• Exam syllabus. The Final Exam will cover the material of the HW and quizzes.
  • Basics of Lie groups: classical groups, Lie algebras, exponential map, coordinates
  • Formulas relating SU₂, SO₃, quaternion ring
  • Representations of finite groups, character theory, symmetric shapes
  • Basics of rings: ideals, quotients, homomorphisms, maximal & prime ideals
  • Polynomial rings: Euclidean Algorithm, principal ideals, C[x]
  • Definitions and algorithms for fields: primitive extensions, finite fields
  • Galois theory: splitting fields, automorphisms, symmetric polynomials, fixed fields, Main Theorem
    Kummer extensions, constructible numbers

Click ⊞ to open sections.

• ⊞ HW 1 - SU(2) and space rotations  
  Artin Ch 9, p 283, #1.1, 1.4, 3.1, M.4. Hints. Also:

  • HW 1-1. Consider the rotation R_θ,p of R³ through angle θ around the pole p = (p₁,p₂,p₃), where |p| = 1. We showed in class that this corresponds to the point (x₀,x₁,x₂,x₃) in the three-sphere with latitude ½θ along the longitude circle through (1,0,0,0) and (0,p): x₀ = cos(½θ),   x₁ = sin(½θ) p₁,   x₂ = sin(½θ) p₂,   x₃ = sin(½θ) p₃. To see the action, we consider the corresponding unitary matrix in SU₂, which acts by conjugation on the matrices corresponding to the equator (0,v₁,v₂,v₃).
    Problem: Work out this conjugation action. Show that for a fixed (θ,p), it gives a linear transformation of (v₁,v₂,v₃). Write down the 3×3 orthogonal matrix of this transformation, which is an explicit matrix for the desired rotation R_θ,p.
    Hint: The matrix entries are quadratic polynomials in x_i.
    Note: We can also consider this action as conjugation within the quaternion algebra. (See Prob M.4 and 3.1 above.)
  • HW 1-2. Recall the Hermitian inner product (named after Hermite) on Cⁿ:
    〈x,y〉_H  =  〈(x₁,...,x_n), (y₁,...,y_n)〉_H  =  x₁y₁ + ... + x_ny_n,
    where y means complex conjugate.
    1. We can regard a vector x ∈ Cⁿ as a real vector in R²ⁿ, and take its real dot product 〈x,x〉. Show that this agrees with the Hermitian inner product: 〈x,x〉_H = 〈x,x〉. (Why is it obvious that 〈x,y〉_H ≠ 〈x,y〉 in general?)
    2. Show that if a linear transformation P : R²ⁿ → R²ⁿ preserves the function q(x) = 〈x,x〉, meaning q(P(x)) = q(x), then P also preserves the dot product: 〈P(x),P(y)〉 = 〈x,y〉. Hint: Write 〈x,y〉 purely in terms of q(x), q(y), and q(x+y).
    3. Show that U_n ⊂ O_2n. (Extra Credit: In fact, U_n ⊂ SO_2n.)
    4. Consider a complex number a + bi as a 1×1 matrix acting by multiplication on C¹ = R². Write this as a 2×2 real matrix.
    5. Given a 2×2 complex matrix in SU₂, determine the corresponding 4×4 orthogonal matrix in SO₄.
• ⊞ HW 2 - Lie groups  
  • 2-1. A manifold M inside a large vector space R^N has a space of tangent vectors T_x(M) ⊂ R^N at each point x ∈ M (imagine the vectors along the tangent plane of a sphere S² ⊂ R³). A vector field is a smoothly varying choice of tangent vector at each point, which you can picture as a flow on the surface M. The vectors could vanish to zero at some points, but we will consider non-vanishing vector fields: famously, it is impossible to construct a non-vanishing vector field on S².
    A Lie group G ⊂ M_n(R) is a manifold, so it has vector fields. An invariant vector field on G is generated from a single tangent vector v ∈ T_I(G) at the identity I ∈ G, by multiplying v by P ∈ G to get a tangent vector at P, namely Pv ∈ T_P(G).
    1. Artin 7.6: Sketch the invariant vector field of the multiplicative group G = GL₁(C) = C−{0} generated from the tangent vector v = 1+i at I = 1 ∈ G. That is, from each sample point a+bi in the complex plane, draw the vector (a+bi)(1+i).
    2. Compute a non-vanishing vector field for the 3-sphere S³ ⊂ R⁴ as follows. (You can comb the hair on a coconut, provided it's in 4 dimensions − those are extremely filling.) The tangent space of S³ at a point x = (x₀,x₁,x₂,x₃) is the 3-space of vectors v orthogonal to the radius x, that is ⟨v,x⟩ = 0. Pick any vector v in the tangent space of SU₂ ≅ S³ at x = I, multiply by an arbitrary x = P, and give a formula for the desired vector Pv at each x ∈ S³. (You can compute the multiplication equivalently in the quaternions H = R⁴ or in SU₂.)
  • 2-2. Find the Lie algebra 𝔤 and dimension d of the following groups. Note that even though the groups and their Lie algebras have complex number entries, we are asking for the dimension over R, the number of real degrees of freedom at a point of 𝔤 or G.
    1. G = SL₂(C)
    2. G = SU_n
  • 2-3. Exponential mapping for SU₂
    1. Verify Euler's Formula exp(a+bi) = exp(a) exp(bi) = e^a(cos(b) + sin(b) i), where a,b ∈ R, by explicitly computing the Taylor series for exp(bi).
    2. Now consider a unit imaginary quaternion, on the equator of S³ ≅ SU₂, denoted p = p₁i + p₂j + p₃k, with p* = −p and p² = −pp* = −|p|² = −1. For a,b ∈ R, explicitly compute exp(a + bp) as a quaternion, then translate to an element of SU₂. Use this to give an explicit formula for exp : 𝔤 → SU₂.
  • Extra: Artin 7.6. Find a connected, non-abelian Lie group of dimension 2 inside GL₂(R).
• ⊞ HW 3 - Spatial symmetries  
  • 3-1. We saw in class that the irreducible complex representations of the 2-element group G = {1,σ} are V₊ = Ce₊ with σ.e₊ = e₊ and V₋ = Ce₋ with σ.e₋ = −e₋, and all other representations are isomorphic to direct sums of these.
    For each possible representation V of dimensions d = 1,2,3, construct a set X ⊂ V whose linear symmetries are precisely G. Don't forget the trivial representation in each dimension. (For d = 1,2, draw a picture; for d = 3, make a figure out of pipe-cleaners or other materials and hand in the physical object or a photo.)
    Example: For d = 3, there are 4 possible representations V_i = V₋ⁱ ⊕ V₊³⁻ⁱ. The trivial representation V₀ corresponds to any 3-dimensional object X_o with no symmetry, such as half a pair of glasses (one lens and stem). The symmetries of V₁, V₂, V₃, where σ acts in 3 different ways, correspond to the following symmetrizations X = X_o ∪ σX_o :            
  • 3-2. Consider the symmetric group G = S₃ with its natural permutation representation on V = R³ with basis {e₁, e₂, e₃}. (The group elements permute the basis vectors.)
    1. Consider the permutations x = (123), y = (12), using the cycle notation from Artin Sec 1.5; these elements generate G. Write the 3×3 representation matrices of these elements.
    2. The representation V splits into an invariant direction v₁ = e₁+e₂+e₃ = (1,1,1) and its orthogonal complement: V = W ⊕ W^⊥, where W = Rv₁. Find an orthonormal basis {v₁, v₂, v₃} of R³ compatible with this splitting, meaning each basis vector is contained in one of the two subspaces. Hint: Start with (1,1,1), (0,1,0), (0,0,1), and use the Gram-Schmidt process.
    3. Change the matrices from (a) into the basis of (b), obtaining block-diagonal matrices which show the splitting of the 3-dimensional representation into 1 and 2-dimensional subrepresentations.
    4. Show that the 2-dimensional subrepresentation W^⊥ is irreducible, so that we have decomposed V into its irreducible components.
    5. Let W˜ denote the complexification of G acting on W^⊥, meaning the same 2×2 real matrices act on C² instead of just R². Show that W˜ is still irreducible: we cannot find any common eigenvectors of the G-matrices, even if we allow complex eigenvectors. Hint: We can find common complex eigenvectors of the cyclic subgroup C₃ = {1,x,x²}, using a kind of weighted averaging process. But these eigenspaces are not preserved by the other generating permutation y.
  • Get access to Mathematica. You can download it for free via the MSU Computer Store. It is also available at MSU computer labs.
    Experiment with it a bit. I intend to use ConvexHullMesh on next week's HW to construct interesting polyhedra from representations. If you know of other available software which can generate a picture of a convex polyhedron, let me know.
  • Extra. As in Artin Sec 6.4, consider the dihedral group D_n with its natural representation as the symmetries of a regular n-gon in V = R². That is, D_n consists of the identity 1, the rotations x, x²,..., xⁿ⁻¹, and the reflections y, xy,..., xⁿ⁻¹y, where y is reflection across the x-axis, and with the relations xⁿ = y² = 1 and yx = x⁻¹y.
    1. Find the matrix for each group element, and show that V is an irreducible representation (no real vector space subrepresentations).
    2. Show that the complexified representation V_C = C², in which we take the same representation matrices acting on complex space, is a reducible representation of the cyclic group C_n, but an irreducible representation of D_n.
  • Extra. We showed that the invariant projector P = ¹⁄_|G| ∑_g∈G g takes any representation V to its component P.V of invariant vectors, that is, the component which is a sum of trivial representations. Now, suppose a finite group G has a one-dimensional representation V_ρ given by numbers ρ_g (1×1 representation matrices).
    Problem: Find a projector P_ρ which takes any representation V to its component P_ρ.V which is a sum of copies of V_ρ
• ⊞ HW 4 - Tetrahedral symmetry S₄ & A₄ from R⁴, permutahedron & skew icosahedron  
  Tetrahedron symmetry shapes.x Repeat Prob 3-2, but in one dimension higher.
  Consider the symmetric group G = S₄ acting on R⁴ by permuting the basis vectors {e₁,...,e₄}. As before, this splits into a one-dimensional and a three-dimensional subrepresentation:

  R⁴  =  Rp ⊕ V ,

  where p = (1,1,1,1) is the invariant vector, and V = p^⊥ ≅ R³ is its orthogonal complement.

  1. Irreducibility of V
     1. Write representatives for the conjugacy classes of S₄, and determine how many elements of each. Example: there are (4 choose 2) = 6 transpositions (ij).
     2. Determine the character of G on R⁴. Hint: When do you get a diagonal entry in a permutation matrix?
     3. Determine the characters of G on Rp and V. Hint: Obvious for Rp, and the character of V follows immediately.
     4. Use the Main Theorem on Characters to determine that V is irreducible. Hint: We must have χ_V = ∑ n_iχ_i, where χ_i runs over the irreducible representations, and n_i ∈ Z_≥0 are multiplicities.
  2. Implement the action of G on V in Mathematica (demo).
     1. Enter the 4! = 24 permutation matrices of G on R⁴ into Mathematica, and make them into a list.
        Hints: It is enough to enter a generating set, and multiply to get the rest. For example, enter σ = (1234), and you get σ² = (13)(24) and σ³ = (4321) for free. Also, conjugating permutes the entries in the cycles of σ, so that ρσρ⁻¹ is the cycle (ρ(1),ρ(2),ρ(3),ρ(4)).
     2. Find an orthonormal basis {v₁,v₂,v₃} for V.
     3. Change basis from {e₁,...,e₄} to {p,v₁,v₂,v₃}, and extract the 3×3 matrices of G acting on V.
     4. Define a sub-list of G containing the subgroup H = A₄ of even permutations, which in this case is the identity and the permutations with cycle structure (123) or (12)(34).
     5. You can picture the G-action on V ≅ R³ as permuting the 4 vertices of a regular tetrahedron (triangle-based pyramid) in R³. These vertices are the orthogonal projections of {e₁,...,e₄} down to V (subtracting off the p-component):
        a_i  =  e_i − Proj_p(e_i)  =  ⟨e_i,v₁⟩ v₁ + ... + ⟨e_i,v₃⟩ v₃,
        since the v_i are an orthonormal basis of V. Write these vertices in v_i-coordinates.
  3. Symmetric polyhedra.
     1. Take the orbit of one of the tetrahedron vertices: G.a₁ = {g.a₁ for g ∈ G}. Display the convex hull of G.a₁, and verify that it is a tetrahedron.
     2. Pick a random point a ∈ V ≅ R³, and take its orbit G.a = {g.a for g ∈ G}. The convex hull of G.a is a polyhedron whose symmetries are precisely G, a shape known as the permutahedron. Can you see how this is obtained from the tetrahedron by slicing off some parts?
     3. Take the convex hull of a random orbit of the subgroup H = A₄. The resulting polyhedron has a name: what is it?
        Hint: We usually see a regular, even-length version of this polyhedron, having more symmetry.
     For Problems 2 and 3, turn in an annotated printout of your Mathematica input/output, including comments to say what you are doing. Also write in pen on the printout, circling and labeling the answer to each HW question. I will mainly grade the answer, not the code.

     Note: Can also realize the tetrahedron using non-adjacent corners of cube, i.e. root system embedding R(A₄) ⊂ R(B₃). Vertices (1,1,1), (1,−1,−1), (−1,1,−1), (−1,−1,1). See MTH 411 HW 3 & 4.

• ⊞ HW 5 - Characters and Quaternion group reps  
  1. One-dimensional representations. Suppose G is a finite group acting on a one-dimensional space C = C_ρ, corresponding to a representation homomorphism ρ : G → GL₁(C) = C−{0}. For example, we have seen that the cyclic group C_n has n such representations, and the symmetric group S_n has the sign representation ρ(g) = sgn(g) = determinant of the permutation matrix of g.
     1. Show that for any g ∈ G, we have ρ(g) = exp(2πi^k⁄_n) for some integers k,n.
     2. Let V be any G-representation acting by g.v = R(g)v for representation matrices R(g). Recall that the averaging operator P = ¹⁄_|G| ∑_g∈G g ∈ C[G] acts on V by the matrix R(P) = ¹⁄_|G| ∑_g∈G R(g), which projects V onto its subrepresentation of invariant vectors: P.V = V₁ = {v ∈ V | g.v = v for all g}.
        Find a similar element P_ρ ∈ C[G] corresponding to the one-dimensional representation ρ, so that P_ρ.V = V_ρ = {v ∈ V | g.v = ρ(g) v}. Note that the vectors in V_ρ are common eigenvectors of all g ∈ G, with eigenvalues ρ(g), and the sub-representation V_ρ is isomorphic to C_ρ^k for some k. We call V_ρ the ρ-isotypic component of V, and P_ρ the ρ-isotypic projector.
  2. Twisted representations
     1. Let V be a G-representation with representation homomorphism R : G → GL(V), and let ρ : G → C−{0} be a one-dimensional representation. Define new representation matrices by R'(g) = ρ(g) R(g), a scalar times a matrix. Show that R' : G → GL(V) defines a new representation, the "ρ-twist" of V, denoted ρ⊗V.
     2. Let G = S₄, and let V = C³ be the standard representation considered in HW 4 (permuting the 4 vertices of the tetrahedron), and consider the twisted irreducible representation V' = sgn⊗V. Using your previous Mathematica file, find the 3×3 matrices of G acting on V', and explain why they must all be rotations. Draw the 24-vertex convex polyhedron defined by a general G-orbit in V', and identify it among the Archimedean solids.
  3. Recall the 8-element quaternion group G = {±1,±i,±j,±k} ⊂ H, which we can also consider as G ⊂ SU₂.
     1. Determine the conjugacy classes of G.
     2. Determine the dimensions of the irreducible representations of G. Hint: Use the known formulas involving these dimensions.
     3. Determine the character table and the representation matrices for all the irreducible representations.
     4. Which of the irreducible representations are faithful?
     5. Classify all the three-dimensional (reducible) representations of G, up to isomorphism. (This is analogous to HW 3, where we dealt with the 4 possible symmetry types of objects in three-space, except here we allow G to act by complex symmetries on C³.)
     6. Can the orbit of some faithful three-dimensional representation produce an 8-vertex solid in R³ with G as its rigid symmetry group?
  4. Extra. Prove the converse of Schur's Lemma: If V is a G-representation, and the only G-invariant linear transformations T : V → V are scalar multiplication T(v) = λv, for constant λ, then V must be irreducible. Hint: Prove the contrapositive: if V is reducible, then V has a non-scalar G-invariant linear transformation.
  5. Extra. Show that a group is simple by looking at its character table
  6. Extra. Explain how to modify the character table to produce a unitary matrix, and show columns are orthogonal.
  Solution
• ⊞ HW 6 - F₈ splitting field of x³+x+1  
  1. Recall the Euclidean Algorithm in a polynomial ring F[x], for F a field. We start with arbitrary polynomials f(x), g(x) ∈ F[x] having deg g(x) ≤ deg f(x). Long division gives f(x) = q₁(x) g(x) + r₁(x) with deg r₁(x) < deg g(x) or r₁(x) = 0. Iterating, we get:
     f(x) = q₁(x) g(x) + r₁(x)
     g(x) = q₂(x) r₁(x) + r₂(x)
     r₁(x) = q₃(x) r₂(x) + r₃(x)
     ⋮
     r_k−2(x) = q_k(x) r_k−1(x) + r_k(x)
     r_k−1(x) = q_k+1(x) r_k(x) + 0,
     with deg f(x) ≥ deg g(x) > deg r₁(x) > ... > deg r_k(x).
     1. Show by decreasing induction that r_k(x) | r_j(x) for all j ≤ k, meaning r_k(x) is a polynomial divisor of r_j(x), including g(x) = r₀(x) and f(x) = r₋₁(x).
     2. Show by decreasing induction that r_k = a_j(x) r_j−1(x) + b_j(x) r_j(x) for all j ≤ k, all the way back to r_k(x) = a(x) f(x) + b(x) g(x) for some a(x), b(x) ∈ F[x].
     3. Prove that d(x) = r_k(x) is the greatest common divisor of f(x), g(x), meaning d(x) | f(x), g(x), and any common divisor c(x) | f(x), g(x) also has c(x) | d(x).
     4. Show that any ideal I ⊂ F[x] is a principal ideal, the multiples of some f(x): that is, I = (f(x)) = {a(x) f(x) for a(x) ∈ F[x]}. Hint: Given I, take f(x) to be a non-zero polynomial of lowest degree in I.
  2. Field F₈ of order 8. Consider the polynomial p(x) = x³ + x + 1 in the ring F₂[x] of polynomials with coefficients in the field F₂ = Z/(2) = {0,1}.
     1. Show that p(x) = x³ + x + 1 is irreducible in F₂[x].
     2. Consider the quotient ring F₈ = F₂[x]/(p(x)), and denote the coset of x by α = x = x + (p(x)) ∈ F₈, so that p(α) = p(x) = 0. Find a basis for F₈ as a vector space over F₂, and show that F₈ indeed has 8 elements. Hint: Any f(α) ∈ F₈ can be reduced to standard form a₀ + a₁α + a₂α², by taking a quotient f(x) = q(x) p(x) + r(x) in F₂[x], with deg r(x) < 3. Note: The dimension of the extension field as a vector space over the subfield is called the degree of the extension, denoted [F₈ : F₂] = dim_F2(F₈).
     3. Find the reciprocal ¹⁄_β for every element 0 ≠ β ∈ F₈. Hint: Write β = f(α) and apply the Euclidean algorithm to the polynomials p(x), f(x).
     4. For each element of F₈, determine the irreducible polynomial in F₂[x] which it satisfies (its minimal polynomial). For example an element a ∈ F₂ ⊂ F₈ satisfies the polynomial f(x) = x − a. The other elements β ∈ F₈ will satisfy degree 3 polynomials, since there is an F₂-linear combination of 1, β, β², β³ which is zero.
     5. Show that F₈ does not contain any subfield with 4 elements.
    Solution
• ⊞ HW 7 (Extra) - Reps of SU(2). Factorization in number rings. Lagrange interpolation. 
  These two assignments are optional, but each will count as an extra 10 points on your HW grade. Do as many problems as you feel like.

  • HW 7A. Read Artin Ch 10.9 on Representations of SU(2).   Hint
    Do Ch 10 Exercises, p. 320: #9.2, 9.3, 9.4, 9.5, 9.6 .
  • HW 7B. Assume the Euclidean Algorithm (gcd(n,m) = an+bm), the Prime Lemma (p|ab ⇒ p|a or p|b), and the Fundamental Theorem of Arithmetic (unique factorization into primes), and other results from Artin Ch 12 if needed.
    Do Ch 12 Exercises, p. 378: #1.1, 1.2, 2.4, 2.5, 2.8, 5.1, 5.2, 5.3, 5.4, 5.5, 5.10.
    • Correction to #2.5(a): C[x] x should be C[x].
    • Ex 7B-1: Consider the ring R = Z[α] for α = √(−5) = i√5 (Artin p. 360, 364). This consists of all integer polynomials in the element α, which reduce via α² = −5 to elements a+bα for a,b ∈ Z. This is the simplest number ring where unique factorization fails. In fact, consider:
      6 = (2)(3) = (1+α)(1−α).
      Problem: Prove that the units of R are ±1; also that the elements 2, 3, 1+α, 1−α are irreducible, i.e. non-factorable. Explain how this shows the failure of the Prime Lemma and Unique Factorization. (According to Artin 12.2.1, p. 360, the above elements are irreducible but not prime. This is the beginning of Algebraic Number Theory.)
  • Lagrange interpolation, Artin Ex 4.13. Given a₁, . . . ,a_n,b₁, . . . ,b_n ∈ K, find the unique f(x) ∈ K[x] of degree n−1 such that f(a_i) = b_i for all i.
    Solution: Vandermonde matrix of a_i's, augmented by first column (1, x, x², . . . , xⁿ⁻¹)^T and top row (f(x), b₁, . . . , b_n). This det = d(x) = 0 since deg d(x) ≤ n−1 and d(a_i) = 0. Now take cofactor expansion along top row, solve for f(x) in terms of Vandermonde dets. Result for n = 2 is 2-point form of linear fun: f(x)  =  ^{b₁(x−a₂)}⁄_(a1−a2) + ^{b₂(x−a₁)}⁄_(a2−a1) .
• ⊞ HW 8 - Splitting field of x³−3x−1 over Q, no mult roots, primitive elt of Q(√2,√3)  
  1. Splitting field. Consider the polynomial:
     p(x)  =  x³ − 3x − 1  ∈  Q[x],
     having three real roots α, β, γ ∈ R. Let K = Q(α) ≅ Q[x]/(p(x)), a field where we know how to calculate +,−,×,÷ based on p(α) = 0.
     Warning: Discriminant √81 = 9, so splitting field L = Q(α,β) = K and Gal = C₃. Use x³ − 4x − 1 instead.
     1. Although p(x) is irreducible in Q[x], it is divisible by x−α in K[x]. Use long division to write p(x) = (x−α)g(x) for g(x) ∈ K[x].
     2. Consider the field L = Q(α,β) = K(β) ≅ K[x]/(g(x)), where we can compute based on g(β) = 0. Use long division in L[x] to write g(x) = (x−β)h(x) for h(x) ∈ L[x]. Use this to write γ in terms of α and β. Since L contains all three roots, we say it is a splitting field for p(x).
     3. Multiply out (x−α)(x−β)(x−γ) and check that it reduces in L[x] to p(x).
     Note: It is irrelevant which of the three roots is labeled α: all computations proceed the same way.
  2. Mulitple roots
     1. Prove the Proposition: Let f(x) ∈ F[x] be an irreducible polynomial over a field F of characteristic zero, and let K ⊃ F be a splitting field containing all the roots of f(x). Suppose f '(x) is not the zero-polynomial. Then f(x) does not have any multiple roots, meaning (x−α)² does not divide f(x) in K[x].
        Hint: Since f(x) is irreducible, we have gcd(f(x),f '(x)) = 1. Could we also have f(x) = g(x)(x−α)² in K[x]?
        An important point: for any f(x),g(x) ∈ F[x], the Euclidean algorithm gives the same gcd computed in F[x] or in K[x]: the coefficient field is relevant for factoring, but irrelevant for gcd.
     2. Prove that an irreducible f(x) ∈ F[x], with coefficients in the finite field F = F_p = Z/(p), cannot have multiple roots.
        Hint: Show that if f(x) has a multiple root α, then f '(α) = 0, and this can only happen for irreducible f(x) if f '(x) is the zero-polynomial. Then use the theorem that a^p = a for all a ∈ F_p, together with (a+b)^p = a^p+b^p, to find that f(x) = g(x^p) = (g(x))^p.
  3. Primitive element. Consider the field:
     K = Q(α,β)   with elements   α = √2,   β = √3,   γ = √2+√3.
     1. Find a polynomial q(x) ∈ Q[x] satisfied by γ. Hint: You can guess that the roots of q(x) are all sums α_i + β_j, where α_i are the roots of x²−2 and β_j are the roots of x²−3. That is, q(x) = (x−√2−√3)(x−√2+√3)···
        Note: Even if we are not sure that q(x) is irreducible, we can still compute in Q[γ] using q(γ) = 0. The only caveat is that an element of Q(γ) might have more than one expression as a₀+a₁γ+a₂γ²+a₃γ³.
     2. Show that K = Q(γ) by writing α and β as polynomials in γ. Use the method from the proof of Lemma 15.8.2, p. 463.
        Hint: The polynomials f(x) = x²−2 and h(x) = (γ−x)²−3 have x = α as their only common root in any field, so their gcd is x−α. On the other hand, for L = Q(γ), we have f(x), h(x) ∈ L[x], and the Euclidean algorithm computes gcd(f(x),h(x)) while staying inside the ring L[x], so the result of the gcd computation is an expression for x−α ∈ L[x], i.e. for α ∈ L = Q(γ). (Keep all computations in terms of γ; don't expand in terms of α, β.)
• ⊞ HW 9 - Compass constructions, finite field properties  
  1. Let K,K' be fields, and let φ : K → K' be a non-zero, surjective homomorphism of rings, respecting addition and multiplication. Prove that φ is an isomorphism of fields, meaning it has zero kernel and respects 0, 1, negatives, and reciprocals.
  2. Prove that a polynomial of degree n has at most n roots in any field.
  3. Geometric constructions (Artin Ch 15.5).
     1. Give a step-by-step compass-and-straightedge construction of a regular pentagon. Starting with the center (0,0) and the vertex (1,0), it suffices to construct the next vertex (cos(θ), sin(θ)) for θ = ^2π⁄₅. Make your constructions algebraically transparent: compute the coordinates (cos(θ), sin(θ)) algebraically, and produce the corresponding lengths by standard constructions. Carry out your constructions to physically draw a pentagon with a compass.   ⊞ Hint
        Write cos(θ) = ½(ζ+ζ⁻¹) for ζ = e^iθ, and adapt the minimal polynomial ζ⁴+ζ³+ζ²+ζ+1 = 0 (cf. Artin Thm 12.4.9), obtaining a quadratic equaton for cos(θ). Solve the equation in terms of a square root, and use the standard constructions for this number. Finally, construct sin(θ) = √(1−cos²(θ)).
     2. Show that an 18-gon is not constructible, meaning it is impossible to trisect an angle of ^2π⁄₆.  ⊞ Hint
        As in part (a), find the minimal polynomial of cos(θ) = ½(ζ+ζ⁻¹) for θ = ^2π⁄₁₈, ζ = e^iθ, starting with ζ⁹ + 1 = (ζ³+1)(ζ⁶−ζ³+1) = 0. Now use Theorem 15.5.6 to show this cannot be constructed.
  4. Finite fields. Prove the following using elementary arguments and previous homework problems, not the general results of Galois Theory from Ch 15 and 16. However, for the later parts, try to avoid brute force calculation, giving arguments which generalize to any finite field q = pⁿ.
     1. Let K be a field of characteristic p, meaning it contains the prime field F = F_p = Z/(p). Define the Frobenius mapping Φ : K → K by Φ(a) = a^p. Show that Φ is an injective homomorphism of fields, fixing F. If K is finite, then Φ is an F-isomorphism.
     2. Construct a field K = F_q with q = pⁿ = 2⁶ = 64 elements as F₂[x]/(f(x)) for an explicit irreducible f(x), and let F = F_p. Briefly describe the elements of K (without listing them all!) and sketch the algorithms for multiplication and division.
     3. Prove K^× = K−{0} is a cyclic group of order q−1 = 63.
     4. Show every element of K is a root of the polynomial x^q − x, and K is a splitting field over F.
     5. Show that the Galois group G = G(K/F) is the cyclic group of order n = 6 generated by the Frobenius isomorphism Φ, namely G = {1,Φ,Φ²,..., Φ⁵} with Φ⁶ = 1. Thus, F ⊂ K is a Galois extension.
     6. Consider each subgroup H ⊂ G generated by Φ^m for m = 1,2,3,6 the divisors of n = 6. Then prove the fixed field L = K^H has p^m elements, and every subfield of K is of this form.
     7. Explain how the theorems of Galois Theory in Ch 16 would imply some of the above results.
  5. Extra: Prove that a finite integral domain (no zero-divisors) is a field.
  6. Extra: Splitting Field ⇔ Galois extension
  7. Extra: Symmetric functions generated by elementary sym funs
• ⊞ HW 10 - Galois theory of x³−4x−1, discriminant, cubic formula.   Solution  
  Lect:

  • Kummer extensions
  • Solvable group <=> solvable by radicals
  • Quadratic formula via Gal(Q(t1,t2)/Q(t1+t2,t1 t2))
  • Normal basis thm, regular representation

  1. Irreducible polynomial of ζ = e^2πi/p, for prime p.
     1. Eisenstein Criterion: Let f(x) = a_nxⁿ + ··· + a₁x + a₀ ∈ Z[x] be an integer polyomial whose coefficients satisfy the following divisibility by p: p ∤ a_n ,  p ∣ a_n−1 , . . . ,  p ∣ a₁ ,  p ∣ a₀ , p² ∤ a₀ . Prove that f(x) is irreducible in Z[x]. ⊞ Hint 
        Any factorization f(x) = g(x)h(x) could be reduced mod p to give a factorization in F_p[x]. Examine the possible coefficients of these factors.
     2. Show that the minimal polynomial of ζ = e^2πi/p is: f(x)  =  x^p−1 + x^p−2 + ··· + x + 1  =  ^(xp−1)⁄_(x−1) . ⊞ Hint 
        The polynomial f(x) is irreducible whenever f(x+1) is irreducible. Use the Binomial Theorem to show f(x+1) satisfies the Eisenstein Criterion.
  2. Let K be the splitting field over F = Q of the polynomial f(x) = x³ − 4x − 1 ∈ Q[x], and let its three real roots be α, β, γ, so that K = Q(α,β,γ) and f(x) = (x−α)(x−β)(x−γ), and the coefficients of f(x) are ± the elementary symmetric polynomials s_i(α,β,γ). In particular, note that s₁ = α+β+γ = 0.
     1. Read Ch 16.2, pp. 481−2, and compute the discriminant Δ of f(x) as an explicit rational number, where we define: Δ = (α−β)²(α−γ)²(β−γ)². Hint: The book explains how to write the symmetric polynomial Δ in terms of the elementary symmetric polynomials s_i(α,β,γ).
     2. Let δ = √Δ = (α−β)(α−γ)(β−γ) ∈ K. Show Q(δ) ≠ Q, [K:Q] = 6, and the Galois group G = G(K/Q) = S₃, the group of all permutations of α, β, γ.
     3. For each L = Q(ψ) for ψ = α, β, γ, δ, write it as the fixed field L = K^H of a subgroup H ⊂ G. Show that these are all of the intermediate fields Q ⊂ L ⊂ K.
     4. Show that δ can be obtained by applying the isotypic projector of the one-dimensional sign representation of G = S₃ to the smallest asymmetric monomial α²β: that is, δ = ∑_π∈G sgn(π) π(α²β).
  3. Normal forms in Mathematica. Continuing the previous problem, the above splitting field K of f(x) = x³−4x−1 has Q-basis {1, α, α², β, αβ, α²β}, so that every element has a unique normal form as a linear combination of these 6 elements.
     1. For an element h(α) ∈ Q[α], with h(x) ∈ Q[x] an arbitrary polynomial, its normal form is the remainder r(α), where polynomial division gives h(x) = q(x)f(x) + r(x). Implement this in Mathematica (or similar software) using the function PolynomialRemainder[h,f,x], where h = h(x), f = f(x). To remind us of the substitution x = α, write h = h(a), f = f(a) as polynomials in the variable a, and use PolynomialRemainder[h,f,a]. Try this for a few specific polynomials h(a).
     2. Find the irreducible polynomial of β over Q(α), namely the polynomial quotient g(x) = f(x) ÷ (x−α) ∈ Q(α)[x]. We can consider g(β) as g(α,β), a two-variable polynomial in α and β with rational coefficients.
     3. Given an element h(α,β) ∈ K, for an arbitrary polynomial h(x,y) ∈ Q[x,y], define a procedure to compute its normal form. First, consider b as the variable and a as a constant coefficient, and divide to get h(a,b) = q(a,b)g(a,b) + r(a,b), where r(a,b) = c₀(a) + c₁(a) b. Next, take the normal form of each coefficient c_i(a), dividing by f(a).
        You can define a procedure in Mathrematica for example as: normform(h_) := Module[{r}, r = PolynomialRemainder[h,g,b]; etc. etc.], where h_ declares an input variable, {r} declares a local variable, and g is g(a,b) as above.
     4. Apply your normal form procedure to Δ = δ² in terms of α, β, and γ = −α−β; and verify in this case the value of δ² from 2(a). (In fact, Mathematica allows you to work with a general polynomial f(x) = x³ + px + q to obtain the normal form δ²  =  −4p³−27q² from p 482.)
  4. Let F' = Q(ζ) for ζ = e^2πi/3, a cube root of unity. Consider the same polynomial as above, f(x) = x³ − 4x − 1 ∈ F'[x], and let K' be the splitting field of f(x) over F'. All the previous calculations for the extension Q = F ⊂ K = F(α,β,γ) are valid for F' ⊂ K' = F'(α,β,γ). In particular, the normal form algorithm is valid, except that you need another variable z to represent ζ, and take one more remainder upon division by its irreducible polynomial z²+z+1. (Alternatively, you can just work with z = Exp[2*Pi*I/3], but this will not automatically simplify expressions in ζ.)
     The extension L' = F'(δ) ⊂ K' again has cyclic Galois group H = G(K'/L') = ⟨τ⟩ for τ = (α β γ). Kummer's Theorem (16.11.1) states that K' = L'(ε), for some ε with ε³ ∈ L'. Find such ε as follows.
     1. We can think of K' as a three-dimensional vector space over L', with basis 1,α, β, and a representation of the group H = ⟨τ⟩. Show that V = L'α ⊕ L'β ⊂ K' is a 2-dimensional H-subrepresentation, determine its character χ_V, and split χ_V into the basis defined by the irreducible characters χ_i(τ) = ζⁱ.
     2. Suppose ε is an eigenvector of τ with eigenvalue λ ≠ 1. Show that K' = L'(ε), and ε³ ∈ L'.
     3. Each eigenvector ε of τ defines a one-dimensional H-subrepresentation L'ε ⊂ K'. Find the desired eigenvector ε using the isotypic projector of the character χ_i, the twisted averaging operator from HW 5 #1b applied to any element of K', such as α: ε_i  =  P_χi(α) = (1 + χ_i(τ⁻¹)τ + χ_i(τ⁻²)τ²)(α), for i = 1,2. Use #3 to find the normal forms in K' of ε_i³, and write each in the form r + sδ for r,s ∈ F'.
        Hint: Compare coefficients of α²β in the normal forms of δ, and ε_i³. Or in Q[α,β], perform the polynomial division ε_i ÷ δ = s rem r.
     4. Use the relations ζ²+ζ+1 = 0 and s₁ = α+β+γ = 0 to show: α = ¹⁄₃(s₁ + ε₁ + ε₂) = ¹⁄₃(ε₁ + ε₂). Also express β, γ in terms of ε₁, ε₂.
     5. Finally, write the roots α, β, γ in terms of Q(ζ), field operations, and square and cube roots: this is the Cubic Formula. To verify these values, compare them to the approximate roots you can see on a graph of f(x).
• ⊞ Extra - Cyclotomic Q[ζ₁₇] ∋ √17, solve by √  
  1. Let K = Q(ζ) for ζ = e^2πi/17 be the splitting field of the polynomial x¹⁷−1. We showed in class (following Artin Ch 16.1) that the Galois group G = G(K/Q) consists of mappings σ_i : K → K defined by σ_i(ζ) = ζⁱ, for i = 1,2,...,16. In fact, G is isomorphic to the multiplicative group of non-zero integers F₁₇^× = (Z/17)^×, a cyclic group of order 16 with generator σ = σ₃.
     The group G = ⟨σ⟩ has subgroups ⟨σ²⟩ with 8 elements; ⟨σ⁴⟩ with 4 elements; ⟨σ⁸⟩ with 2 elements; and the trivial subgroup 1. By the Main Theorem, these three subgroups correspond to intermediate fields: Q ⊂ L₁ ⊂ L₂ ⊂ L₃ ⊂ K. Also, we showed L₃ = Q(cos^2π⁄₁₇) = K^σ2 = Q(^(ζ+ζ−1)⁄₂).
     1. Show that L₁ = Q(√17). ⊞ Hint 
        L₁ = K^σ2 ∋ α,β, where α = ζ + σ²(ζ) + σ⁴(ζ) + ··· + σ¹⁴(ζ) and β = σ(α). Now compute the minimal polynomial (x−α)(x−β) = x² − (α+β)x + αβ = x² + x − 4, by expanding ζ¹⁶ = −ζ¹⁵ − ··· − ζ − 1, and writing everything in the Q-basis 1, ζ,..., ζ¹⁵. Cf Lemma 16.10.6, pp. 498.
     2. Explicitly write L₂ = K^σ4 = L₁(√γ), where γ ∈ L₁.
     3. Explicitly write L₃ = K^σ8 = L₂(√ε), where ε ∈ L₂, and write cos(^2π⁄₁₇) in terms of ε. This completes the algebra needed to construct the 17-gon.
  2. Other cyclotomic fields Q ⊂ Q(ζ) for ζ = exp(^2πi⁄_n)
     • G = Gal(K/Q) ≅ (Z/n)^×.
     • Intermediate fields Q(α) for α = non-zero elem sym function of orbit of H ⊂ G on primitive roots.
     • n = 8, G ≅ C₂ × C₂
     • n = 9, G ≅ C₆.

• Old Math 419H, Math 411
• Outline of Algebraic Geometry
• ⊞ Example of complex Riemann surface 
  • Let V ⊂ A² be the vanishing locus of the polynomial:
    f(x,y)  =  x³ − 2xy + y + 1  ∈  C[x,y].

    

  • Consider the projection onto the y-axis, V → C given by (x,y) ↦ y. The points of V projecting to a given y = b are the roots of the polynomial g(x) = f(x,b) = x³ − 2bx + (b+1), of which there are generically three: x = α₁, α₂, α₃ ∈ C. In the real picture, we see three points of V along each horizontal line y = b > 2.
  • The exceptional case is when g(x) has a double root, which happens when its discriminant vanishes: Δ = −4p³ − 27q² = −4(−2y)³ − 27(y+1)² = 0. This has a real solution y = b₁ ≈ 1.9, and two complex solutions b₂, b₃ ≈ −0.6 ± 0.4i. At each height y = b_j, the polynomial g(x) = f(x,b_j) has a double root x = a_j.
  • We can picture V as a threefold covering space of the y-axis C, except near the three branch points (a_j, b_j) where two of the sheets come together. One of these is real: (a₁, b₁) ≈ (1.4, 1.9), the valley-point in the picture above. At the branch points, the projection mapping looks like the complex square mapping C → C, z ↦ z², which wraps the complex plane twice around the origin.
  • If V has no singularities, it is a Riemann surface, a smooth manifold of two real dimensions, also thought of as one complex dimension (an "algebraic curve"). Like all connected 2-manifolds, it is topologically equivalent to a 2-sphere with some number of handles attached, minus some missing points (which correspond to points of V at infinity). The number of handles g is called the genus of the surface.
    The Riemann-Hurwitz Formula says that g = 1 − d + ^b⁄₂, where d is the generic multiplicity of the projection and b is the number of double-branch points. In our case, g = 1 − 3 + ⁴⁄₂ = 0, since d = 3 (threefold covering), and b = 4, counting the 3 finite branch points plus 1 at infinity.
  • Our polynomial f(x,y) has y-degree 1, so it is easier to picture using the x-projection V → C, (x,y) ↦ x. This is a one-to-one mapping, except for values x = a where h(y) = f(a,y) = (1−2a)y + (a³+1) is a constant, namely the vertical asymptote x = ½. Thus, our V is isomorphic to the complex plane minus one point, or the 2-sphere minus two points, and V does indeed have genus g = 0.
• ⊞ Good books for further reading 
  • Fulton & Harris, Representation Theory.
  • Fulton, Algebraic Curves
  • Cox, Little, O'Shea, Ideals, Varieties, and Algorithms.