Ok, I didn't change much. This looks ok to me. Tell me if you disagree:

Here's a start: 1.6, 3.8, 3.21: Show by example that if hypothesis (1) is deleted from the statement of Eisenstein's Criterion, the resulting statement is false.Here's an end: 4.4, 4.7, 4.8, 4.9.

How about [5.3] and 5.7? 6.13, 6.15, 6.20 (using 6.15): Show that every subextension (use the natural definition) of a simple algebraic extension over an infinite field is simple.

Prove: if z is in P_n, then the conjugate of z is also in P_n.

[As referenced at the very end of §7.2, solve for z and its conjugate.]

7.10, 7.17

I think I'm keeping everything above. To that add [8.2], 8.14. 8.14 is kinda long to typeset. I'll try. Please ask me if there are questions about what it means. 8.14 Let L:K be any field extension. (a) Suppose H1 and H2 are subgroups of the Galois group of L over K, and let H1 v H2 denote the subgroup of the Galois group generated by the union of these two groups. Show that (H1 v H2)-dagger = the intersection between H1-dagger and H2-dagger. (b) Suppose M1 and M2 are intermediate fields of L:K, and let M1M2 denote the intermediate field generated by the union of these two fields. Show that (M1M2)* = the intersection of M1* and M2*.

[Bonus optional question: Prove that the discriminant of the quadratic polynomial ax^2+bx+c using our definition in §8.7 is b^2-4ac, as you expect. Also, express the discriminant of a depressed monic cubic, x^3 + ax +b, in terms of a and b.]

draft form, may be changed. I will try to finalise by 24 March:

9.6, 9.8, 10.3, 10.6 Show that if G is an infinite group of automorphisms of a field K, then [K:G-dagger] is infinite.

One from chapter 11. In Theorem 11.12, Stewart says that [L:K]=[L:K0] and K a subset of K0 implies K = K0. Justify this. If you used anything particular to fields, does the generalisation to vector spaces also hold? (I mean, if the dimension of V over F1 is equal to the dimension of V over F2, and F1 is a subset of F2, is it true that F1=F2?) Either prove or refute.

11.8 Suppose L:K is a normal algebraic extension, E any field containing L, and A any subset of E. Show that the Galois group of L(A):K(A) is isomorphic to a subgroup of the Galois group of L:K, the isomorphism being given by restricting elements of the Galois group of L(A):K(A) to L. (You should begin by indicating why this restriction operation takes elements of the Galois group of L(A):K(A) to elements of the Galois group of L:K.) At least to me - the interesting part of the question - show by example that this result becomes false if L:K is not assumed normal.

Aside from that, also please complete 11.5. I think it's likely that 12.2 will be added.

draft form, may be change (probably by problems being added, not removed). I will try to finalise by 17 April:

13.8 (including 12.5),

14.1 and 8 as one question.Show as many quintic polynomials as you can that cannot be solved by radicals over the rationals. Hint: See 15.3. Note: infinitely many is guaranteed to satisfy "as many as you can". Argue for or against the statement "most quintic polynomials cannot be solved by radicals over the rationals."

15.6, 15.10

We'll see if something from 24 or 20 gets added. Let's try these two curious details:

8.4 which is referenced in 18.10 which is used in the π transcendental proof. We would need to generalise this, but it gives the idea.

In the proof that π is irrational, there are a few steps skipped. Complete the justifications for equations (24.1) and (24.2) that are suggested in the text.

Now this seems more serious. I don't think I'll add anything from 20, but maybe allow for something during the oral final.

On any problem set, full completion of either of these problems will earn an additional +2. No partial credit.

6.15d Prove that m_M determines M uniquely.

Find an extension L:K and K ≤ M ≤ L, and \tau a K-monomorphism M -> L which does not extend to a K-automorphism \sigma of L with \sigma|_M=\tau.