THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION: DIVISORIAL AND TOPOLOGICAL PROPERTIES

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1 ARCHIVUM MATHEMATICUM (BRNO) Tomus 40 (2004), THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION: DIVISORIAL AND TOPOLOGICAL PROPERTIES JAN SNELLMAN Abstract. We study (A, +, ), the rng of arthmetcal functons wth untary convoluton, gvng an somorphsm between (A, +, ) and a generalzed power seres rng on nfntely many varables, smlar to the somorphsm of Cashwell-Everett [4] between the rng (A, +, ) of arthmetcal functons wth Drchlet convoluton and the power seres rng C [x 1, x 2, x 3,... ] on countably many varables. We topologze t wth respect to a natural norm, and show that all deals are quas-fnte. Some elementary results on factorzaton nto atoms are obtaned. We prove the exstence of an abundance of non-assocate regular non-unts. 1. Introducton The rng of arthmetcal functons wth Drchlet convoluton, whch we ll denote by (A, +, ), s the set of all functons N + C, where N + denotes the postve ntegers. It s gven the structure of a commutatve C-algebra by component-wse addton and multplcaton by scalars, and by the Drchlet convoluton (1) f g(k) = r k f(r)g(k/r). Then, the multplcatve unt s the functon e 1 wth e 1 (1) = 1 and e 1 (k) = 0 for k > 1, and the addtve unt s the zero functon 0. (2) Cashwell-Everett [4] showed that (A, +, ) s a UFD usng the somorphsm (A, +, ) C [x 1, x 2, x 3,... ], where each x corresponds to the functon whch s 1 on the th prme number, and 0 otherwse Mathematcs Subject Classfcaton: 11A25, 13J05, 13F25. Key words and phrases: untary convoluton, Schauder Bass, factorzaton nto atoms, zero dvsors. Receved January 28, 2002.

2 162 J. SNELLMAN (3) Schwab and Slberberg [9] topologsed (A, +, ) by means of the norm f = 1 mn k f(k) 0 }. They noted that ths norm s an ultra-metrc, and that ( (A, +, ), ) s a valued rng,.e. that 1. 0 = 0 and f > 0 for f 0, 2. f g max f, g }, 3. fg = f g. They showed that (A, ) s complete, and that each deal s quas-fnte, whch means that there exsts a sequence (e k ) k=1, wth e k 0, such that every element n the deal can be wrtten as a convergent sum k=1 c ke k, wth c k A. In ths artcle, we treat nstead (A, +, ), the rng of all arthmetcal functons wth untary convoluton. Ths rng has been studed by several authors, such as Vadyanathaswamy [11], Cohen [5], and Yocom [13]. We topologse A n the same way as Schwab and Slberberg [9], so that (A, +, ) becomes a normed rng (but, n contrast to (A, +, ), not a valued rng). We show that all deals n (A, +, ) are quas-fnte. We show that (A, +, ) s somorphc to a monomal quotent of a power seres rng on countably many varables. It s présmplfable and atomc, and there s a bound on the lengths of factorzatons of a gven element. We gve a suffcent condton for nlpotency, and prove the exstence of plenty of regular non-unts. Fnally, we show that the set of arthmetcal functons supported on square-free ntegers s a retract of (A, +, ). 2. The rng of arthmetcal functons wth untary convoluton We denote the ntegers by Z, the non-negatve ntegers by N, and the postve ntegers by N +. Let p be the th prme number. Denote by P the set of prme numbers, and by PP the set of prme powers. The nteger 1 s not a prme, nor a prme power. Let ω(r) be the number of dstnct prme factors of r, wth ω(1) = 0. Defnton 2.1. If k, m are postve ntegers, we defne ther untary product as km gcd(k, m) = 1 (4) k m = 0 otherwse If k m = p, then we wrte k p and say that k s a untary dvsor of p. The so-called untary convoluton was ntroduced by Vadyanathaswamy [11], and was further studed Eckford Cohen [5]. Defnton 2.2. A = f : N + C}, the set of complex-valued functons on the postve ntegers. We defne the untary convoluton of f, g A as (5) (f g)(n) = f(m)g(n) = f(d)g(n/d) d n m p=n m,n 1

3 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION 163 and the addton as (f + g)(n) = f(n) + g(n). The rng (A, +, ) s called the rng of arthmetc functons wth untary convoluton. Defnton 2.3. For each postve nteger k, we defne e k A by 1 k = n (6) e k (n) = 0 k n We also defne 1 0 as the zero functon, and 1 as the functon whch s constantly 1. Lemma s the addtve unt of A, and e 1 s the multplcatve unt. We have that 1 n = k 1 k 2 k r and gcd(k, k j ) = 1 (7) (e k1 e k2 e kr )(n) = for j 0 otherwse hence (8) e k1 e k2 e kr = e k1k 2 k r f gcd(k, k j ) = 1 for j 0 otherwse Proof. The frst assertons are trval. We have [10] that for f 1,..., f r A, (9) (f 1 f r )(n) = f 1 (a 1 ) f r (a r ) Snce (7) follows. a 1 a r=n e k1 (a 1 )e k2 (a 2 ) e kr (a r ) = 1 ff : k = a, Lemma 2.5. For n N +, e n can be unquely expressed as a square-free monomal n e k k PP } (we use the conventon that the empty product corresponds to the multplcatve unt e 1 ). Proof. By unque factorzaton, there s a unque way of wrtng n = p a1 1 and (8) gves that e n = e a p 1 p ar = e a 1 r p 1 e p ar. 1 r p ar r, Theorem 2.6. (A, +, ) s a quas-local, non-noetheran commutatve rng havng dvsors of zero. The unts U(A) conssts of those f such that f(1) 0. 1 In [10], 1 s denoted e, and e 1 denoted e 0.

4 164 J. SNELLMAN Proof. It s shown n [10] that (A, +, ) s a commutatve rng, havng zerodvsors, and that the unts conssts of those f such that f(1) 0. If f(1) = 0 then (f g)(1) = f(1)g(1) = 0. Hence the non-unts form an deal m, whch s then the unque maxmal deal. We wll show (Lemma 3.10) that m contans an deal (the deal generated by all e k, for k > 1) whch s not fntely generated, so A s non-noetheran. 3. A topology on A The results of ths secton are nspred by [9], were the authors studed the rng of arthmetcal functons under Drchlet convoluton. We ll use the notatons of [3]. We regard C as trvally normed. Defnton 3.1. Let f A \ 0}. We defne the support of f as (10) supp(f) = n N + f(n) 0 }. We defne the order 2 of a non-zero element by (11) We also defne the norm of f as (12) and the degree as (13) N(f) = mn supp(f). f = N(f) 1 D(f) = mn ω(k) k supp(f) }. By defnton, the zero element has order nfnty, norm 0, and degree. Lemma 3.2. The value semgroup of (A, ) s a dscrete subset of R +. A \ 0} = 1/k k N + }, Lemma 3.3. Let f, g A \ 0}. Let N(f) =, N(g) = j, so that f() 0 but f(k) = 0 for all k <, and smlarly for g. Then, the followng hold: () N(f g) mn N(f), N(g)}. () N(cf) = N(f) for c C \ 0}. () N(f) = 1 ff f s a unt. (v) N(f g) = N(f)N(g) N(f g), wth equalty ff gcd(, j) = 1. (v) N(f g) max N(f), N(g)}, wth strct nequalty ff both f and g are non-unts. (v) D(f + g) mn D(f), D(g)}. (v) D(f) = 0 f and only f f s a unt. (v) D(f g) D(f) + D(g) max D(f), D(g)}, wth D(f) + D(g) > max D(f), D(g)} f f, g are non-unts. 2 In [10] the term norm s used.

5 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION 165 Proof. (), (), and () are trval, and (v) s proved n [10]. If ω(s) < mn D(f), D(g)} then so s supp(f) supp(g), (f + g)(s) = f(s) + g(s) = 0. Ths proves (v). Snce f s a unt ff f(1) 0, (v) follows. For any a n the support of f and any b n the support of g, such that a b 0, we have that ω(a b) = ω(a) + ω(b) D(f) + D(g). Ths proves the frst nequalty of (v). Usng (v) the other asserton follows. (v) s proved smlarly. Corollary 3.4. f g f g = f g. Proposton 3.5. s an ultrametrc functon on A, makng (A, +, ) a normed rng, as well as a fathfully normed, b-separable complete vector space over C. Proof. ((A, +, ), ) s a valuated rng, and a fathfully normed complete vector space over C [9]. It s also separable wth respect to bounded maps [3, Corollary 2.2.3]. So (A, +) s a normed group, hence Corollary 3.4 shows that (A, +, ) s a normed rng. Note that, unlke ((A, +, ), ), the normed rng ((A, +, ), ) s not a valued rng, snce e 2 e 2 = 0 = 0 < e 2 2 = 1/4. In fact, defnng f n to be the n th untary power of n, we have that Lemma 3.6. If f s a unt, then 1 = f n = f n for all postve ntegers n. If n s a non-unt, then f n < f n for all n > 1. Proof. The frst asserton s trval, so suppose that f s a non-unt. From Corollary 3.4 we have that f n f n. If f = 1/k, k > 1,.e. f(k) 0 but f(j) = 0 for j < k, then f 2 (k 2 ) = 0 snce gcd(k, k) = k > 1. It follows that f 2 > f 2, from whch the result follows. Recall that n a normed rng, a non-zero element f s called topologcally nlpotent f f n 0, power-multplcatve f f n = f n for all n, multplcatve f fg = f g for all g n the rng. Theorem 3.7. Let f ((A, +, ), ), f 0. Then the followng are equvalent: (1) f s topologcally nlpotent, (2) f s not power-multplcatve, (3) f s not multplcatve 3 n the normed rng (A, +, ), ), 3 Ths s not the same concept as multplcatvty for arthmetcal functons,.e. that f(nm) = f(n)f(m) whenever gcd(n, m) = 1. However, snce the latter knd of elements satsfy f(1) = 1, they are unts, and hence multplcatve n the normed-rng sense.

6 166 J. SNELLMAN (4) f s a non-unt, (5) f < 1. Proof. Usng [3, 1.2.2, Prop. 2], ths follows from the prevous Lemma, and the fact that for a unt f, 3.1. A Schauder bass for (A, ). 1 = f 1 = f 1. Defnton 3.8. Let A denote the subset of A consstng of functons wth fnte support. We defne a parng (14) A A C f, g = f(k)g(k) k=1 Theorem 3.9. The set e k k N + } s an ordered orthogonal Schauder base n the normed vector space (A, ). In other words, f f A then (15) f = c k e k, c k C k=1 where () e k 0, () the nfnte sum (15) converges w.r.t. the ultrametrc topology, () the coeffcents c k are unquely determned by the fact that (16) (v) (17) f, e k = f(k) = c k max c k e k } = k N + c k e k. The set e 1 } e p p PP } generates a dense subalgebra of ((A, +, ), ). Proof. It s proved n [9] that ths set s a Schauder base n the topologcal vector space (A, ). It also follows from [9] that the coeffcents c k n (3.9) are gven by c k = f(k). It remans to prove orthogonalty. Wth the above notaton, f = c k e k = 1/j, k=1 where j s the smallest k such that c k 0. Recallng that C s trvally normed, we have that e k = 1/k f c k 0 c k e k = 0 f c k = 0 k=1

7 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION 167 so max k N + c k e k } = 1/j, wth j as above, so (17) holds. By Lemma 2.5 any e k can be wrtten as a square-free monomal n the elements of e p p PP }. The set e k k N + } s dense n A, so e p p PP } generates a dense subalgebra. Let J m denote the deal generated by all e k, k > 1. Lemma J s not fntely generated. Proof. The followng proof was provded by the anonymous referee. Consder the followng deal I n A: I = f A f(1) = 0, p P : f(p) = 0}. Then the unts of A/I are precsely the elements of the form g + I, where g A, g(1) 0. Moreover, for any f, g A such that f(1) = a C, g(1) = 0, we have (f + I) (g + I) = (ag) + I = a(g + I). Assume that J s fntely generated deal, say J = (b 1,..., b r ). Then b 1 (1) = = b r (1) = 0 and any element of J s of the form r =1 f b for sutable f 1,... f r A. We have ( r ) r r f b + I = (f + I) (b + I) = a (b + I), =1 =1 where a = f (1) C, whch belongs to the fntely dmensonal lnear subspace of A/I generated by b 1 + I,..., b r + I. Ths s a contradcton wth the fact that the lnear subspace of A/I generated by e k + I, k > 1, s of nfnte dmenson. Defnton An deal I A s called quas-fnte f there exsts a sequence (g k ) k=1 n I such that g k 0 and such that every element f I can be wrtten (not necessarly unquely) as a convergent sum (18) f = a k g k, a k A. k=1 Lemma m s quas-fnte. Proof. By Theorem 3.9 the set e k k > 1 } s a quas-fnte generatng set for m. Snce all deals are contaned n m, t follows that any deal contanng e k k > 1 } s quas-fnte. Furthermore, such an deal has m as ts closure. In partcular, J s quas-fnte, but not closed. Theorem All (non-zero) deals n A are quas-fnte. In fact, gven any subspace I we can fnd (19) such that for all f I, (20) G(I) := (g k ) k=1 c 1, c 2, c 3, C, f = So all subspaces possesses a Schauder bass. =1 c g. =1

8 168 J. SNELLMAN Proof. We construct G(I) n the followng way: for each k N(f) f I \ 0} } =: N(I) we choose a g k I wth N(g k ) = k, and wth g k (k) = 1. In other words, we make sure that the leadng coeffcent s 1; ths can always be acheved snce the coeffcents le n a feld. For k N(I) we put g k = 0. To show that ths choce of elements satsfy (20), take any f I, and put f 0 = f. Then defne recursvely, as long as f 0, n := N(f ), C a := f (n ), A f +1 := f a g n. Of course, f f = 0, then we have expressed f as a lnear combnaton of g n1,..., g n 1, and we are done. Otherwse, note that by nducton f I, so n N(I), hence g n 0. Thus N(f +1 ) > N(f ), so f +1 < f, whence But so f 0 > f 1 > f 2 > 0. f +1 = f F := whch shows that j=1 a jg j = f. a j g nj, j=1 a j g nj f, j=1 4. A fundamental somorphsm 4.1. The monod of separated monomals. Let } (21) Y = y (j), j N + be an nfnte set of varables, n bjectve correspondence wth the nteger lattce ponts n the frst quadrant mnus the axes. We call the subset } (22) Y = y (j) j N + the th column of Y. Let [Y ] denote the free abelan monod on Y, and let M be the subset of separated monomals,.e. monomals n whch no two occurrng varables come from the same column: } (23) M = y (j2) y (jr) r 1 < 2 < r. y (j1) 1 2

9 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION 169 We regard M as a monod-wth-zero, so that the multplcaton s gven by m m mm mm M (24) = 0 otherwse Note that the zero s exteror to M,.e. 0 M. The set M 0} s a (noncancellatve) monod f we defne m 0 = 0 for all m M. Recall that PP denotes the set of prme powers. It follows from the fundamental theorem of arthmetc that any postve nteger n can be unquely wrtten as a square-free product of prme powers. Hence we have that Φ : Y PP, (25) y (j) p j s a bjecton whch can be extended to a bjecton (26) Φ : M N +, 1 1, y (j1) 1 y (jr) r p j1 1 p jr r. If we regard N + as a monod-wth-zero wth the operaton of (4), then (26) s a monod-wth-zero somorphsm The rng A as a generalzed power seres rng, and as a quotent of C [Y ]. Let R be the large power seres rng on [Y ],.e. R = C [Y ] conssts of all formal power seres c α y α, where the sum s over all mult-sets α on Y. Let S be the generalzed monod-wth-zero rng on M. By ths, we mean that S s the set of all formal power seres (27) f(m)m, f(m) C m M wth component-wse addton, and wth multplcaton ( ) ( ) ( ) f(m)m g(m)m = h(m)m, (28) m M m M m M h(m) = (f g)(m) = (29) (30) (31) Defne Let furthermore supp( m [Y ] supp( m M c m m) = m [Y ] c m 0 }, c m m) = m M c m 0 }. D = f R supp(f) M = }. Theorem 4.1. S and R D and A are somorphc as C-algebras. s t=m f(s)g(t).

10 170 J. SNELLMAN Proof. The bjecton (26) nduces a bjecton between S and A whch s an somorphsm because of the way multplcaton s defned on S. In detal, the somorphsm s defned by S c m m f A, (32) m M f(φ(m)) = c m. For the second part, consder the epmorphsm φ Clearly, ker(φ) = D, hence S m [Y ] φ : R S, c m m = c m m. R ker(φ) = R D. m M Let us exemplfy ths somorphsm by notng that e n, where n has the square-free factorzaton n = p a1 1 par r, corresponds to the square-free monomal y (a1) 1 y r (ar), and that 1 = (33) m = 1 +. m M =1 What does ts nverse µ correspond to? Defnton 4.2. For m M, we denote by ω(m) the number of occurrng varables n m (by defnton, ω(1) = 0). For S f = c m m we put (34) m M j=1 y (j) D(f) = mn ω(m) c m 0 } f f 0 and D(0) =. Then ω(φ(m)) = ω(m), and f f and g correspond to each other va the somorphsm (32), then D(f) = D(g). (35) It s known (see [10]) that We then have that µ corresponds to (36) 1 1 = 1 =1 (1 + ) = j=1 y(j) µ (r) = ( 1) ω(r). = j=1 y(j) = ( 1) ω(m) m. m M Recall that f A s a multplcatve arthmetc functon f f(nm) = f(n)f(m) whenever gcd(n, m) = 1. Regardng f 0 as an element of S we have that f s

11 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION 171 multplcatve f and only f t can be wrtten as (37) f = 1 + c,j y (j). =1 It s now easy to see that the multplcatve functons form a group under multplcaton The contnuous endomorphsms. In [9], Schwab and Slberberg characterzed all contnuous endomorphsms of (A, +, ), the rng of arthmetcal functons wth Drchlet convoluton. We gve the correspondng result for A = (A, +, ): Theorem 4.3. Every contnuous endomorphsm θ of the C-algebra S A s defned by (38) where (39) and (40) j=1 θ(y (j) ) = γ,j, γ,j γ,k = 0 for all, j, k γ a1(n),b 1(n)... γ ar(n),b r(n) 0 as n = p b1(n) a 1(n)... pbr(n) a r(n). Proof. Recall that S R D, where R = C [Y ] and D s the closure of the deal generated by all non-separated quadratc monomals y (j) y (k). Snce the set of square-free monomals n the y (j) s form a Schauder base of S, any contnuous C- s, and must -algebra endomorphsm θ of S s determned by ts values on the y (j) fulfll (40). Snce y (j) y (k) = 0 n S, we must have that θ(0) = θ(y (j) y (k) ) = θ(y (j) )θ(y (k) ) = γ,j γ,k = Nlpotent elements and zero dvsors Defnton 5.1. For m N +, defne the prme support of m as (41) psupp(m) = p P p m } and (when m > 1) the leadng prme as (42) lp(m) = mn psupp(m). For n N +, put A (n) = k N + p n k but p k for < n } = k N + } (43) lp(k) = p n. Then N + \ 1} s a dsjont unon (44) N + \ 1} = A (). =1

12 172 J. SNELLMAN Defnton 5.2. If f A s a non-unt, then the canoncal decomposton of f s the unque way of expressng f as a convergent sum (45) f = f, f = f(k)e k. =1 k A () The element f s sad to be of polynomal type f all but fntely many of the f s are zero. In that case, the largest N such that f N 0 s called the fltraton degree of f. Lemma 5.3. If f A s a non-unt wth canoncal decomposton (45), then (46) f = e p j g,j, j=1 where r, p r n mples that g,j (n) = 0. For any n there s at most one par (, j) such that ( ) e p j g,j (n) 0. More precsely, f n = p j1 1 p jr r, 1 < < r, then ( e p b a g a,b ) (n) may be non-zero only for a = 1, b = j 1. Defnton 5.4. For k N, defne (47) I k = f A f(n) = 0 for every n such that gcd(n, p 1 p 2 p k ) = 1 }. Lemma 5.5. I k s an deal n (A, +, ). Proof. It s shown n [8] that the I k s form an ascendng chan of deals n (A, +, ). They are also easly seen to be deals n (A, +, ): f then f I k, g A and gcd(n, p 1 p 2 p k ) = 1 (f g)(n) = d n f(d)g(n/d) = 0, snce gcd(d, p 1 p 2 p k ) = 1 for any untary dvsor of n. For any h A, the annhlator ann(h) A s the deal consstng of all elements g A such that gh = 0. Theorem 5.6. Let N N +, then I N = ann(e p1 p N ) = 0} f A f s a non-unt of polynomal type = A e p a a, N +, N }, and has fltraton degree at most N} where AW denotes the topologcal closure of the deal generated by the set W.

13 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION 173 Proof. If f I N then for all k (48) (f e p1 p N )(k) = = a p 1 p N =k a p 1 p N =k f(a)e p1 p N (p 1 p N ) f(a) = 0, so f ann(e p1 p N ). Conversely, f f ann(e p1 p N ) then (f e p1 p N )(k) = 0 for all k, hence f gcd(n, p 1 p N ) = 1 then (49) 0 = (f e p1 p N )(np 1 p N ) = f(n)e p1 p N (p 1 p N ) = f(n) hence f I N. If f I N then for j > N we get that f j = 0, snce 0 f k A (j) f j (k) = f(k) = 0 f k A (j) Hence f = N =1 f. Conversely, f f can be expressed n ths way, then f(k) = f j1 (k) = 0 for k = p a1 j 1 p ar j r wth N < j 1 < < j r. The last equalty follows from Theorem 3.9. Theorem 5.7. Let f A be a non-unt. The followng are equvalent: () f s of polynomal type. () f k=0 I k, () There s a fnte subset Q P such that f(k) = 0 for all k relatvely prme to all p Q. (v) f N=1 ann(e p 1p 2 p N ). (v) There s a postve nteger N such that f s contaned n the topologcal closure of the deal generated by the set ep a a, N +, N }. If f has fnte support, then t s of polynomal type. If f s of polynomal type, then t s nlpotent. Proof. Clearly, a fntely supported f s of polynomal type. The equvalence () () () (v) (v) follows from the prevous theorem. If f s of polynomal type, say of fltraton degree N, then N (50) f = =1 and we see that f f N+1 s the N + 1 st untary power of f, then f N+1 s the lnear combnaton of monomals n the f s, and none of these monomals s square-free. Snce f f = 0 for all, we have that f N+1 = 0. So f s nlpotent. Lemma 5.8. The elements of polynomal type forms an deal. f

14 174 J. SNELLMAN Proof. By the prevous theorem, ths set can be expressed as I n, n=1 whch s an deal snce I n form an ascendng chan of deals. Queston 5.9. Are all [nlpotent elements, zero dvsors] of polynomal type? If one could prove that the zero dvsors are precsely the elements of polynomal type, then by Lemma 5.8 t would follow that Z(A) s an deal, and moreover a prme deal, snce the product of two regular elements s regular (n any commutatve rng). Then one could conclude [6] that (A, +, ) has few zero dvsors, hence s addtvely regular, hence s a Marot rng. Theorem (A, +, ) contans nfntely many non-assocate regular nonunts. Proof. Step 1. We frst show that there s at least one such element. Let f A denote the arthmetcal functon 1 f k PP f(k) = 0 otherwse Then f s a non-unt, and usng a result by Yocom [13, 8] we have that f s contaned n a subrng of (A, +, ) whch s a dscrete valuaton rng somorphc to C [t], the power seres rng n one ndetermnate. Ths rng s a doman, so f s not nlpotent. We clam that f s n fact regular. To show ths, suppose that g A, f g = 0. We wll show that g = 0. Any postve nteger m can be wrtten m = q a1 1 qar r, where the q are dstnct prme numbers. If r = 0, then m = 1, and g(1) = 0, snce 0 = (f g)(2) = f(2)g(1) = g(1). For the case r = 1, we want to show that g(q a ) = 0 for all prme numbers q. Choose three dfferent prme powers q a1 1, qa2 2, and qa3 3. Then 0 = f g(q a q aj j ) = f(qa )g(q aj j ) + f(qaj j )g(qa ) = g(q aj j ) + g(qa ), when j,, j 1, 2, 3}. In matrx notaton, these three equatons can be wrtten as g(qa1 1 ) g(q a2 2 ) = g(q a3 0 from whch we conclude (snce the determnant of the coeffcent matrx s nonzero) that 0 = g(q a1 1 ) = g(qa2 2 ) = g(qa3 3 ). Now for the general case, r > 1. We need to show that that (51) whenever q a1 1,..., qar r 3 ) g(q a1 1 qar r ) = 0 are par-wse relatvely prme prme powers.

15 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION 175 Choose N par-wse relatvely prme prme powers q a1 1,..., qan N. For each r subset q s1,..., q sr+1 of ths set we get a homogeneous lnear equaton (52) 0 = f g(q s1... q sr+1 ) = g(q s2 q sr+1 ) + g(q s1 q s3 q sr+1 ) + + g(q s1 g sr ). The matrx of the homogeneous lnear equaton system formed by all these equatons s the ncdence matrx of r-subsets (of a set of N elements) nto r+1-subsets. It has full rank [12]. Snce t conssts of ( ( N r+1) equatons and N ) r varables, we get that for suffcently large N, the null-space s zero-dmensonal, thus the homogeneous system has only the trval soluton. It follows, n partcular, that (51) holds. Thus, g(m) = 0 for all m, so f s a regular element. Step 2. We construct nfntely many dfferent regular non-unts. Consder the element f, wth c k k PP f(k) = 0 otherwse and where the c k s are suffcently generc non-zero complex numbers, then we clam that f, too, s a regular non-unt. Wth g, m, r as before, we have that, for r = 0, 0 = f g(p a ) = f(p a )g(1) = c p ag(1). We demand that c p a 0, then g(1) = 0. For a general r, we argue as follows: the ncdence matrces that occurred before wll be replaced wth generc matrces whose elements are c k s or zeroes, and whch specalze, when settng all c k = 1, to full-rank matrces. They must therefore have full rank, and the proof goes through. Step 3. Let g be a unt n A, and f as above. We clam that f g f s of the above form,.e. supported on PP, then g must be a constant. Hence there are nfntely many non-assocate regular non-unts of the above form. To prove the clam, we argue exactly as before, usng the fact that g f s supported on PP. For m = q a1 1 qar r as before, the case r = 0 yelds nothng: nether does the case r = 1: 0 = g f(1) = f(1)g(1) = 0g(1) = 0, w = g f(q a ) = f(q a )g(1), so g(1) may be non-zero. But for r = 2 we get and also a1 0 = g f(q a1 0 = g f(q a2 0 = g f(q 1 qa2 2 1 qa3 3 2 qa3 3 ) = f(q a1 1 )g(qa2 ) = f(q a1 1 )g(qa3 ) = f(q a2 2 )g(qa3 2 ) + g(qa1 1 3 ) + g(qa1 1 3 ) + g(qa1 1 ) f(q a2 2 ), ) f(q a3 3 ), ) f(q a3 3 ),

16 176 J. SNELLMAN whch means that f(q a2 2 ) a f(q 1 1 ) 0 f(q a3 3 ) 0 a f(q 1 a 0 f(q 3 3 ) a f(q 2 1 ) 2 ) g(qa1 g(q a2 = 0 0 g(q a3 0 By our assumptons, the coeffcent matrx s non-sngular, so only the zero soluton exsts, hence g(q a1 1 ) = 0. An analyss smlar to what we dd before shows that g(q a1 1 qar r ) = 0 for r > 1. Wth the same method, one can easly show that the characterstc functon on P s regular. 1 ) 2 ) 3 ) 6. Some smple results on factorsaton Cashwell-Everett [4] showed that (A, +, ) s a UFD. We wll brefly treat the factorsaton propertes of (A, +, ). Defntons and facts regardng factorsaton n commutatve rngs wth zero-dvsors from the artcles by Anderson and Valdes- Leon [1, 2] wll be used. Frst, we note that snce (A, +, ) s quas-local, t s présmplfable,.e. a 0, a = r a mples that r s a unt. It follows that for a, b A, the followng three condtons are equvalent: (1) a, b are assocates,.e. A a = A b. (2) a, b are strong assocates,.e. a = u b for some unt u. (3) a, b are very strong assocates,.e. A a = A b and ether a = b = 0, or a 0 and a = r b = r U(A). We say that a A s rreducble, or an atom, f a = b c mples that a s assocate wth ether b or c. Theorem 6.1. (A, +, ) s atomc,.e. all non-unts can be wrtten as a product of fntely many atoms. In fact, (A, +, ) s a bounded factoral rng (BFR),.e. there s a bound on the length of all factorsatons of an element. Proof. It follows from Lemma 3.3 that the non-unt f has a factorsaton nto at most D(f) atoms. Example 6.2. We have that e 2 (e 2 k +e 3 ) = e 6 for all k, hence e 6 has an nfnte number of non-assocate rreducble dvsors, and nfntely many factorsatons nto atoms. Example 6.3. The element h = e 30 can be factored as e 2 e 3 e 5, or as (e 6 + e 20 ) (e 2 + e 5 ). These examples show that (A, +, ) s nether a half-factoral rng, nor a fnte factorsaton rng, nor a weak fnte factorsaton rng, nor an atomc df-rng.

17 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION The subrng of arthmetcal functons supported on square-free ntegers (53) Let SQF N + denote the set of square-free ntegers, and put C = f A supp(f) SQF }. For any f A, denote by p(f) C the restrcton of f to SQF. Theorem 7.1. (C, +, ) s a subrng of (A, +, ), and a closed C-subalgebra wth respect to the norm. The map (54) p : A C, f p(f) s a contnuous C-algebra epmorphsm, and a retracton of the ncluson map C A. Proof. Let f, g C. If n N + \ SQF then (f + g)(n) = f(n) + g(n) = 0, and cf(n) = 0 for all c C. Snce n N + \ SQF, there s at least on prme p such that p 2 n. If m s a untary dvsor of m, then ether m or n/m s dvsble by p 2. Thus (f g)(n) = m n f(m)g(n/m) = 0. If f k f n A, and all f k C, let n supp(f). Then there s an N such that f(n) = f k (n) for all k N. But supp(f k ) SQF, so n SQF. Ths shows that C s a closed subalgebra of A. It s clear that p(f + g) = p(f) + p(g) and that p(cf) = cp(f) for any c C. If n s not square-free, we have already showed that 0 = (p(f) p(g))(n) = p((f g))(n). Suppose therefore that n s square-free. Then so s all ts untary dvsors, hence p(f g)(n) = (f g)(n) = m n f(m)g(n/m) = m n p(f)(m)p(g)(n/m) = (p(f) p(g))(n). We have that p(f) = f f and only f f C, hence p(p(f)) = p(f), so p s a retracton to the ncluson : C A. In other words, p = d C. Corollary 7.2. The multplcatve nverse of an element n C les n C. Proof. If f C, f g = e 1 then e 1 = p(e 1 ) = p(f g) = p(f) p(g) = f p(g), hence g = p(g), so g C. Alternatvely, we can reason as follows. If f s a unt n C then we can wthout loss of generalty assume that f(1) = 1. By Theorem 3.7, g = f + e 1 s topologcally nlpotent, hence by Proposton of [3] we have that the nverse of

18 178 J. SNELLMAN e 1 g = f can be expressed as =0 g. It s clear that g, and every power of t, s supported on SQF, hence so s f 1. Corollary 7.3. (C, +, ) s sem-local. Proof. The unts conssts of all f C wth f(1) 0, and the non-unts form the unque maxmal deal. Remark 7.4. More generally, gven any subset Q N +, we get a retract of (A, +, ) when consderng those arthmetcal functons that are supported on the ntegers n = p a1 1 par r wth a Q 0}. Ths property s unque for the untary convoluton, among all regular convolutons n the sense of Narkewcz [7]. In partcular, the set of arthmetcal functons supported on the exponentally odd ntegers (those n for whch all a are odd) forms a retract of (A, +, ). It follows that the nverse of such a functon s of the same form. Let T = C [x 1, x 2, x 3,... ], the large power seres rng on countably many varables, and let J denote the deal of elements supported on non square-free monomals. Theorem 7.5. (C, +, ) T/J. Ths algebra can also be descrbed as the generalzed power seres rng on the monod-wth-zero whose elements are all fnte subsets of a fxed countable set X, wth multplcaton A B f A B = (55) A B = 0 otherwse Proof. Defne η by (56) η( m η : T A c m m) = m square-free c m e m, where for a square-free monomal m = x 1 x r wth 1 1 < < r we put e m = e p1 p r. Then η(t ) = C, ker η = J. It follows that C T/J. Acknowledgement. I am grateful to the anonymous referee for suggestng several correctons and mprovements, and for the proof of Lemma References [1] Anderson, D. D. and Valdes-Leon, S., Factorzaton n commutatve rngs wth zero dvsors, Rocky Mountan J. Math. 26(2) (1996), [2] Anderson, D. D. and Valdes-Leon, S., Factorzaton n commutatve rngs wth zero dvsors, II, In Factorzaton n ntegral domans (Iowa Cty, IA, 1996), Marcel Dekker, New York 1997, [3] Bosch, S., Güntzer, U. and Remmert, R., Non-Archmedean analyss, A systematc approach to rgd analytc geometry, Sprnger-Verlag, Berln, [4] Cashwell, E. D. and Everett, C. J., The rng of number-theorethc functons, Pacfc J. Math. 9 (1959),

19 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION 179 [5] Cohen, E., Arthmetcal functons assocated wth the untary dvsors of an nteger, Math. Z. 74 (1960), [6] Huckaba, J. A., Commutatve rngs wth zero dvsors, Marcel Dekker Inc., New York, [7] Narkewcz, W., On a class of arthmetcal convolutons, Colloq. Math. 10 (1963), [8] Schwab, E. D. and Slberberg, G., A note on some dscrete valuaton rngs of arthmetcal functons, Arch. Math. (Brno) 36 (2000), [9] Schwab, E. D. and Slberberg, G., The valuated rng of the arthmetcal functons as a power seres rng, Arch. Math. (Brno) 37(1) (2001), [10] Svaramakrshnan, R., Classcal theory of arthmetc functons, Pure and Appled Mathematcs, volume 126, Marcel Dekker, [11] Vadyanathaswamy, R., The theory of multplcatve arthmetc functons, Trans. Amer. Math. Soc. 33(2) (1931), [12] Wlson, R. M. The necessary condtons for t-desgns are suffcent for somethng, Utl. Math. 4 (1973), [13] Yocom, K. L., Totally multplcatve functons n regular convoluton rngs, Canad. Math. Bull. 16 (1973), Department of Mathematcs, Stockholm Unversty SE Stockholm, Sweden E-mal: Jan.Snellman@math.su.se