Math 100C homework  
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HW #8 (for June 2)
Problems 1-3 at the end of section 8.5, and
Problem A:   Let f_1, f_2, f_3, ... be a
sequence of symmetric polynomials
in Q[x_1, ..., x_n]  with the property that for each j, the leading 
term in f_j is greater than the leading term of f_(j+1) whenever
f_j is nonconstant.
(By "greater", we mean greater in the sense of lexicographic order,
for example, x^4y^2z^0 is greater than x^2y^3z^1).
Explain why f_t is a constant polynomial for some positive integer t.

Here is a solution to Problem A, but try solving it yourself before looking.

Solution:  f_1 is associated with a vector v_1=(i_1, ..., i_n),
where the leading term of f_1 is x_1^i_1 x_2^i_2 ...x_n^i_n.
Recall that i_1 is greater than or equal to i_2 which is greater or equal
to i_3, etc.   

Similarly, for each j>0, f_j is associated with a vector v_j.

Suppose for the purpose of contradiction that none of the f_t's are constant.
Since there exist fewer than (i_1 + 1)^n different vectors v_j, the box
principle implies that there are integers r < s for which
v_r = v_s.  This contradicts the fact (due to transitivity of lexicographic
ordering) that the leading term of
f_r is greater than the leading term of f_s.
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HW #7 (for May 19)
Problems 7 and 8, page 392.
Also, prove that x^4 + x - 1 has Galois gp iso to S_4
and that x^4 + 4x - 1 has Galois grp iso to D_4.
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HW #6 (for May 12)

Problem A:  Show that the polynomial x^3 -3x -1 over Q has roots g, 2-g^2,
and g^2 -g -2, where g=2*cos(Pi/9).  Explain why the splitting field
of this polynomial is NOT a radical extension of Q, and then explain why it
becomes a radical extension once a complex cube root of unity is adjoined.
HINT:  The minimal polynomial of a primitive 18th root of unity is
x^6-x^3+1.

Problem B:  Prove that a factor group and a subgroup of a solvable group
are both solvable groups.
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HW #5 (for May 5)

Problem A:  For the proof of the FTGT part (d), consider the map T
which maps an auto to the restriction of that auto to E.
Prove that T is "into" by showing that the auto restricted to E is onto E.
Also prove that T is onto.

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HW #4  (for April 21)

Problem A:  Prove that (x^p -1)/(x-1)  is irreducible over Q when p is prime.

Problem B:  Show that the degree of
u=sqrt(1+sqrt2) + sqrt(3+sqrt2)  over Q is 8,
and construct all 8 conjugates of u over Q.

Problem C:  Let u= 2^(1/3) and let w = (-1 + i*sqrt3)/2.
The minimal poly of u  over Q is  x^3 -2.
What is the minimal poly of u  over the larger field Q(uw)?

Challenge problem:  Let p(x) be the Min Poly of u over Q.
Let F be the splitting field of p(x) over Q.  How many elements
are in the group Gal(F/Q)?

Note:  F *properly* contains Q(u), contrary to what I said in class.
Thanks to Nathan for pointing this out.
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HW#3  (for April 14)

Problem A:  Let T: K ---> L  be a field iso.  Show that for a poly f(x) in K[x],
T(f') = (T(f))'.

Problem B:  Generalize the challenge problem above to a sum of n
terms, where n > 3. Allow squarefree numbers inside the square roots, 
not just primes.

Problem C:  Explicitly describe all of the conjugates of
sqrt(6)+sqrt(10)+sqrt(15)+sqrt(5)+sqrt(7)+sqrt(22)  over the rationals Q.

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HW #2  (for April 7)

Problem A: Let p(x) be a poly over K of degree n  with root u.  Prove that

1, u, u^2,..., u^(n-1)  spans K[u]  over   K.

Problem B:  Prove that sqrt(2) + sqrt(3) + sqrt(5) is irrational.

Challenge problem:  Prove that sqrt(3)+sqrt(5)+sqrt(7) has
degree 8 over the rationals.  (Since we know it is irrational,
its degree cannot be 1, but degrees 2 and 4 have to be ruled out as well.)
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