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10 Oct 2012: Show that L = K(α) = {a bα |a, b K} for some α F with α2 K. ... ii) Show that if L/K is a finite extension with Q K, for which there exist only finitelymany intermediate subfields K K′ L, then L = K(α) for some α

22 Oct 2012: Show that they have the samesplitting field if and only if b = cnar for some c K and r N with gcd(r, n) = 1. ... for some divisor d > 1 of n.

29 Feb 2012: α for some α.(iv) Any limit ordinal can be written in the form α. ... Given a subset b Pα withmore than one element, consider the least β belonging to some but not all members of b.].

30 Oct 2012: α for some α.(iv) Any limit ordinal can be written in the form α. ... Given a subset b Pα withmore than one element, consider the least β belonging to some but not all members of b.].

3 Dec 2012: first part. 1. The square matrices A and B over the field F are congruent if B = P T AP for some invertible matrixP over F. ... for some λ,µ,ν R (which should turn out in this example to be integers).

28 Feb 2012: where α = 12(1. 71). Find an element of OK with norm 2a 3b 5 for some. ... Hint: Use that some power of a is principal.]. The following extra questions are on cyclotomic fields.

5 Nov 2012: 3.17. (i) Let f K(X). Show that K(X) = K(f) if and only if f = (aX b)/(cX d)for some a, b, c, d K with ad ... Deduce that every finite abelian group is the Galois group of some Galois extension K/Q.[It is a long-standing unsolved problem (inverse Galois

18 Oct 2012: Some peopledon’t require v(K) 6= {0} (so allow the “trivial valuation”).Examples. ... p 1}. for some N. In either case, vp(x) = min{n | an 6= 0}.

6 Feb 2012: Suppose that both Φ and ΦB are alternating for some B Mat5(K).Then PΦ = PΦB = 0 and ΦPT = ΦBPT = 0. ... for some A,B GL5(K) and µ K. Commutativity of this diagramgives PT1 = µA.

24 Jul 2012: exp(iNαj) λj| < ǫ. for each 1 j M. Our construction requires some preliminary results. ... 14. 5 Some simple geometry of numbers. We need the following extension of Theorem 35.