Garrett Z p. Few explicit parametrizations of algebraic closures of fields are known: not Q, for sure. But we do also know

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1 Garrett Examples (cont d): Function fields in one variable... as algebraic parallels to Z and Q. Theorem: All finite field extensions of C((X z)) are by adjoining solutions to Y e = X z for e = 2, 3, 4,.... [Done] Thus, Gal ( C((X))/C((X)) ) = lim d Z/d = Ẑ p Z p Few explicit parametrizations of algebraic closures of fields are known: not Q, for sure. But we do also know Gal(F q /F q ) = lim d Z/d = Ẑ p Z p

2 Garrett In anticipation: Newton polygons over Q p This is the assertion for Z p [T ] corresponding to C[[X]][T ] above. The Newton polygon of a polynomial f(t ) = T n + a n T n a o Z p [T ] is the (downward) convex hull of the points (0, 0), (, ord p a n ), (2, ord p a n 2 ),... (n, ord p a o ) When we extend ord p (p n a b ) = n to algebraic extensions of Q p, we will prove that the slopes of the line segments on the Newton polygon are the ords, with multiplicities, of the zeros. The extreme case that ord p a 0 = is Eisenstein s criterion. This device is one of few human-accessible computational means. We will get to this...

3 Garrett Returning to finite scalars in place of C... a key point is the finiteness of residue fields o/p. Infinitude of primes: Because the algebraic closure of F q is of infinite degree over F q, by separability there are single elements α of arbitrarily large degree, whose minimal polynomials in F q [X] give prime elements of arbitrarily large degree, thus, infinitelymany. Also, we can mimic Euclid s proof. Use the fact that F q [X] is a PID. Given any finite collection P,..., P n of monic irreducibles in F q [X], the element N = X P... P n + is of positive degree, so has some irreducible factor, but is not divisible by any P j. /// One should contemplate what it would take to prove an analogue of Dirichlet s Theorem on primes in arithmetic progressions.

4 Garrett The finiteness of residue fields allows definition of the zeta function of o = F q [X]: Z(s) = = = = = = 0 a ideal F p [X] 0 a ideal F p [X] monic f monic f degrees d degrees d (Na) s (#F p [X]/a) s (#F p [X]/ f ) s (q deg f ) s #{monic f : deg f = d} q ds q d q ds = q s

5 Since F q [X] is a PID, there is an Euler product Z(s) = = = d 0 p prime monic irred f ( (Np) s convergent for R(s) >. Observe that = d #irred monics deg d = (q d prime p d q d/p + q s deg f Garrett q sd ) #monic irred f deg=d distinct p,p 2 d # elements degree d over F q #each Galois conjugacy class q d/p p 2 ) q d/p p 2 p distinct p,p 2,p 3 d The fact that Z(s) = /( q s ) is not obvious from the Euler factorization.

6 Garrett Example: in F 3 [x], monic irreducibles of low degrees are x, x +, x + 2 (3 (irred) monic linear) x 2 +, x 2 + 2x + 2, x 2 2x + 2 x 3 x +, x 3 x + 2,... (all x 3 a s are reducible!?!) ( = 3 irred monic quadratics) ( = 8 irred monic cubics) x 4 2x +,... ( = 8 irred monic quartics) (all x 4 a s are reducible!?!)??? ( = 48 irred monic quintics) (all x 5 a s are reducible!?!) No simple conceptual argument, but some reusable tricks... :

7 Garrett Since F 3 is a cyclic 2-group, there is no 4th root of unity, so the 4 th cyclotomic polynomial x 2 + is irreducible. Then (x + j) 2 + is irreducible for j =, 2. This happens to give all 3 irreducible monic quadratics. Since x 3 a = (x a) 3 for a F 3, none of these cubics is irreducible. The two cubics x 3 x + a with a 0 are Artin-Schreier polynomials over F 3. Since α 3 α = 0 for α F 3, these have no linear factors, so are irreducible. With j F 3, x x + j leaves these unchanged! No quartic x 4 a F 3 [x] is irreducible: F 3 is cyclic of order = 80 = 2 4 5, so every a F 3 is an 8th power. Since (3 2 )/4 = 2, fourth powers of α F 3 have order 2, so are 2 in F 3. Thus, α4 aα + b for non-zero a, b F 3. Thus, the four polynomials x 4 ax b with non-zero a, b F 3 are irreducible.

8 Garrett Artin-Schreier polynomials: Taking p th roots is problematical in characteristic p... Already the quadratic formula fails in characteristic 2. A root of x 2 + x + = 0 in F 2 2 cannot be expressed in terms of square roots! Over F p with prime p, the Artin-Schreier polynomials are x p x + a, with a F p. Claim: Artin-Schreier polynomials are irreducible, with Galois group cyclic of order p. Proof: For a root α F p of x p x + a = 0, (α + ) p (α + ) + a = α p α + a = 0 Thus, any field extension containing one root contains all roots. That is, the splitting field is F p (α) for any root α. But the Frobenius automorphism α α p generates the Galois group, whatever it is, and α p = α a, which is of order p. Thus, the Galois group is cyclic of order p. ///

9 Garrett For o = F p [x], completions are x-adic completion of o = F p [[x]] (x + )-adic completion of o = F p [[x + ]] (x 2 + )-adic completion of o = F p [[x 2 + ]][x] = {(a o x + b o ) + (x 2 + )(a x + b ) + (x 2 + ) 2 (a 2 x + b 2 ) +...} Generally, for P irreducible monic P -adic completion of o = c o (x) + c (x) P + c 2 (x) P (deg c j < deg P ) Also, corresponding to the point at infinity and its local ring F p [[/x]] F p (x) inside F p (x), x adic completion of o = F p[[/x]]

10 Garrett In his 92 thesis, E. Artin considered hyperelliptic curves over a finite field (of odd characteristic, for simplicity): y 2 = f(x) (with monic f(x) F q [x]) These are the quadratic extensions K of k = F q (x)... other than constant field extensions going from F q (x) to F q 2(x). We saw that the integral closure of o = F p [x] in K is F p [x, y]. How do primes in o = F q [X] behave in these extensions? The algebra computation can be applied: for P degree d monic prime in F q [x], and for O = F q [x, y], letting α be the image of x in F q [x]/p F q d, O/ P F q [x, t]/ P, t 2 f F q d[t]/ t 2 f(α) Thus, apart from the ramified prime f(x) F q [x], which becomes a square, there are split primes and inert primes: O/ P F q d F q d and P O P P 2 (if f(α) (F q d) 2 ) O/ P F q 2d and P O = prime in O (if f(α) (F q d) 2 )

11 Garrett Example: for y 2 = x 2 + over F 3, O/ x F 3 [x, t]/ x, t 2 x 2 F 3 [t]/ t 2 F 3 F 3 O/ x + F 3 [x, t]/ x +, t 2 x 2 F 3 [t]/ t 2 2 F 3 2 O/ x F 3 [x, t]/ x, t 2 x 2 F 3 [t]/ t 2 2 F 3 2 O/ x 2 + F 3 [x, t]/ x 2 +, t 2 x 2 F 3 2[t]/ t 2 not product That is, unsurprisingly, the prime x 2 + is ramified. Ok. O/ x 2 + 2x + 2 F 3 [x, t]/ x 2 + 2x + 2, t 2 x 2 F 3 (α)[t]/ t 2 α 2 Is α 2 + a square in F 3 (α) F 3 2 brute-force computation? where α 2 + 2α + 2 = 0? Some

12 Garrett O/ x 3 x + F 3 [x, t]/ x 3 x +, t 2 x 2 F 3 (α)[t]/ t 2 α 2 (with α 3 α + = 0) Is α 2 + a square in F 3 (α) F 3 3? More brute-force computation? Or,... a clear pattern of whether f(α) is a square in F p (α)? F p (α) is cyclic, and Euler s criterion applies: f(α) F p (α) 2 f(α) qd 2 = What should quadratic reciprocity be here? Why should there be a quadratic reciprocity? What about quadratic reciprocity over extensions of Q, like Q(i), too!?! A preview... and example of the way that more classical reciprocity laws are corollaries of fancier-looking things... :