaranteed existence of x relies on the well-foundedness of N and it is natural to ask whether we obtain a larger class of principal ideal domains by replacing N by any infinite well-ordered set (i.e. infinite ordinal number). This natural question goes back to the algebraists Motzkin [Mot49] and Samuel [Sam71, page 283], who studied the different Euclidean norms that can exist on a given ring upon a suggestion of Zariski. More recently, logicians and algebraists have become interested in analyzing the complexity of (transfinite) constructions in this class of rings [DK11, Cla14]. Definition 1.1. If the ring R satisfies (1) above with a well-founded set (i.e. ordinal number) α replacing N, then we say that R is a transfinite Euclidean domain. Since we will be working exclusively with transfinite Euclidean domains throughout the rest of this article, we will refer to them simply as Euclidean domains. Let O denote the class of ordinal numbers. We will use lower case Greek letters to refer to elements of O. It is well-known that every non-empty subclass of O contains a minimal element. The same fact holds for any α O. Definition 1.2. Let R be a Euclidean domain with Euclidean norm ϕ : R α, α O. Then it follows that there is a unique minimal α 0 such that R is Euclidean with witness ϕ 0 : R α 0. We call this unique ordinal α 0 the Euclidean rank of R. There is a well-known theorem regarding minimal Euclidean functions Mathematics Subject Classification. Primary 13G05 Secondary 03D80. The author would like to thank Rod Downey and Asher Kach for introducing him to the problem of the Main Theorem, as well as Hiblot s work [Hib75]. 1

2 2 CHRIS J. CONIDIS Theorem 1.3. [Sam71, Proposition 9] Let R be a Euclidean domain with Euclidean norm ϕ : R α, α O. Now let E be the class of all Euclidean norms for R and, for each x R, define ϕ 0 (x) = min ψ E ψ(x). Then ϕ 0 is the unique (point-wise) minimal Euclidean norm for R. Motzkin gave the following constructive method for obtaining the unique minimal Euclidean norm. Theorem 1.4. [Mot49, page 1142] Let R be a ring. Define an increasing sequence of subsets of R, {S β : β O}, as follows: (1) S 0 consists of zero and the units of R; (2) given S β, define S β+1 = {x R : ( y R)( q R, r S β )[y = qx + r]}; (3) S λ = β<λ S β for all limit ordinals λ. Then R is a Euclidean domain if and only if there exists some limit ordinal λ such that R = S λ and moreover the unique minimal Euclidean function on R is given by ϕ(x) = min{β O : x S β }. In 1975 Hiblot [Hib75] gave a construction of a Euclidean domain with Euclidean rank α, ω < α ω 2, where ω = {0, 1, 2,...} denotes the ordinal with order type N (i.e. the smallest infinite ordinal number), thus answering the question of Motzkin and Samuel mentioned above. One glaring problem with Hiblot s proof, from the point of view of Mathematical Logic which aims to classify the complexity of mathematical constructions, is that it is not relativizable. In other words, one cannot repeat Hiblot s basic construction (on top of itself) to construct rings of arbitrarily large Euclidean ranks. There is an elementary explanation of this fact, as follows. Recall that Hiblot begins with a base ring R 0 of Euclidean rank ω, and then constructs his (transfinite) Euclidean domain R as a localization of the power series ring R = R 0 [[X]]. He then goes on to prove that, in his ring, the Euclidean norm of X R must always be at least ω. Now, if we wanted to repeat this argument, relative to itself, we would construct a Euclidean domain localization of R = R 0 [[X, Y ]], with irreducible elements X and Y, such that for every Euclidean norm ϕ on R we have that ϕ(x) ω (as before) and ϕ(y ) ω 2. However, note that, since R is a Euclidean ring, then it must be a principal ideal domain, from which it follows that there exist ring elements a, b such that ax + by = 1. But then we have that by = 1 ax, and ax = 1 by, and hence X and Y are invertible in R = R 0 [[X, Y ]], which implies that ϕ(x) = ϕ(y ) = 0 for the unique minimal Euclidean norm on R, a contradiction. The purpose of this article is to construct Euclidean domains of arbitrarily large Euclidean ranks by giving a completely new proof/construction of Hiblot s old result. In particular we will show how to construct commutative 1 Euclidean domains with arbitrarily large Euclidean ranks. Our construction will surely be useful to logicians and other mathematicians who seek to better understand the class of (transfinite) Euclidean rings. 1 This has already been done in the noncommutative case. See Lenstra s excellent monograph [Len74] for more details.

3 HIGHER EUCLIDEAN DOMAINS 3 2. Constructing Higher Euclidean Domains We begin this section with an important definition that will be used throughout the rest of this article. Definition 2.1. Let ρ be an ordinal. We say that ρ is terminally-admissible whenever every terminal segment of ρ has the same cardinality as ρ. This is equivalent to saying that the cofinality of ρ is equal to the cardinality of ρ, or that the cofinality of ρ is as large as possible. 2 Remark 2.2. All countable limit ordinals and cardinals are terminally-admissible. See [Kun11] for more details on ordinal and cardinal numbers, as well as the definition of cofinality. More background information on Euclidean rings and domains can be found in [Len74, Sam71]. Let ρ ρ denote the cardinality of ρ O, and if ρ is a successor ordinal we let ρ O denote its predecessor. We now proceed to construct a Euclidean domain of Euclidean rank ω λ, for every terminally-admissible ordinal λ. Theorem 2.3 (Main Theorem). For every terminally-admissible ordinal λ, there is a Euclidean domain R of Euclidean rank ω λ. Constructing R. Recall that O denotes the class of ordinals and {γ k : k O}, is an indexing of the ordinals. Now, let λ O be a fixed terminally-admissible ordinal and let Q λ denote the free polynomial ring with countably many indeterminates indexed by ordinals less than or equal to λ. In other words, Q λ = Q[X β : 0 β < λ, β O]. Recall that Q λ, as well as all of its localizations, are unique factorization domains. Let F λ denote the fraction field of Q λ and for every pair of polynomials p, q Q λ, let (p, q) = (p, q) Qλ Q λ denote the greatest common divisor of p and q. Definition 2.4. Let m = i F λ F < X n i i Q λ0 Q λ be a monomial (with finitely many indeterminate factors) in Q λ. We define the degree of m via deg(m) = n i ω γ i O. i F λ F < If p Q λ is a polynomial then we define the degree of p to be deg(p) = max{deg(m) : m is a nonzero monomial summand of p}. We also define deg(0) = 0. Note that {deg(p) : p Q λ } = ω λ. Now, let α and β be ordinals with Cantor normal form α = n i ω γ i, and β = i F 1 λ F 1 < i F 2 λ F 2 < m i ω γ i. 2 One can easily show that the cofinality of ρ is at most the cardinality of ρ.

4 4 CHRIS J. CONIDIS Recall that the Hessenberg-Brookfield ordinal sum (see [Cla14, Section 1.2] for more details) of α and β, which we will denote as α + ω β, is given by α + ω β = (n i + m i )ω γi. i F 1 F 2 This is essentially the base-ω summation of α and β and is generally different from the usual ordinal sum of α and β. It follows that if m 1 and m 2 are monomials in Q λ then deg(m 1 m 2 ) = deg(m 1 )+ ω deg(m 2 ) (we leave the simple verification of this fact, which follows directly from the definitions of the concepts involved, to the reader). The main reasons why + ω is so important for us is because (1) it is commutative; and moreover (2) it enables us to employ various forms of subtraction. For example, let α, β, γ O, then we have that α + ω β = α + ω γ implies β = γ and also α + ω β α + ω γ implies β γ. Let {α k : k n}, n N, be a finite sequence of ordinal numbers. Then we will write n to denote the sum k=0 ω α k α 0 + ω α 1 + ω + ω α n. Let {a β,n,ρ : β, ρ < λ, n < ω} and {c β,n : β < λ, n < ω} be disjoint sequences of ordinal numbers < λ with no repetitions such that (1) for each fixed β 0 < λ, n 0 < ω we have that {a β0,n 0,ρ } ρ<λ is a strictly increasing sequence of ordinals cofinal in λ; and (2) for each fixed β 0 < λ we have that {c β0,n} n<ω is a strictly increasing sequence between β 0 and < β 0 + ω. The existence of such sequences follows from our hypothesis that λ is terminally-admissible. Moreover, for each β 0 < λ, n < ω, we have that every tail sequence of the form {a β0,n 0,ρ } ρ<λ has the same cardinality as Q λ, namely λ = Q λ. The ring R F will be a localization of Q λ. More specifically, we will construct sets U β = n<ω U β,n Q λ, β < λ, such that R = Q λ [U0 1 ] F and U β = ϕ 1 (β), β < λ, where ϕ : R ω λ is a Euclidean function for R. Let R 0 = Q λ. We now construct R and {U β : β < λ} as the limits of approximations in stages via recursion, R = n<ω R n F λ, U β = lim n U β,n = {x R : ( α 0 )( α 1 α 0 )[x U β,α1 ]} Q λ Q λ, where U β,α = ϕ 1 α (β), and Stage n = 0: First of all, define ϕ = lim α ϕ α, ϕ α : R α ω λ. ϕ 0 (p) = deg(p),

5 HIGHER EUCLIDEAN DOMAINS 5 for each p Q λ. It is not difficult to see that for any p, q Q λ we have that ϕ 0 (pq) = ϕ 0 (p) + ω ϕ 0 (q). Now, for each β < λ, set U β,0 = ϕ 1 0 (β) R 0 = Q λ. This completes our constructions of {U β,0 : β < λ}, ϕ 0, and R 0 = Q λ. We now proceed to stage n + 1 ω and the construction of U n+1, V n+1, ϕ n+1, and R n+1. We will work inside the (given) unique factorization domain R n. Stage n + 1: First we construct ϕ n+1, in substages γ λ as follows Constructing ϕ n+1. This subsection is perhaps the most important in this article, because it gives the key ingredient in our construction of ϕ that will allow us to prove our Main Theorem below. We will refer back to this subsection, and in particular our construction of δ γ,1 below, later on in the proof of our Main Theorem. The following combinatorial fact will be useful later on and essentially lies at the heart of all of our most important arguments. Theorem 2.5 (König s Lemma). If T is a finitely branching tree with no infinite paths, then T is finite i.e. T has finitely many nodes. In other words, every infinite finitely-branching tree has an infinite path. Our construction at stage n+1 proceeds in substages γ < λ. At substage γ < λ we choose the least ordered pair of polynomials p, q Q 2 λ R2 α that we have not yet considered at any previous substage (relative to some previously chosen fixed ordering) and such that: (1) 1 ϕ α (q) ϕ α (p); and (2) (p, q) Qλ = 1. Now, let (a) δ γ,0 = ϕ n (q), if ϕ n (q) is a successor ordinal; (b) δ γ,0 = 1, otherwise. Note that in either case we have that δ γ,0 < ϕ n (q). In this paragraph we will construct another ordinal number, δ γ,1, as follows. To do this we will refer to special indices, special indeterminates, and special factors/polynomials 3 defined later on after our (recursive) construction of ϕ α+1 at substage γ. To construct δ γ,1 we first construct a set/tree of special indices in stages as follows. At stage s = 0 we let S 0 be the indices of all indeterminates appearing in either p or q. Then, at stage s + 1 we are given a set of ordinals S s and we add to it all indices of indeterminates appearing in special factors with special indeterminates already in S s. One can (and should) think of S s as collecting ordinal indices of indeterminates along a tree of special factors, beginning with the indices appearing in either p or q. By the way that we (will) construct special polynomials (see below for more details), we (will) have that the unique special indeterminate of a given special factor is always a strict upper bound for the indeterminates appearing in that factor, and hence via König s Lemma it follows that there is some s N for which S s = S s+1 = S is finite and our construction terminates (our tree of special indices can have no infinite paths since the special indices strictly decrease along paths). Finally, we define δ γ,1 to be the maximum of the finitely many ordinals in S that are strictly less than ϕ n (q). Let ɛ γ = max{δ γ,0, δ γ,1 }. It is not difficult to verify that 1 δ γ,0, δ γ,1 < ϕ n (q), from which it follows that 1 ɛ γ < ϕ n (q). 3 Special indices will be the ordinal indices of special indeterminates corresponding to special factors. Special factors are irreducible polynomials of the form p Xq, for polynomials p, q with (p, q) = 1, and special indeterminate X = X τ with special index τ O strictly larger than any index appearing in either p or q.

6 6 CHRIS J. CONIDIS Now, choose an ordinal ρ < λ large enough so that a = a γ = a ϕn(q),n,ρ is larger than the index of any indeterminate appearing in either p or q, and different from any a γ, γ < γ, and set ϕ n+1 (p X a q) = ɛ γ < ϕ n (q). Note that p X a q is irreducible since (p, q) Qλ = 1 and the ordinal index a was chosen strictly larger than that of any indeterminate appearing in either p or q (see property (3) above). Also note that if ϕ n (p X a q) is a successor ordinal then it follows that ϕ n+1 (p X a q) = ϕ n (p x a q). We declare the irreducible polynomial p X a q to be a special factor, or special polynomial, with (unique) special indeterminate X a larger than any indeterminate appearing in either p or q, and special index a = a γ. These declarations remain valid at all future stages n n+1 and substages γ. Finally, at substage γ = λ we extend ϕ n+1 to all of Q λ by first setting ϕ n+1 (p 0 ) = ϕ n (p 0 ), for any irreducible polynomial p 0 Q λ R such that ϕ n+1 (p 0 ) is currently undefined, and then setting N ϕ n+1 (p) = ϕ n+1 (p i ), whenever is the unique factorization of p Q λ. p = i=0 ω N i= Constructing R n+1 and R. Let W n+1 Q λ consist of all those polynomials p such that ϕ n+1 (p) = 0 (so far at stage n + 1). Now, we define p i R n+1 = R n [W 1 n+1]. By induction it follows that ϕ n+1 (p) = 0 iff p Q λ is a unit in R n+1. We extend ϕ n+1 to all of R n+1 via ϕ n+1 ( p q ) = ϕ n+1(p), as required by any Euclidean function on a localization. Finally, for each β < λ we set Note that for any x, y R n+1 we have that U β,n+1 = ϕ 1 n+1(β). ϕ n+1 (xy) = ϕ n+1 (x) + ω ϕ n+1 (y). This completes our construction of ϕ n+1, R n+1, and {U β,n+1 : β < λ}. Finally, we set R = n<ω R n, and ϕ = lim n ϕ n. The following lemma and its corollary below essentially implies that ϕ is well-defined. Their proofs follow directly from our construction of ϕ 0 = deg and ϕ n+1 above and are left to the reader. Lemma 2.6. Let p 0 Q λ R be an irreducible polynomial, and let ρ < λ be the largest ordinal index of any indeterminant appearing in p 0. Then we can write p 0 = q 0 + X ρ q 1,

7 HIGHER EUCLIDEAN DOMAINS 7 for some unique polynomials q 0, q 1 Q λ. Now, for any polynomial p 0 Q λ R we have that ϕ(p 0 ) < deg(p 0 ) iff there exists a unique n N such that ϕ n+1 (p 0 ) < ϕ n (p 0 ), and this implies that ρ = a ϕn(q 1 ),n,ρ 1 = a deg(q1 ),n,ρ 1, for some ρ 1 < λ and p 0 is a special polynomial (as defined above). Corollary 2.7. (1) For each (fixed) x R, the values ϕ n (x) < ω λ decrease as n ω increases. (2) If x R is irreducible then there is at most one n ω such that ϕ n+1 (x) < ϕ n (x) and such an n exists iff x is a special factor. (3) For all x R the number of n ω such that ϕ n+1 (x) < ϕ n (x) is equal to the number of special factors dividing x. (4) lim n ϕ n (x) exists, for all x R. (5) For each pair of elements x, y R, ϕ(xy) = ϕ(x) + ω ϕ(y). We now verify that ϕ is a Euclidean function for R, making R a Euclidean domain. Verifying that R is a Euclidean Domain: Lemma 2.8. (R, ϕ) is a Euclidean domain. Proof. First of all, note that R is the (transfinite) union of unique factorization domains, and hence is a unique factorization domain itself. So it suffices to show that ϕ is a Euclidean norm for R. First of all, note that, by our construction of R = α<λ R α above it follows that ϕ(u) = 0 if and only if u R is a unit (in this case it follows that u is a product of elements of U 0 R). Now, to show that ϕ is a Euclidean function for R it suffices to show that whenever p, q R are such that ϕ(p) ϕ(q) 1, then there exists X R such that ϕ(p Xq) < ϕ(q). To prove this, first of all note that without any loss of generality we may assume that p and q are in Q λ R. Otherwise, if p = p 1 p 2 and q = q 1 q 2, with p 1, p 2, q 1, q 2 Q λ R and p 2, q 2 U, then it follows from our construction of ϕ above (or any other potential Euclidean function ϕ for R) that ϕ(p 1 q 2 ) = ϕ(p 1 ) = ϕ(p) ϕ(q) = ϕ(q 1 ) = ϕ(q 1 p 2 ). Hence, if we could find an X R such that ϕ(p 1 q 2 Xq 1 p 2 ) < ϕ(q 1 p 2 ) = ϕ(q 1 ) = ϕ(q) then (for this X) we would have that ( p1 ϕ(p Xq) = ϕ X q ) ( ) 1 p1 q 2 Xq 1 p 2 = ϕ = ϕ(p 1 q 2 Xq 1 p 2 ) < ϕ(q), p 2 q 2 p 2 q 2 thus proving the lemma. So we may assume that p, q Q λ R. Let p, q Q λ R be nonunits such that ϕ(p) ϕ(q) 1. To show that ϕ is a Euclidean function for R we need to show that there exists X R such that ϕ(p Xq) < ϕ(q). Without any loss of generality we may assume that (p, q) Qλ = 1 because it always suffices to find such an X for p and q, d = (p, q) d d Q λ, in place of p and q, respectively. Here we are using that fact that ϕ(xy) = ϕ(x) + ω ϕ(y), for all x, y R (see Lemma 2.6 and Corollary 2.7 above for more details).

8 8 CHRIS J. CONIDIS Now, assume that (p, q) Qλ = 1, and let k ω, k > 0, be such that for all k k we have that ϕ(p) = ϕ k (p) ϕ k (q) = ϕ(q) 1. By our construction of ϕ k above, we have that ϕ τ (p X a q) < ϕ τ (q), for some indeterminant X a with a = a ϕk (q),k,ρ, for some ordinal ρ < λ, and a O larger than any indeterminant appearing in either p or q. Now, by our construction of k and ϕ k (above) it follows that as required. Therefore, R is Euclidean via ϕ. ϕ(p X a q) = ϕ k (p X a q) < ϕ k (q) = ϕ(q), For the remainder of this section we turn our attention to proving that R has Euclidean rank ω λ. The Euclidean Rank of R is ω λ : The following definition will help us to prove the last part our Main Theorem asserting that the Euclidean rank of R is ω λ. Definition 2.9. Let ψ 0, ψ 1 be Euclidean functions on a ring T. Then ψ 0 = ψ1 whenever we have that, for all x T, there is no limit ordinal between ψ 0 (x) and ψ 1 (x). In other words, for all x T, ψ 0 (x) and ψ 1 (x) differ by a finite number. We write ψ 0 ψ 1 to indicate that it is not the case that ψ 0 = ψ1. Proposition The Euclidean rank of R is ω λ. To prove the proposition it suffices to prove the following lemma. Lemma Let ψ be the unique minimal Euclidean function on our Euclidean domain R, and let ϕ be the Euclidean function on R constructed above. Then ψ = ϕ. Proof. Suppose, for a contradiction, that ψ and ϕ satisfy the hypotheses of the theorem but ψ ϕ. Then, for some x R, there is a limit ordinal λ x, ψ(x) < λ x ϕ(x). Without any loss of generality we can take ψ(x) to be minimal with respect to this property. In other words, if µ x x is the greatest limit ordinal less than or equal to x, then we have that for all y R, if ϕ(y) µ x, then ψ(y) µ x. Now, since R is Euclidean with respect to ψ, for any ordinal c in the sequence {c ψ(x),n } n ω such that ψ(x) c < λ x we have that (1) wx c + qx = v, or w 0 w n X c + qx = v 0 v m, where w, q, v Q λ R, w = w 0 w 1 w m, w, w 0,..., w m U Q λ R irreducible in Q λ, v = v 0,..., v n, v 0,..., v m irreducible in Q λ \ U, ψ(v) < ψ(x) < ψ(x), and moreover ψ(v i ) = ϕ(v i ), 0 i m (by minimality of ψ(x)). Note that there are infinitely many candidates for the ordinal c and that equation (1) above is actually a (polynomial) equation in Q λ (we have cleared denominators in R, putting us in Q λ ). Note that we have not made any claims about ψ(x c ), although we know that ϕ(x c ) = deg(x c ) = ω c c. By choosing c not appearing in x, we may assume (without any loss of generality) that (wx c, qx) = (qx, v) = (wx c, v) = 1.

9 HIGHER EUCLIDEAN DOMAINS 9 Now, let Q c = Q[X ρ : ρ c], and let h c : Q λ Q c be the unique homomorphism Q λ Q λ that fixes all X a, a c, and maps X c to zero. Then, if we chose c such that the indeterminant X c does not appear in the polynomial x above (which is possible, since there are infinitely many choices for c and only finitely many indeterminants appearing in x), then it follows from equation (1) above that h c (q)x = h c (v). Thus x divides h c (v), which implies that h c (v) v, from which it follows that X c appears in v and hence X c appears in some irreducible v i0 Q λ, 0 i 0 m (possibly a unit in R). We claim that v i0 is a special factor. For suppose not, then we would have that ϕ(v i0 ) = deg(v i0 ) and thus ψ(v) ψ(v i0 ) = ϕ(v i0 ) = deg(v i0 ) c ψ(x), contradicting the fact that ψ(v) < ψ(x). We have established that v i0 is special with X c appearing and so v i0 = a 0 + X β0 b 0, for some a 0, b 0 Q λ R, (a 0, b 0 ) = 1, β 0 > c (note that c β 0 because our sequences {a β,n,l } and {c β,l } above are constructed disjoint from each other β 0 is a member of the former sequence while c is a member of the latter). We can now define a finite sequence of ordinals β 0 < β 1 < < β N, N N, as follows: (1) β 0 is taken from v i0 as above, and (2) Given β i, β i+1 > β i is the largest ordinal index of a special indeterminant in a special factor of the form v i or w j, 0 i m, 0 j n, in equation (1) above, (3) If β i+1 in (2) above does not exist, then terminate the sequence at β i = β N, N N. Now, by our construction of β N above it follows that X βn is the special indeterminant in a special factor of the form v i or w j, 0 i m, 0 j n in equation (1) above. Set β N = β and write z = a+x β b for this special factor. Now, by our construction of special factors and special indeterminants it follows that if z = w j, 0 j n, then X β cannot appear in any v i, 0 i m, and vice versa. Furthermore, without any loss of generality (i.e. by repeating the current proof with sufficiently many candidates for c, and resulting β = β c = β Nc ) we may assume that X β is not the index of any indeterminant appearing in the polynomial x (recall that first we fixed the polynomial x with ψ(x) minimal, and then we chose one of infinitely many candidates for c). 4 Hence, (without loss of generality) the only polynomial in which X β can possibly appear in equation 1 above is q. It follows that X β must appear in q (corresponding to its other appearance in either v i or w j ). But now we can substitute q = q 0 + X β q 1 and z = a + X β b in equation (1) above, with a, q 0, and either v or w (but not both), X β -free. It follows that x must divide b, since (x, v) = (x, w) = 1. Now, if z = z c = w j, 0 j n, then it follows that ϕ(b) = 1 and hence ϕ(x) = 1, a contradiction. Therefore we have that z = v i, 0 i m. In this case, though, by our construction of β = β N > β N 1 > > β 0 above and our construction of ϕ it follows that ϕ(z) depends on β 0 which, in turn, depends upon c (see Corollary 2.7 (ii) along with our choice of ɛ γ and its role in constructing ϕ α+1 in Subsection 2.1 above). More specifically, we have that ϕ(z) c > ψ(x). 4 The pigeonhole argument outlined here is a bit more subtle than we may have let on. One needs to construct finitely many finite trees of special factors/indeterminates/indices, one for each special indeterminate appearing in x. This construction is very similar to the one we used to construct ɛ γ in subsection 2.1 above. Finally, we choose an ordinal index c not mentioned in any of the finitely many special factors corresponding to nodes of these trees.

10 10 CHRIS J. CONIDIS Now, suppose that there exists a valid candidate for c such that ψ(z) = ψ(z c ) ψ(x). Then it follows that ψ(v) ψ(v i ) = ψ(z) ψ(x), a contradiction. So we must have that for all c, µ x ψ(z c ) < ψ(x), where µ x is the greatest limit ordinal ψ(x). Let m ω be such that ψ(x) = µ x + m. We can replace x by any z c, c µ x + m + 1 (our only assumptions on c-values were that there were infinitely many and ϕ(x c ) ψ(x)), and repeat the argument above (m+1)-many times to eventually obtain z = z c such that ϕ(z) µ x but ψ(z) < µ x, a contradiction to the minimality of x 5. Note that in every iteration of z in our argument, X c will always appear in z, and so by our construction of ϕ and δ γ,1 in Section 2.1 above, it follows that the first iteration of z for which ϕ(z) < λ x must also satisfy ϕ(z) c = µ x + m + 1. Furthermore, at all subsequent iterations ϕ(z) is exactly one less than the previous iteration while ψ(z) < ψ(x) < c is at least one less than the previous iteration. This completes our proof of Lemma 2.11 above, thus proving our Main Theorem. References [Cla14] P.L. Clark. Note on euclidean order types. Order, To appear. [DK11] R.G. Downey and A. Kach. Euclidean functions of computable euclidean domains. Notre Dame J. Form. Log., 52: , [Hib75] J.-J. Hiblot. Des anneaux euclidiens dont le plus petit algorithme nest pas valeurs finies. C.R. des Sciences A, 281: , [Kun11] K. Kunen. Set Theory. Studies in Logic. College Publications, [Len74] H.L. Lenstra. Lectures on Euclidean Rings. University of Bielefeld, [Mot49] Th. Motzkin. The euclidean algorithm. Bull. Amer. Math Soc., 55: , [Sam71] P. Samuel. About euclidean rings. Journal of Algebra, 19: , Note that, by our construction (i.e. the minimality) of x above it follows that there is an ordinal τ, µ x τ ψ(x), such that whenever ϕ(y) τ it follows that ψ(y) τ. Note that this part of the proof is similar in spirit to the verification of Euclid s Algorithm, since we construct new divisors from old remainders, as did Euclid.