THE P-ADIC NUMBERS AND FINITE FIELD EXTENSIONS OF Q p

THE P-ADIC NUMBERS AND FINITE FIELD EXTENSIONS OF Q p EVAN TURNER Abstract. This paper will focus on the p-adic numbers and their properties. First, we will examine the p-adic norm and look at some of the more interesting properties that result from it being non-archimedean. We will then try to understand the structure of the p-adic numbers and look at Hensel s Lemma. Finally, we will extend the norm to finite extensions of Q p and try to understand some of the structure behind totally ramified extensions. Contents. Introduction 2. The P-Adic Norm 2 3. The P-Adic Numbers 3 4. Extension Fields of Q p 6 Acknowledgments 0 References 0. Introduction The construction of the real numbers from the rationals relies very much on the usual absolute value. Furthermore, until one has constructed the p-adic numbers, it is not clear how important a role this norm plays in determining the properties of the real numbers. The p-adic norm behaves much differently than our usual concept of distance. For fixed prime p, two numbers are close together if their difference is divisible by a high power of p. The p-adic norm is also non-archimedean, and many unusual properties result from this fact. The completion of Q under the p-adic norm yields an interesting field Q p. We can think of elements in this field simply as infinite sequences of the form a = b m p m + b m p m +... + b 0 + b p +... + b n p n +... where 0 b i p. It s clear Q and Z lie in Q p, and elements of both Q and Z, when expanded p-adically, take on a particular form in Q p. We will explore the p-adic numbers and will be aided in this endeavor by Hensel s Lemma, which simplifies the task of finding roots to polynomials in Z p [x]. Our final task will be to examine extension fields of Q p. We will first have to extend the p-adic norm to these fields. It turns out there is a unique way to do this, and, for α K, where K is a finite extension, the extended p-adic norm will Date: August 26, 20.

2 EVAN TURNER only depend on the constant term in the monic irreducible polynomial of α over Q p. With these tools, it will not be difficult to characterize how one constructs totally ramified extension fields K of Q p I will assume fairly basic knowledge of field theory in this paper. For questions on that, any introductory algebra text will do. See, for instance, A First Course in Abstract Algebra by John Fraleigh. 2. The P-Adic Norm Definition 2.. For a prime number p and integer a, ord p a (the order of p at a) is the highest power of p that divides a, i.e. the greatest integer m such that a 0 mod p m. So ord 5 25 = 2, while ord 5 35 = and ord 3 25 = 0. It is a simple exercise to see that ord p (a a 2 ) = ord p a + ord p a 2. Another useful fact is that ord p (ad + bc) min (ord p (ad), ord p (bc)) simply because if the minimum order is r, then p r ad and bc as well as their sum. For x rational, i.e. for x = a, it makes sense to define b ord p x = ord p a ord p b. Now we introduce the p-adic norm on Q, which has many different properties from the usual absolute value. x p = p ordp x, x 0 0, x =0 It is relatively straightforward to prove that the p-adic norm is in fact a norm. However, while every norm must satisfy the triangle inequality, the p-adic norm satisfies an even stronger property that leads to some interesting results. Definition 2.2. A norm is non-archimedean if x + y max ( x, y ). Theorem 2.3. x p is a non-archimedean norm. Proof. If x = 0, y = 0, or x + y = 0, then it s obvious. For x = a b and y = c ad + bc, x + y = and d bd ord p (x + y) = ord p (ad + bc) ord p b ord p d min (ord p (ad), ord p (bc)) ord p b ord p d = min (ord p a + ord p d, ord p b + ord p c) ord p b ord p d = min (ord p a ord p b, ord p c ord p d) = min (ord p x, ord p y) So x + y p = ordp p (x+y) min(ordp x,ordp p y) = max ( x p, y p )

THE P-ADIC NUMBERS AND FINITE FIELD EXTENSIONS OF Q p 3 One peculiar result of this fact is that, under a non-archimedean norm, every triangle must be isosceles. If we suppose x < y, then x y max ( x, y ) = y But y = x (x y) max ( x, x y ) = x y and y = x y. 3. The P-Adic Numbers Building up the p-adic field Q p from the rational numbers is an interesting task and is quite similar to the way one constructs the real numbers from Q. However, as I would like to talk more about properties of the p-adic numbers, I will skip over this. For more information, see []. It turns out, though, that each p-adic number can be thought of as an infinite sequence of the form a = b m p m + b m p m +... + b 0 + b p +... + b n p n +... where 0 b i p. Furthermore, this construction is unique. Theorem 3.. A series converges in Q p if and only if its terms approach zero. Proof. Note that Q p is complete. I will not prove this fact but it is a straightforward result of the construction of the p-adic field. Suppose {a i } is a sequence in Q p such that a i p 0 as i. Then its partial sums S n = a 0 +... + a n are Cauchy since for n > m S n S m p = a m+ +... + a n p max ( a m+ p,..., a n p ) 0 as n, m by assumption. Q p is complete so this implies convergence. Now suppose then that we have a series that converges in Q p. Then it is Cauchy as well, and given ɛ > 0, there exists N such that for n, m > N, S n S m p < ɛ, where S n denotes the nth partial sum. In particular, S n+ S n p = a n+ p < ɛ. Thus, the terms approach zero as n. We define the p-adic integers to be the subring Z p = {x Q p x p } If a Z p, then clearly a = a 0 + a p + a 2 p 2 +...

4 EVAN TURNER since ord p a 0. Note that in this paper, I will use the word integer to mean elements of Z, while elements of Z p will always be specified as p-adic integers. Similarly, we define the set of p-adic units Z p = {a Z p /a Z p } = {a Z p a p =} The equivalence of these two sets is easy to see since if a p =, then /a p = so /a Z p. We now turn to a very important and useful result that will greatly simplify the task of finding roots in Z p for polynomials in Z p [x]. Theorem 3.2. (Hensel s Lemma) Let f(x) =x n + a n x n +... + a 0 be a polynomial in Z p [x] with formal derivative f (x) =nx n +... + a. Let c 0 be a p-adic integer such that f(c 0 ) 0 mod p and f (c 0 ) 0 mod p. Then there exists a unique p-adic integer c such that f(c) = 0 and c 0 c mod p Proof. The idea behind the proof is to construct a p-adic number c such that f(c) 0 mod p n for all n, which then clearly implies f(c) = 0. We will do this by first constructing a unique sequence of integers {b i } in Z such that the following conditions hold for all n : ) f(b n ) 0 mod p n+ 2) b n b n mod p n 3) 0 b n <p n+ We will prove this by induction on n. We begin with the case n =. There is clearly a unique integer, b 0, s.t. b 0 c 0 mod p and 0 b 0 <p. By 2) and 3), we need b b 0 mod p and 0 b <p 2 ; thus, b = b 0 + c p for some 0 c <p. Now we must check ). f(b )=f(b 0 + c p) = = a i (b 0 + c p) i (a i b i 0 + ia i b i 0 c p + terms divisible by p 2 ) (a i b i 0 + ia i b i 0 c p) mod p 2 = f(b 0 )+f (b 0 )c p We chose b 0 c 0 mod p, and since f(x) is a polynomial it follows that f(b 0 ) f(c 0 ) 0 mod p. Thus, f(b 0 ) βp mod p 2 for some 0 β <p. Similarly, since f (b 0 ) f (c 0 ) 0 mod p, we can uniquely solve for c in the equation βp + f (b 0 )c p 0 mod p 2 so that 0 c <p. This is equivalent to solving β + f (b 0 )c 0 mod p. c is then the unique integer between 0 and p such that c β f (b 0 ) mod p, and, for this choice of c, b = b 0 +c p satisfies properties ), 2), and 3). Now we continue the induction step and assume that we have found b,..., b n which satisfy ), 2), and 3). We need b n b n mod p n and 0 b n <p n+, so b n = b n + c n p n for some c n satisfying 0 c n <p. We will now solve for c n so

THE P-ADIC NUMBERS AND FINITE FIELD EXTENSIONS OF Q p 5 that ) is satisfied as well. f(b n )=f(b n + c n p n ) = = a i (b n + c n p n ) i (a i b i n + ia i b i n c np n + terms divisible by p n+ ) (a i b i n + ia i b i n c np n ) mod p n+ = f(b n )+f (b n )c n p n We know that f(b n ) 0 mod p n by the inductive assumption, so we can write f(b n ) αp n mod p n+. Our equation then becomes αp n + f (b n )c n p n 0 mod p n+ or α + f (b n )c n 0 mod p We constructed b n so that b n b 0 c 0 mod p so, as we argued before, f (b n ) f (c 0 ) 0 mod p. As before then, we can solve for c n uniquely so that 0 c n <p. With this c n, b n satisfies ), 2), and 3), and our claim is proved. Now we let Note that c b n mod p n+ so that c = b 0 + c p + c 2 p 2 +... f(c) f(b n ) 0 mod p n+ for all n which implies that f(c) = 0. The uniqueness of c follows from the uniqueness of the {b i }. Hensel s Lemma has many useful applications, one of which is the existence of square roots. Suppose a Z p, then a has a square root in Q p if and only if there exists b Z p such that f(b) =b 2 a 0 mod p. So if a = a 0 + a p + a 2 p 2 +... and b = b 0 + b p + b 2 p 2 +... then b 2 a mod p iff. b 2 0 a 0 mod p. Thus, we really just need to check if there exists an integer b 0 with 0 b 0 <psuch b 2 0 a 0 mod p. This makes the problem much simpler. The other {b i } can all be solved for once b 0 is found, and the process is similar to that used in the proof of the lemma. Example 3.3. Let s examine 9 in Z 5. 9 = 4 + 5 and 2 2 4 mod 5 so we can immediately see that 9 has a square root in Q 5. 7 = 2 + 5 does not however, since 2 a 2 mod 5 for a {0,, 2, 3, 4}. It is also an interesting exercise to examine the p-adic expansion of rational x. Proposition 3.4. The p-adic expansion of a Q p has repeating digits if and only if a Q Proof. Suppose a Q p has repeating digits, i.e., a = a m p m +... + a 0 +... ( a i p i + a i+ p i+ +... + a i+r p i+r )( +p r + p 2r +... )

6 EVAN TURNER First we need to show that ( +p r + p 2r +... ) converges to norm. p r ( +p r + p 2r +... + p nr) = p which goes to zero as n. So we have a = a m p m under the p-adic pr p (n+)r p r p n+ +... + a 0 +... ( a i p i + a i+ p i+ +... + a i+r p i+r ) ( ) p r which can be evaluated under our normal rules for addition and multiplication to get an element of Q. Now suppose x Q. We really just need to show that any rational number x = a b can be put in some form x = c + d p r for c, d positive integers and some r Z. It is not difficult to see that the p-adic expansion of any positive integer is the same as its base p expansion. Thus, any integer will have a finite p-adic expansion. With that in mind, if we can show that x is of this form, then x = a b will have a p-adic expansion with repeating digits once we expand c,d, and p r p-adically. We can reduce to the situation where p does not divide b since if it does and b = p r b 0, we simply examine a/b 0 and the extra /p r term will not change the fact that there are repeating digits. Finally, we let x = a/b be between - and 0, since if it s not, we can add a constant term that will not change the fact that there are repeating digits. We look at Z/bZ = {0,, 2,..., b }. Clearly for some m, n 0, p m p n ( mod b) since there are only b equivalence classes. It is fine to assume m>n, then p m n ( mod b), which implies that there exists c such that cb = p m n. Thus, we write x = a b = ac bc = and we have our repeating expansion. ac p m n p 4. Extension Fields of Q p Definition 4.. Let V be a finite-dimensional vector space over a field F. A field norm on V is a map v satisfying ) x v = 0 if and only if x = 0, 2) ax v = a x v for a F, x V, and the norm on F, and 3) x+y v x v + y v. If K is a finite extension field of F, K is an n-dimensional vector space over F. Thus, the concept of a field norm applies here. Before we can discuss extensions fields though, we must first extend the p-adic norm to these fields. First, however, we will define a different kind of norm.

THE P-ADIC NUMBERS AND FINITE FIELD EXTENSIONS OF Q p 7 Definition 4.2. Let K = F (α) be a finite extension of a field F and suppose α has monic irreducible polynomial f(x) =x n +... + a 0 for a i F. The norm of α from K to F N K/F (α) = det(a α ) where A α is the corresponding matrix for the F -linear map σ : K K given by σ(x) =αx. Notation 4.3. The monic irreducible polynomial for α over Q p will be denoted irred(α), if the polynomial exists. Proposition 4.4. The definition given above for N K/F (α) is equivalent to the following: ) N K/F (α) = ( ) n a 0, where n is the degree of irred(α) and a 0 is its constant term. n 2) N K/F (α) = α i, where the α i are the conjugates of α = α over F. i= Proof. We choose a convenient basis for K, an n-dimensional vector space over F, namely the set {, α,..., α n }. With this basis, the matrix A α has the following form, which shows that det(a α )=( ) n a 0 0 0 0... 0 a 0 0 0... 0 a 0 0... 0 a 2..................... a n Once factored, n irred(α) =x n + a n x n... + a 0 = (x α i ) Thus, it is clear that n α i =( ) n a 0 i= One can prove that there can only be one extension of the p-adic norm to any finite extension field K of Q p. For the interested reader, see []. I will motivate the extension, but I will not prove that the extended norm is a field norm nor that it is the unique extension. Remarks 4.5. Note that Proposition 4.4 implies N K/F (α) Q p. We can also see that N K/F (α)n K/F (β) =N K/F (αβ) simply because the determinant map satisfies this multiplicative property. Construction 4.6. Suppose irred(α)= x n + a n x n +... + a 0. Let K be the finite extension field of Q p containing α and all its conjugates. Then K is a Galois extension. Let σ be an automorphism of K fixing Q p sending α to a conjugate α i. If is the unique field norm extending p to K, then let : K R be given by x = σ(x). It is not difficult to show that satisfies a field norm as well. i=

8 EVAN TURNER But this means α = α = σ(α) = α i, so conjugates have the same norm. By Remarks 4.5 and Proposition 4.4, N Qp(α)/Q p (α) p = N Qp(α)/Q p (α) n = α i = i= n α i i= = α n So we see that α = N Qp(α)/Q p (α) /n p. If K is a finite extension of Q p containing α, then α = N K/Qp (α) /[K:Qp]. This follows from the fact that we can form a basis for K over Q p from the product of {, α,..., α n } and {x,x 2,..., x m } where {α i } form a basis for Q p (α) over Q p, {x j } form a basis for K over Q p (α), and m =[K : Q p (α)]. The linear map given by multiplication by α then yields a matrix with m blocks along the diagonal given by the matrix A α given in Proposition 4.4. The determinant of this block matrix is then det(a α ) m. Thus, So N K/Qp(α) = ( N Qp(α)/Q p ) [K:Qp(α)] ( NK/Qp(α)) /[K:Qp] = ( ) [K:Qp(α)] [K:Qp] N Qp(α)/Q p = ( N Qp(α)/Q p ) /[Qp(α):Q p] = ( N Qp(α)/Q p ) /n Note that the extension of the p-adic norm will still be non-archimedean. Notation 4.7. From now on, we will denote the extension of the p-adic norm by p too. Definition 4.8. Let K be a finite extension of Q p, and let A be the set of all α K such that α is a root of a polynomial in Z p [x], i.e., f(α) =α n + a n α n +... + a 0 = 0, Then A is called the integral closure of Z p in K. a i Z p Theorem 4.9. Let K be a finite extension of Q p of degree n, and let A = {x K x p } M = {x K x p < } Then A is a ring, the integral closure of Z p in K. M is its unique maximal ideal, and A/M is a finite extension of F p of degree at most n. Proof. A inherits the structure of K so we just need to check that it s closed under addition and multiplication. Clearly if x, y A, then xy p, so xy A. Also, x + y p max( x p, y p), so x + y A, and A is a ring. For a A and m M, am p = a p m p <, so am M. M is an additive subgroup of A, so M is an ideal.

THE P-ADIC NUMBERS AND FINITE FIELD EXTENSIONS OF Q p 9 Now we want to show that A is the integral closure of Z p in K. Suppose α A, then α p. Since α K, α is algebraic over Q p with irred(α) =x m + a m x m +... + a 0 = m (x α i ) where the α i are conjugates of α. We know that all conjugates of α have the same norm, i.e. α i and α i A. But the coefficients a i of irred(α) are sums and products of the α i, which implies that a i p as well. Since the a i Q p already, the a i are actually in Z p and we see that irred(α) Z p [x]. If α K is the root of a polynomial in Z p [x], it s pretty clear that irred(α) Z p [x] too. Let the degree of irred(α) be m so that α m +a m α m +...+a 0 = 0. Suppose α p >. Then i= α m p = α m p = a m α m... a 0 p max( a m α m p,..., a 0 p ) max( α m p,..., ) since a i p = α m p but this is a contradiction since α p >. Thus, α p, and α A. The set A defined in the theorem is then the integral closure of Z p in K. Let s now turn to M being a maximal ideal. Suppose N is an ideal such that M < N < A (this is a strict inequality). So there exists α A such that α N but α M. This implies α p = so that /α p = and /α A. But then α α = N, which is a contradiction since N A. So M is maximal in A. Since M is maximal and A is a ring, we know that A/M is a field. It s not difficult to see that M Z p = pz p. Clearly if a Z p, then a A since a p. If a, b Z p, then a + M and b + M represent the same coset iff. a b M Z p = pz p. But this means that for a + pz p Z p /pz p, there exists a corresponding a + M A/M. Or equivalently, that F p = Z p /pz p lies in A/M. Thus, A/M is an extension field of F p. To show that [A/M : F p ] n, we show that any linear combination of n + elements over F p is linearly dependent. So we take a + M, a 2 + M,..., a n+ + M A/M for a i A The a i are clearly linearly dependent over Q p since each a i K. Thus, (4.0) b a +... + b n+ a n+ = 0 for b i Q p If we let m = min(ord p b,..., ord p b n+ ), then we can multiply 4.0 by p m and obtain a similar equation with all coefficients in Z p and at least one is not in pz p (Just to make sure notation is clear, note that if m<0, then we ll be multiplying by p m ). Thus, if we map 4.0 into A/M, we get an equation of the form (4.) b a +... + b n+ a n+ + M = 0 + M for b i F p Any b i pz p is mapped to 0 + M in A/M, but we know this isn t the case for all the b i. This shows that the {a i } are linearly dependent and the claim is proved. Definition 4.2. The field A/M described in the preceding theorem is called the residue field of K.

0 EVAN TURNER We can extend the p-adic norm even further, to the algebraic closure of Q p, denoted Q p. For α Q p, the norm itself depends only on the constant term in irred(α), so this is fine intuitively. Let K be an extension of Q p of degree n, and let α K. Then we define ord p α = log p α p = log p N K/Qp (α) /n p = n log p N K/Qp (α) p For α Q p, this agrees with the earlier definition, and has the same multiplicative property. The image of K under the ord p map is contained in (/n) Z = {x Q nx Z} The image is an additive subgroup of (/n) Z so it s of the form /e for some integer e dividing n. Definition 4.3. This integer e is called the index of ramification of K over Q p. If e =, K is called an unramified extension of Q p, while if e = n, the extension K is called totally ramified. Theorem 4.4. If K is totally ramified and α K is such that ord p α =/e, then α satisfies an Eisenstein equation x e + a e x e +... + a 0 =0, for a i Z p where a i 0( mod p) for all i, but a 0 0( mod p 2 ). Conversely, if α is a root of an Eisenstein polynomial of degree e over Q p, then Q p (α) is totally ramified over Q p of degree e. Proof. If ord p α =/e, then irred(α) over Q p has degree e based on the definition of ord p α. Thus, we need to show that the coefficients a i are in Z p. In Q p [x], e irred(α) = (x α i ) i= where α i are conjugates of α. Thus, we see that the a i are symmetric in α i, which implies a i p. Also, a 0 p = α e p =/p so we see that a 0 0( mod p 2 ). Now suppose α is a root of an Eisenstein polynomial. It is well-known that an Eisenstein polynomial is irreducible over Q, and the same applies in Q p, although I will not prove it. Thus, irred(α) has degree e and [Q p (α) :Q p ]=e. ord p a 0 = by assumption, so so Q p (α) is totally ramified over Q p. ord p α = log p α p = e log p a 0 p =/e Acknowledgments. I want to thank my mentors, Shawn Drenning and Casey Rodriguez, for their time in helping me with this paper. They were great to work with and excellent resources. I also want to thank the REU program in general for giving me the opportunity to work on math this summer and to study a topic that I may not have been able to experience otherwise. References [] Neal Koblitz p-adic numbers, p-adic analysis, and Zeta-Functions, 2nd ed. Springer-Verlag. 984.