The Multiplicative Structure of the Group of Units of Zp[x]/ where f(x) is Reducible

Rose-Hulman Undergraduate Mathematics Journal Volume 16 Issue 1 Article 5 The Multiplicative Structure of the Group of Units of Zp[x]/ where f(x) is Reducible Erika Gerhold Salisbury University Jennifer Ferralli Salisbury University Jason Jachowski Salisbury University Follow this and additional works at: http://scholarrose-hulmanedu/rhumj Recommended Citation Gerhold, Erika; Ferralli, Jennifer; and Jachowski, Jason (2015) "The Multiplicative Structure of the Group of Units of Zp[x]/ where f(x) is Reducible," Rose-Hulman Undergraduate Mathematics Journal: Vol 16 : Iss 1, Article 5 Available at: http://scholarrose-hulmanedu/rhumj/vol16/iss1/5

Rose- Hulman Undergraduate Mathematics Journal The Multiplicative Structure of the Group of Units of Z p [x]/ f(x) where f(x) is Reducible Erika Gerhold a Jason Jachowski b Jennifer Ferralli (Larson) c Volume 16, No 1, Spring 2015 Sponsored by Rose-Hulman Institute of Technology Department of Mathematics Terre Haute, IN 47803 Email: mathjournal@rose-hulmanedu http://wwwrose-hulmanedu/mathjournal a Salisbury University b Salisbury University c Salisbury University

Rose-Hulman Undergraduate Mathematics Journal Volume 16, No 1, Spring 2015 The Multiplicative Structure of the Group of Units of Z p [x]/ f(x) where f(x) is Reducible Erika Gerhold Jason Jachowski Jennifer Ferralli (Larson) Abstract Factor rings of the form Z p [x]/ f(x), with p prime and f(x) irreducible in Z p [x] form a field, with cyclic multiplicative group structure When f(x) is reducible in Z p [x] this factor ring is no longer a field, nor even an integral domain, and the structure of its group of units is no longer cyclic In this paper we develop concise formulas for determining the cyclic group decomposition of the multiplicative group of units for Z p [x]/ f(x) that is only dependent on the multiplicities and degrees of the irreducible factors of f(x), and p Acknowledgements: The authors wish to thank Dr Donald Spickler for all of his advice and guidance on this project

RHIT Undergrad Math J, Vol 16, No 1 Page 89 1 Introduction If R is a ring with identity, we will let U(R) denote that group of units of that ring We will also let G represent the group of units of Z p [x]/ f(x) That is, G = U (Z p [x]/ f(x) ) with Z p [x] the polynomials in x with coefficients in Z p It is well known that factor rings of the form Z p [x]/ h(x)g(x), where h(x) and g(x) are relatively prime, are isomorphic to Z p [x]/ h(x) Z p [x]/ g(x) [3] This reduces the problem of finding the decomposition of G to finding the decomposition of the unit group for rings of the form Z p [x]/ f n (x) where f(x) is irreducible Throughout the paper we will let r = deg(f(x)), the degree of f(x) Since G is a finite Abelian group, G is isomorphic to a direct product of cyclic groups That is, G = Z c 1 a 1 Z c 2 a 2 Za cm m, where Z c i a i represents the Cartesian product of Z ai with itself, having c i factors [2] Our goal is to determine the size of the cyclic factors of this decomposition In general, if two finite groups have equal numbers of elements of equal orders, it does not follow that the two groups are isomorphic In the case of finite Abelian groups, however, this condition is sufficient for isomorphism [4] This allows us to take a combinatorial approach to the problem In Section 2, we will develop formulas for the number of elements of each order in G Then in Section 3, we consider the number of elements of each order in the cyclic group decomposition Combining these in Section 4, we will be able to determine the factors in the decomposition of G, which gives us our main theorem Theorem 1 Let p be a prime and f(x) an irreducible polynomial of degree r in Z p [x] Let n be any positive integer and define s = log p (n) n Define q i = 1 for i = p (s 1) i 0,, (s 1) and q i = 0 for i < 0 The group of units G is isomorphic to where c i = r(q i 2q i 1 + q i 2 ) G = U (Z p [x]/ f(x) ) = Z p r 1 Z c s 1 p Z c s 2 p 2 Z c 0 p s We will conclude with generalizations and examples in Section 5 2 Element Orders in Z p [x] / f n (x) Throughout the paper we will slightly abuse notation Since we are examining elements of Z p [x] / f n (x) exclusively, we will say that g(x) is an element of Z p [x] / f n (x), meaning that g(x) is a coset representative, that is, g(x)+ f n (x) Z p [x] / f n (x) The continual use of the coset notation makes the content of the paper more difficult to read Also, throughout the remainder of the paper we will assume that f(x) is an irreducible polynomial in Z p [x], unless otherwise stated The number of units in Z p [x]/ f n (x) where f(x) is irreducible of degree r is the total number of elements of Z p [x]/ f n (x) minus the number of zero-divisors in Z p [x]/ f n (x) The total number of elements in Z p [x]/ f n (x) is p rn Since f(x) is irreducible, any zero divisor, g(x), must be of the form g(x) = h(x)f(x) where the degree of h(x) is at most

Page 90 RHIT Undergrad Math J, Vol 16, No 1 r(n 1) 1, and there are p r(n 1) such elements So in our notation, G = p rn p r(n 1) = (p r 1)p r(n 1) The following discussion will show that G = Z p r 1 H where H is a p-group of order p r(n 1) Lemma 1 G = U(Z p [x]/ f n (x) ) has a cyclic subgroup of order p r 1 Proof Let the map φ : Z p [x] / f n (x) Z p [x] / f(x) be defined by φ(g(x) + f n (x) ) = g(x) + f(x) It is an easy exercise to show that φ is a surjective ring homomorphism Let k(x) be a generator for the multiplicative group of units in Z p [x]/ f(x) Since φ is surjective, there exists a g(x) Z p [x]/ f n (x) with φ(g(x)) = k(x) Let the order of g(x) be m Then 1 = φ(g m (x)) = φ(g(x)) m = k m (x) So m is a multiple of the order of k(x), that is, (p r 1) m Since g(x) is a cyclic subgroup of order m and (p r 1) m there is a cyclic subgroup of g(x) of order p r 1 Consequently this is also a cyclic subgroup of order p r 1 in G This allows us to write G = Z p r 1 H where H is a p-group of order p r(n 1) and hence has the form, H = Z c s 1 p Z c s 2 p Z c 2 0 p s We may now concentrate our efforts on H Since H is a p-group we need only consider elements of G of prime power order Lemma 2 The set K = { g(x) g(x) = 1 + a 1 (x)f(x) + a 2 (x)f 2 (x) + + a n 1 (x)f n 1 (x) } is the entire set of elements of G with order a power of p Proof Let g(x) be any polynomial in G of order p b We can write g(x) = a 0 (x)+a t (x)f t (x)+ a t+1 (x)f t+1 (x) + + a n 1 (x)f n 1 (x) where t is the smallest positive power of f(x) in g(x) and a i (x) is a polynomial with degree less than deg(f(x)) = r, for all i First observe that if g(x) = 1 + a t (x)f t (x) + a t+1 (x)f t+1 (x) + + a n 1 (x)f n 1 (x), then g(x) is a unit in Z p [x] / f n (x) and hence g(x) G To see this, rewrite g(x) as g(x) = 1 h(x)f t (x) for some t 1 We can build (g(x)) 1 as follows Let h 1 (x) = 1 + h(x)f t (x) Then g(x)h 1 (x) = 1 h 2 (x)f 2t (x) If 2t n, then g(x)h 1 (x) = 1 which implies that h 1 (x) = (g(x)) 1 If 2t < n, let h 2 (x) = 1 + h 2 (x)f 2t (x) Then g(x)h 1 (x)h 2 (x) = 1 h 4 (x)f 4t (x) and if 4t n, we can stop with h 1 (x)h 2 (x) = (g(x)) 1 If not, continue until 2 α t n, then (g(x)) 1 = α i=1 h i(x) This shows that g(x) G Furthermore, g(x) = 1 + a t (x)f t (x) + a t+1 (x)f t+1 (x) + + a n 1 (x)f n 1 (x) has order a power of p To see this, let us first consider g p (x) By the multinomial theorem, g p (x) = ( 1 + a t (x)f t (x) + a t+1 (x)f t+1 (x) + + a n 1 (x)f n 1 (x) ) p ( ) p ( = ai (x)f i (x) ) k i k 0, k t,, k n 1 k 0 +k t+ +k n 1 =p i=0,t,,n 1

RHIT Undergrad Math J, Vol 16, No 1 Page 91 where we consider a 0 (x) = 1 Since ( ) p k 0,k t,,k n 1 = p! k 0!k t! k n 1 is divisible by p, unless k! i = p, all but these terms vanish from the sum This simplifies the expression to g p (x) = 1 p + (a t (x)f t (x)) p + (a t+1 (x)f t+1 (x)) p + + (a n 1 (x)f n 1 (x)) p = 1 + a p t (x)f tp (x) + a p t+1(x)f (t+1)p (x) + + a p n 1(x)f (n 1)p (x) Repeated applications of this gives an expression for g pb (x), g pb (x) = 1 + a pb t (x)f tpb (x) + + a pb n 1(x)f (n 1)pb (x) Let b be large enough so that tp b n, then all of the terms, except for 1, will vanish Therefore, g pb (x) = 1 and the order of g(x) must divide p b Hence, the order of g(x) is a power of p So g(x) will correspond to an element in H via the isomorphism G = Z p r 1 H We will now show that the set of all the elements of this form constitutes all of H, via the isomorphism Let K = { g(x) g(x) = 1 + a 1 (x)f(x) + a 2 (x)f 2 (x) + + a n 1 (x)f n 1 (x) } Counting the elements gives us K = (p r ) n 1 = p r(n 1) = H Since the order counts match, the isomorphism maps K onto H One corollary of this is that if we let t = 1, and the order of g(x) be p b, we find that the maximum order of any polynomial in K is p log p (n), where log p (n) denotes the ceiling of log p (n) This can be seen as follows As we saw above, if g(x) = 1+a 1 (x)f(x)+a 2 (x)f 2 (x)+ + a n 1 (x)f n 1 (x) and p b is the order of g(x) then 1 = g pb (x) = 1 + a pb 1 (x)f pb (x) + a pb 2 (x)f 2pb (x)+ +a pb n 1(x)f (n 1)pb (x) So b must be the smallest positive number such that p b n, that is, b log p (n) and so b = log p (n) Consequently, the maximum order of any element in K is p log p (n) The number log p (n) plays a prominent role in the calculations that follow, so throughout the paper we will denote it as s Since the remainder of the decomposition is a p-group, all the elements are of order p y with y s By our above observations, the order of any element in K is completely dependent on the smallest positive power t of f(x) Our next result establishes a range for t in which all polynomials in K with smallest positive power t of f(x) in that range have the same order Lemma 3 The order of g(x) = 1 + a t (x)f t (x) + a t+1 (x)f t+1 (x) + + a n 1 (x)f n 1 (x) for n n 1 + 1 t 1 is p s i for 0 i s 1 p s i p s i 1 Proof We know g(x) has order p s i if s i is the smallest integer such that t n Since p s i n n t Z, we have t = 1 + 1 Also since s i is the smallest integer such that p s i p s i t n we must have t < n n, which implies t 1 p s i p s i 1 p s i 1

Page 92 RHIT Undergrad Math J, Vol 16, No 1 This lemma sets up intervals of t where the order of any polynomial with lowest power of f(x) in that interval are all the same These intervals will also play a prominent role in our n calculations, so throughout the remainder of the discussion we will let q i = 1 p (s 1) i We will also define q i = 0 when i < 0 In this notation, our above lemma can be restated as follows The order of g(x) = 1 + a t (x)f t (x) + a t+1 (x)f t+1 (x) + + a n 1 (x)f n 1 (x) where q i 1 + 1 t q i is p s i Given the ranges of t from the above lemma it is now fairly easy to determine the number of elements of K with order p s i for all i Theorem 2 The number of units in Z p [x]/ f n (x) with order p s i is where r = deg(f(x)) O i = (p r 1) q i j=1+q i 1 p r(n 1 j) = p r(n 1 qi) ( p r(qi qi 1) 1 ) Proof By Lemma 3, g(x) = 1+a t (x)f t (x)+ +a n 1 (x)f n 1 (x) where 1+q i 1 t q i will have order p s i Let the number of elements with order p s i be denoted as O i First assume that a qi 1 +1(x) 0, there are p r 1 possibilities for a qi 1 +1(x) and p r possibilities for each of a qi 1 +2(x), a qi 1 +3(x),, a n 1 (x) The total number of these is (p r 1)p r((n 1) (1+q i 1)) By the same method, assuming a qi 1 +1(x) = 0 and a qi 1 +2(x) 0 we have (p r 1)p r((n 1) (2+q i 1)) elements Continuing this process through q i gives O i = (p r 1)p r((n 1) (1+q i 1)) + (p r 1)p r((n 1) (2+q i 1)) + + (p r 1)p r((n 1) q i) q i = (p r 1) j=1+q i 1 p r(n 1 j) = (p r 1)p ( r(n 1 q i) 1 + p r + + p ) r(q i q i 1 1) ( = (p r 1)p r(n 1 q i) 1 p r(q i q i 1 ) ) 1 p r = p r(n 1 q i) ( p r(q i q i 1 ) 1 ) The proofs of the main results of this paper are simplified considerably by using sums of the elements of particular orders instead of the number of elements of one order Specifically, we will count of the number of elements of order p k where s m k s, for fixed m Corollary 1 The number of units with order p k in Z p [x]/ f n (x) where s m k s is p r(n 1) p r(n 1 qm)

RHIT Undergrad Math J, Vol 16, No 1 Page 93 Proof By Theorem 2, the number of elements of order p s i is O i = (p r 1) q i j=1+q i 1 p r(n 1 j) Now simply take the sum of the orders from i = 0 to i = m, m O i = (p r 1) i=0 = (p r 1) m i=0 q i j=1+q i 1 p r(n 1 j) q m p r(n 1 j) j=1 = (p r 1)p r(n 1 qm) (1 + p r + p 2r + + p r(qm 1) ) = (p r 1)p r(n 1 qm) prqm 1 p r 1 = p r(n 1 qm) (p rqm 1) = p r(n 1) p r(n 1 qm) 3 Element Orders in H We will now consider element orders in the cyclic decomposition In the last section we have established formulas for the number of units of order p s i in the polynomial factor ring In this section we wish to find similar formulas for elements of order p s i in Z c s 1 p Z c s 2 p 2 Z c 0 p s Then from there we will be able to determine each c i in the cyclic decomposition If a, b and c are positive integers with b a and p prime, then the number of elements of Z c p of order a pb is p bc p (b 1)c If we consider the set of all the elements of Z c pa with order p b or less we obtain a subgroup isomorphic to Z c Similarly, if we consider the set of all the p b elements of Z c p with order strictly less than a pb we obtain a subgroup isomorphic to Z c So p b 1 the number of elements of Z c p of order exactly a pb will be the difference of the orders of these two subgroups, specifically p bc p (b 1)c A more detailed discussion of the element counts can be found in the Jones and Finch paper [1] So if we consider the Z p s factors in our decomposition, Z c 0 ps, the number of elements of order p s will be p sc 0 p (s 1)c 0 where we recall that s = log p (n) Generalizing this slightly gives us the following result

Page 94 RHIT Undergrad Math J, Vol 16, No 1 Theorem 3 The number of elements of order p s i in H = Z c s 1 p Z c s 2 p 2 Z c 0 p s is O i = p r(n 1) i 1 j=0 (i j)c j p r(n 1) i j=0 (i j+1)c j, where H = p r(n 1) Proof From our above discussion, the number of elements of order p s i in Z c 0 p s Zc i p s i is p (s i)( i j=0 c j) p (s i 1)( i j=0 c j) Also the number of elements in the remainder of the decomposition of H is This yields p r(n 1) sc 0 (s 1)c 1 (s i)c i O i = p r(n 1) sc 0 (s 1)c 1 (s i)c i ( p (s i)( i j=0 c j) p (s i 1)( i j=0 c j) ) = p r(n 1) sc 0 (s 1)c 1 (s i)c i ( p sc 0 +(s 1)c 1 + +(s i)c i ic 0 (i 1)c 1 (i i)c i p sc 0+(s 1)c 1 + +(s i)c i (i+1)c 0 ic 1 (i i+1)c i ) = p r(n 1) i 1 j=0 (i j)c j p r(n 1) i j=0 (i j+1)c j As with the orders of the polynomial elements it will be easier to compare the sums of the O i Corollary 2 The number of elements of order p k with s m k s is m O i = p r(n 1) p r(n 1) m j=0 (m j+1)c j Proof By Theorem 3, i=0 O m = p r(n 1) m 1 j=0 (m j)c j p r(n 1) m j=0 (m j+1)c j O m 1 = p r(n 1) m 2 j=0 (m j 1)c j p r(n 1) m 1 j=0 (m j)c j O m 2 = p r(n 1) m 3 j=0 (m j 2)c j p r(n 1) m 2 j=0 (m j 1)c j O 1 = p r(n 1) m m j=0 (m (m 1) j)c j p r(n 1) m (m 1) j=0 (m (m 1) j+1)c j = p r(n 1) c 0 p r(n 1) 2c 0 c 1 O 0 = p r(n 1) p r(n 1) c 0 This turns into a nice telescoping sum giving, m O i = p r(n 1) p r(n 1) m j=0 (m j+1)c j i=0

RHIT Undergrad Math J, Vol 16, No 1 Page 95 4 Equating the Counts Using Corollary 1 and Corollary 2, we simply equate the expressions to construct a recursive formula for c i i 1 Lemma 4 c i = rq i (i j + 1)c j j=0 Proof From Corollaries 1 and 2, p r(n 1) p r(n 1) i j=0 (i j+1)c j = p r(n 1) p r(n 1 q i) which implies that i j=0 (i j + 1)c j = rq i, and therefore i 1 c i = rq i (i j + 1)c j j=0 With a little manipulation, this recursive formula can be converted into a closed form calculation of c i that is only dependent on the numbers q j, which in turn are only dependent on p, n, and r (the degree of f(x)) Theorem 4 c i = r(q i 2q i 1 + q i 2 ) Proof We will proceed by induction Our base cases will be the verifications of c 0, c 1, and c 2 Recall that we define q i = 0, and hence c i = 0, if i < 0 Using c i = rq i i 1 j=0 (i j +1)c j we obtain, c 0 = rq 0 c 1 = rq 1 (1 + 1)c 0 = rq 1 2c 0 = rq 1 2rq 0 = r(q 1 2q 0 ) c 2 = rq 2 [(2 0 + 1)c 0 + (2 1 + 1)c 1 ] = rq 2 3rq 0 2(rq 1 2rq 0 ) = rq 2 2rq 1 + rq 0 = r(q 2 2q 1 + q 0 )

Page 96 RHIT Undergrad Math J, Vol 16, No 1 For the inductive step, assume that c j = r(q j 2q j 1 + q j 2 ) for all j i, then c i+1 = rq i+1 i ((i + 1) j + 1)r(q j 2q j 1 + q j 2 ) j=0 = rq i+1 [(i + 2)rq 0 + (i + 1)r(q 1 2q 0 ) + ir(q 2 2q 1 + q 0 ) + (i 1)r(q 3 2q 2 + q 1 ) + 4r(q i 2 2q i 3 + q i 4 ) + 3r(q i 1 2q i 2 + q i 3 ) + 2r(q i 2q i 1 + q i 2 )] = rq i+1 [(i + 2)rq 0 2(i + 1)rq 0 + irq 0 + (i + 1)rq 1 2irq 1 + (i 1)rq 1 + 4rq i 2 6rq i 2 + 2rq i 2 + 3rq i 1 4rq i 1 + 2rq i ] = r(q i+1 2q i + q i 1 ) 5 The General Case These formulas, along with our original reductions, give us a quick way to compute the multiplicative group structure of the group of units of Z p [x]/ f(x) for all polynomials f(x) Say f(x) factors in Z p [x], as f(x) = f α 1 1 (x)f α 2 2 (x) f α k (x) where f i (x) are all irreducible and the f i (x) are relatively prime to each other Then we have a ring isomorphism, Z p [x]/ f(x) = Z p [x]/ f α 1 1 (x) Z p [x]/ f α 2 2 (x) Z p [x]/ f α k (x) where our results may be applied to each of the factors Example 1 Say we wanted to determine the structure of the group of units of Z 3 [x]/ (x 2 + 1) 7 (x 3 + x 2 + x + 2) 15 (x + 2) 28 k k

RHIT Undergrad Math J, Vol 16, No 1 Page 97 We know that Z 3 [x]/ ( (x 2 + 1) 7 (x 3 + x 2 + x + 2) 15 (x + 2) 28) = Z 3 [x]/ (x 2 + 1) 7 Z 3 [x]/ (x 3 + x 2 + x + 2) 15 Z 3 [x]/ (x + 2) 28 For the first factor, s = log p (n) = log 3 (7) = 2 giving 7 q 0 = 3 1 = 2 2 1 0 7 q 1 = 3 1 = 6 2 1 1 By Theorem 4 we have So, c 0 = rq 0 = 4 c 1 = r(q 1 2q 0 ) = 2(6 (2)2) = 4 U ( Z 3 [x]/ (x 2 + 1) 7 ) = Z8 Z 4 3 2 Z4 3 For the second factor, s = log p (n) = log 3 (15) = 3 giving 15 q 0 = 3 1 = 1 3 1 0 15 q 1 = 3 1 = 4 3 1 1 15 q 2 = 3 1 = 14 3 1 2 and thus, c 0 = rq 0 = 3 c 1 = r(q 1 2q 0 ) = 3(4 (2)1) = 6 c 2 = r(q 2 2q 1 + q 0 ) = 3(14 (2)4 + 1) = 21 So, U ( Z 3 [x]/ (x 3 + x 2 + x + 2) 15 ) = Z26 Z 3 3 3 Z6 3 2 Z21 3 For the final factor, s = log p (n) = log 3 (28) = 4 giving 28 q 0 = 3 1 = 1 4 1 0 28 q 1 = 3 1 = 3 4 1 1 28 q 2 = 3 1 = 9 4 1 2 28 q 3 = 3 1 = 27 4 1 3

Page 98 RHIT Undergrad Math J, Vol 16, No 1 and thus, So, c 0 = rq 0 = 1 c 1 = r(q 1 2q 0 ) = 1(3 (2)1) = 1 c 2 = r(q 2 2q 1 + q 0 ) = 1(9 (2)3 + 1) = 4 c 3 = r(q 3 2q 2 + q 1 ) = 1(27 (2)9 + 3) = 12 U ( Z 3 [x]/ (x + 2) 28 ) = Z2 Z 3 4 Z 3 3 Z 4 3 2 Z12 3 Combining the three gives us the isomorphism, U ( Z 3 [x]/ (x 2 + 1) 7 (x 3 + x 2 + x + 2) 15 (x + 2) 28 ) = Z 26 Z 8 Z 2 Z 3 4 Z 4 3 3 Z14 3 2 Z37 3 Example 2 In the special case where p n > 1, the situation simplifies significantly Here s = log p (n) = 1, giving q 0 = n 1, so the group of units of Z p [x]/ f n (x) will be isomorphic to Z p r 1 Z r(n 1) p Example 3 If we apply our result to Z p [x]/ f(x), where f(x) is irreducible in Z p [x] We have, s = log p (1) = 0, which implies that there is no decomposition for H, hence there is no H So the group of units of Z p [x]/ f(x) will be isomorphic to Z p r 1, as expected 6 Open Questions and Future Work One possible continuation of this work would be to examine the structure of Z m [x]/ f n (x) for m composite Although we do not have a conjecture to the multiplicative group structure of units of this ring at present, our preliminary analysis shows that there are some similarities to the prime case, but there is a definite splitting of cases that did not happen with the prime modulus

RHIT Undergrad Math J, Vol 16, No 1 Page 99 References [1] C Finch and L Jones, A curious connection between fermat numbers and finite groups, American Mathematical Monthly, 109 (2002), pp 517 524 [2] J A Gallian, Contemporary Abstract Algebra, New York: Houghton Mifflin Company, 2002 [3] S Mac Lane and G Birkhoff, Algebra, New York: MacMillan Publishing Co, Inc, 1979 [4] J J Rotman, An Introduction to the Theory of Groups, Newton: Allyn and Bacon, Inc, 1984