Witt Vectors

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Witt Vectors Last revised 12:50 p.m. April 12, 2018 Witt vectors Bill Casselman University of British Columbia [email protected] A unipotent algebraic group is one that possesses a filtration whose associated graded module is a direct sum of copies of Ga. In characteristic 0 all commutative ones are isomorphic to products of Ga, but in characteristic p> 0 there are others. Most important are the groups associated to Witt vectors. Contents 1. Introduction 2. Unramified p•adic extensions 3. The Witt scheme 4. The ring of power series 5. The Artin•Hasse exponential 6. References 1. Introduction Suppose p to be prime. The additive group of Z/p2 has the same number of elements as the vector space of dimension two over the field Fp = Z/p, but of course is not isomorphic to it as a group. The following curious question however arises: can we introduce Fp•coordinates on the first group so that the group 2 structure becomes algebraic? Or, equivalently, can we find polynomials defining on Fp the structure of an 2 algebraic group whose set of rational points over Fp is isomorphic to Z/p ? In fact, we shall require and obtain something a bit stronger, as I shall explain soon. We can begin in an elementary fashion. Every element x of Z/p2 can be written uniquely in the form x = a + bp where 0 ≤ a,b < p; thus one natural coordinate system is to associate to x the vector (a,b). I define this coordinate map a little more precisely, Let π be the canonical projection from Z/p2 to Z/p, and let θ be the splitting of π which takes an element a of Z/p to the image in Z/p2 of the unique integer m in [0,p) 2 with π(m)= a. For any x in Z/p its first coordinate is x0 = π(x). The difference x − θ(x0) is a multiple of p, and its second coordinate is the image in Z/p of the quotient: x − θ(x ) x = π 0 . 1 p 2 Suppose x and y to be elements of Z/p with coordinates (x0, x1) and (y0,y1). What are the coordinates of the sum x + y? Its first coordinate is z0 = π(x + y)= x0 + y0, and according to the prescription above the second is x + y − θ(x + y ) z = π 0 0 1 p x + y − θ(x ) − θ(y ) θ(x )+ θ(y ) − θ(x + y ) = π 0 0 + π 0 0 0 0 p p x − θ(x )+ y − θ(y ) θ(x )+ θ(y ) − θ(x + y ) = π 0 0 + π 0 0 0 0 p p θ(x )+ θ(y ) − θ(x + y ) = x + y + π 0 0 0 0 . 1 1 p At first sight the question of algebraicity therefore amounts to the question: does there exist a polynomial f(x0,y0) with coefficients in Fp such that θ(x )+ θ(y ) − θ(x + y ) f(x ,y )= π 0 0 0 0 ? 0 0 p Witt vectors 2 It is eventually more illuminating to consider a more general question: suppose q to be a power of p, o a discrete valuation ring with maximal ideal p = (p) such that o/p is Fq. Let θ be any splitting of the canonical 2 2 projection π, and we ask the same question about the group o/p of q elements, except that now f(x0,y0) is allowed to have coefficients in Fq. q n Since x = x in Fq, every polynomial over Fq determines the same function on Fq as one which has only monomials of degree < q in any one variable. I call this a reduced polynomial. We then have this result, n which provides an answer to the previous question: Any Fq•valued function on Fq may be expressed by a unique reduced polynomial over Fq in n variables. Proving this remark is straightforward. It suffices to follow the method of Lagrange interpolation, and construct for each α in Fq the reduced indicator polynomial Pα(x) in one variable x, satisfying the conditions 0 (β = α) P (β)= . α 1 (β = α) This is because we can take a product of such one•variable polynomials to get indicator polynomials in several variables. The formula of Lagrange is valid in any field: β=α(x − β) Pα(x)= β=α(α − β) The situation is not yet very satisfactory, since the polynomial we get depends strongly on the splitting θ we have chosen. In fact, there is in every case a canonical splitting: For any element x0 of o/p there exists a unique element x 2 2 of o/p with the two properties (a) π(x)= x0; (b) x is a p•th power in o/p . 1/p This element is called the Teichmuller¨ representative τ(x0) of x0. Tofindit: let x0 be given, and let a = x0 . Choose α in o/p2 with π(α) = a, and let x = αp. It turns out that x does not depend on the choice of α, and we can therefore define τ(a) to be x. Why doesn’t x depend on α? If α and α∗ are two choices, then α∗ = α + cp for some c in o, and p p 2 p−1 p 2 α∗ = α + cp α + · · · = α (mod p ). The map τ is clearly a multiplicative homomorphism. With this choice of splitting, we can do calculations explicitly. Suppose x in o/p with (1.1) x = τ(x0)+ τ(X1)p . (We’ll see later why I use X instead of x. Be assured, it is only temporary.) How do we specify x0 and X1? p 2 p Well, x0 is (still) the image of x modulo p. Say α0 = x0, and α in o/p has image α0. Then α = τ(x0) and hence x − αp X = π . 1 p Similarly write y = βp. For the second coordinate of the sum x + y we have αp + βp − (α + β)p Z = X + Y + π 1 1 1 p (1.2) p−1 p−1 = X1 + Y1 − π(α β + · · · + αβ ) p−1 p−1 = X1 + Y1 − (α0 β0 + · · · + α0β0 ) Witt vectors 3 This is still not quite what we would like, because the last part of this expression is not a polynomial in x0 1/p 1/p and y0, but only one in α0 = x0 and β0 = y0 . There is a simple trick to solve thsi problem. Apply the Frobenius automorphism to both sides of this last equation. We get p p p p−1 p−1 Z1 = X1 + Y1 − (x0 y0 + · · · + x0y0 ) . p 1/p This suggests choosing the second coordinate of x to be x1 = X1 rather than X1. Since X1 = x1 we now have 1/p (1.3) x = τ(x0)+ τ(x1 )p , and (1.2) becomes p−1 p−1 z1 = x1 + y1 − (x0 y0 + · · · + x0y0 ). We shall see later that the entire ring structure of o can be recovered similarly. At any rate, we now have a candidate for a commutative algebraic group structure on affine space of dimension two in characteristic p: 1.4. Proposition. The formula p−1 p−1 (x0, x1) + (y0,y1)= x0 + y0, x1 + y1 − (x0 y0 + · · · + x0y0 ) defines a group law. Proof. The second coordinate of the sum x + y + z is (formally) αp + βp + γp − (α + β + γ )p x + y + z + 0 0 0 0 0 0 1 1 1 p from which associativity is transparent. The algebraic group W2 defined by the formula fits into a short exact sequence 1 −→ Ga −→ W2 −→ Ga −→ 1 that doesn’t split. 2. Unramified p-adic extensions m Suppose that q = p , let f(x) be an irreducible polynomial with coefficients in Fp defining the extension Fq, and let F (x) be any polynomial with coefficients in Zp reducing to f(x) modulo p. Hensel’s Lemma implies that the field extension of Qp obtained by adjoining a root of F (x) is Galois and unramified. The Galois group is cyclic of degree m, generated by the Frobenius F. Let o be its ring of integers, π the canonical projection o → Fq. 2.1. Proposition. There exists a unique multiplicative splitting τ of π. The following are equivalent: (a) the element x lies in the image of τ; (b) the element x is a pn•th power for every n> 0; (c) F(x)= xp; (d) xq = x. The map τ is called the Teichmuller¨ map. Proof. All these assertions will follow from the simple claim that given a in Fq there exists a unique x in o q such that π(x)= a; (2) x = x. To see this, suppose x0 to be any element of o with π(x)= a. I claim that the sequence q xn+1 = xn Witt vectors 4 converges in o to some x with xq = x. This is immediate from k k 2.2. Lemma. In any ring, if x ≡ y (mod p)) then xp ≡ yp (mod pk). To prove this lemma, it suffices to show that if x ≡ y (mod pk) then xp ≡ yp (mod pk+1). This is immediate from the binomial theorem. Define τ(a) to be the limit x. Any element x of o may be expressed as a unique series 2 x = τ(x0)+ τ(x1)p + τ(x2)p + · · · and since Fq is perfect it may in fact be expressed as some other series 2 1/p 1/p 2 (2.3) x = τ(x0)+ τ(x1 )p + τ(x2 )p + · · · .
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