## Math 250      Higher Algebra

 Harvard Fall 2006 Instructor: Thomas Lam
Course Assistant: Thanos Papaioannou (email: apap@fas)
 Lectures: Monday, Wednesday, Friday 2-3      Science Center 109
 Office Hours: Monday 3:15-4:30, Thursday 11:00-12:00.      Science Center 435

Prerequisites:
Mathematics 123 or equivalent.

Problem sets: There will be problem sets roughly once a week. Collaboration on homework is permitted, but you are not allowed to just copy someone else's work. You have to mention on your problem set who you worked with.
Pset1 (Due 09/27)
Pset2 (Due 10/04)
Pset3 (Due 10/11)
Pset4 (Due 10/20)
Pset5 (Due 11/1) Updated 10/25
Pset6 (Due 11/10) You may assume that char F = 0 in Problem 1 d,e
Pset7 (Due 11/17) sl(2) denotes the lie algebra of matrices with trace 0

Grading: Problem sets (70%), Paper (30%).

Notes:
(12/05) Office hours : I won't be able to make it to office hours Thursday 12/07. Please email me for an appointment.
(09/18) When I classified the maps phi: F[x] -> F[u], I assumed that F[u] was a subring of a field. This implies that it is an integral domain, so that the kernel of phi is a prime ideal and generated by an irreducible polynomial. This is not true for an arbitrary ring F[u] which is generated by F and an element u.
(09/22) The polynomial f(x) = x^(p^n) - x has no multiple roots over F_p, since its derivative is -1 which is relatively prime to f(x). In class, I mistakedly used the fact that F_p is perfect.
(10/06) The equality (in the proof of Lemma 2b) sigma(w^i,c) = w^{-i}(w^i,c) is not quite sufficient to show that (w^i,c) generates: one also needs to know that (w^i,c) is non-zero. So perhaps in the end the determinant still has to be calculated.
(11/1) Michael pointed out the following. To show that a' = a + n - 2an is not nilpotent, we should not raise a' to a high power but instead raise a = a' - n + 2an to a high power. If a' is nilpotent, then this shows that a is also nilpotent, a contradiction.
(11/21) Happy Thanksgiving! Thanks for checking the webpage, but there won't be any more problem sets :).