
An introduction to p-adic numbers Tudor Ciurca June 22, 2020 Introduction : let p be a positive integer prime There are 3 main ways to think about the p-adic numbers. I will go through them starting with the most elementary method. Note that this is a 5-adic integer. 1 : Example A p-adic number can be thought of as a base-p number where we allow digits to go infinitely far to the left. We can represent 1=3 as a 5-adic integer as follows 1 ::: 1313132:0 = 2 3 Z5 1 : Example A p-adic number can be thought of as a base-p number where we allow digits to go infinitely far to the left. We can represent 1=3 as a 5-adic integer as follows 1 ::: 1313132:0 = 2 3 Z5 Note that this is a 5-adic integer. A p-adic number can be written as 1 X k akp k=n where n is an integer and each \digit" ak is one of 0; 1 : : : p − 1. If n is a non-negative integer then we call the p-adic number above a p-adic integer. 1 : p-adic numbers as power series in p One way to think about the p-adic numbers is as power series in p, just as the real numbers can be thought of as power series in 10−1 [Decimal expansions]. If n is a non-negative integer then we call the p-adic number above a p-adic integer. 1 : p-adic numbers as power series in p One way to think about the p-adic numbers is as power series in p, just as the real numbers can be thought of as power series in 10−1 [Decimal expansions]. A p-adic number can be written as 1 X k akp k=n where n is an integer and each \digit" ak is one of 0; 1 : : : p − 1. 1 : p-adic numbers as power series in p One way to think about the p-adic numbers is as power series in p, just as the real numbers can be thought of as power series in 10−1 [Decimal expansions]. A p-adic number can be written as 1 X k akp k=n where n is an integer and each \digit" ak is one of 0; 1 : : : p − 1. If n is a non-negative integer then we call the p-adic number above a p-adic integer. Let Zp be the p-adic integers. 1 : p-adic numbers as power series in p For a prime p let Qp be the p-adic numbers. 1 : p-adic numbers as power series in p For a prime p let Qp be the p-adic numbers. Let Zp be the p-adic integers. To get the field of p-adic numbers Qp, we just add the fractions s pn where s is a p-adic integer and n a positive integer. 2 : p-adic integers as an algebraic limit Another way to construct a p-adic integer is by giving a sequence of elements 2 3 a1 2 Z=pZ; a2 2 Z=p Z; a3 2 Z=p Z ::: m which are compatible, so that ak ≡ am (mod p ) if k ≥ m. 2 : p-adic integers as an algebraic limit Another way to construct a p-adic integer is by giving a sequence of elements 2 3 a1 2 Z=pZ; a2 2 Z=p Z; a3 2 Z=p Z ::: m which are compatible, so that ak ≡ am (mod p ) if k ≥ m. To get the field of p-adic numbers Qp, we just add the fractions s pn where s is a p-adic integer and n a positive integer. so the bk which are the partial sums of a, are also the components of a modulo pm. Relation with the first definition (from 1 to 2) k k Let us think of Z=p Z as the set of representatives 0; 1 : : : p − 1 P1 k in Z, and suppose we are given a p-adic integer a = k=0 akp . Then we have m−1 X k m bm := akp ≡ a (mod p ) k=0 Relation with the first definition (from 1 to 2) k k Let us think of Z=p Z as the set of representatives 0; 1 : : : p − 1 P1 k in Z, and suppose we are given a p-adic integer a = k=0 akp . Then we have m−1 X k m bm := akp ≡ a (mod p ) k=0 so the bk which are the partial sums of a, are also the components of a modulo pm. th ak−ak−1 To get the k coefficient bk we just use pk−1 = bk when k k ≥ 1, and we set b0 = a0. Here we again identify Z=p Z with the representatives 0; 1 : : : pk − 1. Relation with the first definition (from 2 to 1) Suppose we are given a compatible sequence k a = fak 2 Z=p Zgk≥1. We want to give the power series P1 k expansion a = k=0 bkp of the corresponding p-adic integer. Relation with the first definition (from 2 to 1) Suppose we are given a compatible sequence k a = fak 2 Z=p Zgk≥1. We want to give the power series P1 k expansion a = k=0 bkp of the corresponding p-adic integer. th ak−ak−1 To get the k coefficient bk we just use pk−1 = bk when k k ≥ 1, and we set b0 = a0. Here we again identify Z=p Z with the representatives 0; 1 : : : pk − 1. It can also be written using a power series expansion in 5 1 = 2 + 3 · 5 + 1 · 52 + ::: = (::: 132:0) 3 5 1 Example: 3 in Z5 Since 3 is coprime to 5 we can find its multiplicative inverse modulo 5k for all positive integers k. For example 1 1 1 ≡ 2 (mod 5); ≡ 17 (mod 25); ≡ 42 (mod 125) ::: 3 3 3 1 Example: 3 in Z5 Since 3 is coprime to 5 we can find its multiplicative inverse modulo 5k for all positive integers k. For example 1 1 1 ≡ 2 (mod 5); ≡ 17 (mod 25); ≡ 42 (mod 125) ::: 3 3 3 It can also be written using a power series expansion in 5 1 = 2 + 3 · 5 + 1 · 52 + ::: = (::: 132:0) 3 5 It can also be written using a power series expansion in 7 2 = 6 + 2 · 7 + 1 · 72 + ::: = (::: 126:0) 5 7 2 Example: 5 in Z7 Since 5 is coprime to 5 we can find its multiplicative inverse modulo 5k for all positive integers k. For example 2 2 2 ≡ 6 (mod 7); ≡ 20 (mod 49); ≡ 69 (mod 343) ::: 5 5 5 2 Example: 5 in Z7 Since 5 is coprime to 5 we can find its multiplicative inverse modulo 5k for all positive integers k. For example 2 2 2 ≡ 6 (mod 7); ≡ 20 (mod 49); ≡ 69 (mod 343) ::: 5 5 5 It can also be written using a power series expansion in 7 2 = 6 + 2 · 7 + 1 · 72 + ::: = (::: 126:0) 5 7 We select one p of the representatives for 7: p p p 7 ≡ 1 (mod 3); 7 ≡ 4 (mod 9); 7 ≡ 13 (mod 27) ::: It can also be written using a power series expansion in 3 p 2 7 = 1 + 1 · 3 + 1 · 3 + ::: = (::: 111:0)3 p Example: 7 in Z3 We will see later that to verify the existence of algebraic numbers in p-adic number systems, it suffices to verify their existence modulo p (and some other conditions). It can also be written using a power series expansion in 3 p 2 7 = 1 + 1 · 3 + 1 · 3 + ::: = (::: 111:0)3 p Example: 7 in Z3 We will see later that to verify the existence of algebraic numbers in p-adic number systems, it suffices to verify their existence modulo p (and some other conditions). We select one p of the representatives for 7: p p p 7 ≡ 1 (mod 3); 7 ≡ 4 (mod 9); 7 ≡ 13 (mod 27) ::: p Example: 7 in Z3 We will see later that to verify the existence of algebraic numbers in p-adic number systems, it suffices to verify their existence modulo p (and some other conditions). We select one p of the representatives for 7: p p p 7 ≡ 1 (mod 3); 7 ≡ 4 (mod 9); 7 ≡ 13 (mod 27) ::: It can also be written using a power series expansion in 3 p 2 7 = 1 + 1 · 3 + 1 · 3 + ::: = (::: 111:0)3 Picture of Z3 1 jj · jj is positive definite. This means jjqjj ≥ 0 for all q 2 Q, and equality holds if and only if q = 0. 2 jj · jj is multiplicative. This means jjqjj · jjrjj = jjq · rjj for all q; r 2 Q. 3 jj · jj satisfies the triangle inequality. This means jjq + rjj ≤ jjqjj + jjrjj for all q; r 2 Q. Absolute values on Q An absolute value is a nice way of measuring size. It is a function jj · jj : Q ! Q satisfying the following conditions: 2 jj · jj is multiplicative. This means jjqjj · jjrjj = jjq · rjj for all q; r 2 Q. 3 jj · jj satisfies the triangle inequality. This means jjq + rjj ≤ jjqjj + jjrjj for all q; r 2 Q. Absolute values on Q An absolute value is a nice way of measuring size. It is a function jj · jj : Q ! Q satisfying the following conditions: 1 jj · jj is positive definite.
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