Note: All assignments in Hungerford, unless otherwise noted. Also, assignments are to be turned in and will be graded, unless otherwise noted.

1. Let R be the ring Z[sqrt(10)] (of Exercise 3, Section III, 3 in Hungerford). Show that the ideal P of R generated by 2 and sqrt(10) is a projective R-module that is not free. (Hint: If it were free, then since P embeds in R, it would have to have rank 1, and thus would have to be a principal ideal, which it is not. Why? On the other hand, show that P is projective by considering the map p:R² --> P sending (x, y) to 2x+sqrt(10)y. This map is a surjective R-module homomorphism. Show that it has a splitting map s defined by sending x to (-2x, x sqrt(10)/2), and thus that P embeds as a direct summand in R². But you need to show that s really sends P into R², and not just into Q[sqrt(10)]².)
2. Show that in the notation of the problem above, P² is free (and is isomorphic to R²). The isomorphism is given by right multiplication by the matrix    -2,      -sqrt(10)/2  
     sqrt(10)/2,     1  . Hint: the inverse map is given by right multiplication by the inverse matrix.

1. Show that Ext¹(M, A) = 0 if M is R-projective.
2. Show that one gets an exact sequence 0 --> Hom(M, A) --> Hom(M, B) --> Hom(M, C) --> Ext¹(M, A). The proof is a long diagram chase.
3. (This part is hard. If you get stuck, you can consult a book on homological algebra or Jacobson's Basic Algebra II.) Show that Ext¹(M, A) is independent of the choice of P₁, P₂, and P₃, so that the notation Ext¹(M, A) makes sense. Ext¹ is the simplest example of what is called a derived functor; there are many other examples in algebra.
4. Do this part even if you got stuck on part 3; just take the result there for granted. Let R = Z, M = Z/k, and choose P₁ = P₂ = Z. Show that Ext¹(Z/k, A) is just A/kA, and check the exact sequence 0 --> Hom(Z/k, A) --> Hom(Z/k, B) --> Hom(Z/k, C) --> Ext¹(Z/k, A) directly.

1. Let u = 2 cos (2pi/17). Show that [Q(u):Q] = 8 and that Q(u) is a Galois extension of Q with cyclic Galois group. (Hint: write u = w + w^-1, where w = exp(2pi i/17), and show that Q(w) is a Galois extension of Q with cyclic Galois group of order 16 generated by g : w |--> w³. Then show that Q(u) is the fixed field for g⁸.)
2. Find an explicit tower of fields Q = E₁, E₂, E₃, E₄ = Q(u) with [E_i+1:E_i] = 2 for i = 1, 2, 3.
3. Deduce that a regular 17-gon is constructible with straightedge and compass, and outline a scheme (you don't need to include all the steps) for an explicit such construction.

1. Let K be a field of characteristic not equal to 2, and let D be an element of K that is not a perfect square. Let L = K(sqrt(D)), a cyclic extension of K of degree 2. Write down explicitly the formulas for N^L_K and T^L_K, and verify Theorem V.7.6 in Hungerford directly for this case.
2. Let K be a finite field with q elements, and let L be a finite extension field with [L: K] = r. (Thus L has q^r elements.) Recall that the multiplicative groups L^* and K^* are cyclic, and that G = Gal(L/K) is also cyclic. Compute the norm map N^L_K: L^* --> K^* explicitly, and show that it is surjective. Show that your calculation agrees with the prediction of Hilbert's Theorem 90.
3. Let K be the splitting field over Q(w), w a primitive cube root of unity, of the polynomial x³ - 3x + 1. Show that K is a cyclic extension of Q(w) of degree 3, and use the Lagrange resolvant method to show it's obtained by adjoining a cube root of something. Again write down the norm map K^* --> Q(w)^* explicitly and verify the conclusion of Hilbert's Theorem 90 for this case.

1. Let G = Q₈ be the quaternion group of order 8. Recall that it has generators i and j satisfying i⁴ = j⁴ = 1, i² = j² central, iji^-1 = j^-1. Find representatives for all the equivalence classes of irreducible (complex) representations of G, and compute their characters. Hint: You can construct a representation of dimension 2 by letting G act on the algebra H of quaternions by left multiplication, and choosing a basis of H over C to write the action in terms of 2-by-2 matrices with complex entries.
2. Let G = A₄, the alternating group on 4 letters, which has order 12. Find representatives for all the equivalence classes of irreducible (complex) representations of G, and compute their characters. Hint: G has a normal subgroup H of order 4.
3. Let G = S₄, the symmetric group on 4 letters, which has order 24. Find representatives for all the equivalence classes of irreducible (complex) representations of G, and compute their characters. Using your answer to #2, determine how the representations restrict to A₄. (Hints: You can construct a representation of dimension 3 by letting G act on the subspace {(x₁,x₂,x₃, x₄) in C⁴ : x₁ + x₂ + x₃ + x₄ = 0} by permutation of the coordinates. The restriction of each irreducible representation to A₄ will either be irreducible or will be a direct sum of two irreducible representations of the same dimension.)

Let R be a commutative ring (with unit). The space X = Spec R is defined to be the set of prime ideals of R. Call a subset Y of X closed if Y is of the form {P in X : P contains J} for some ideal J of R.
1. 1. Show that Ø and X are closed.
   2. Show that a finite union of closed subsets of X is closed.
   3. Show that an arbitrary intersection of closed subsets of X is closed.
   These are the axioms for a topological space. A subset of X is called open if its complement is closed.
2. Show that the closed points of X correspond to the maximal ideals. Thus X is T₁ (all points are closed) exactly if all prime ideals are maximal.
3. Show that X is always a T₀ space; that is, given distinct P and Q in X, there is a closed set containing one but not the other.
4. Show that X is quasi-compact. That means that given a set F of closed subsets of X such that F has the finite intersection property (any finite subset of F has non-empty intersection), the intersection of all the sets in F is non-empty.
5. Suppose X is disconnected, i.e., that X is the union of two disjoint closed subsets. Deduce that R contains an idempotent e (an element with e² = e) other than 0 and 1. (Hint: Chinese Remainder Theorem.) Also prove the converse (this is easier). Solutions to this part of the assignment are available here in DVI format and here in PDF format.