Questions & explanations
1. Give an example of a normal ring that is not Cohen-Macaulay.
Consider the cone over an elliptic curve: R = k[x,y,z]/(x^3 + y^3 + z^3). This ring is normal but not Cohen-Macaulay. It is a 2-dimensional domain, normal because it is a hypersurface with isolated singularity, but depth(R)=1 < 2, so it fails S2? Actually, normal implies S2, so it must satisfy S2. Wait, this ring is actually Cohen-Macaulay? Hypersurfaces are Cohen-Macaulay. Need a non-Cohen-Macaulay normal ring. Example: the ring of invariants of a finite group acting on a regular ring may be normal but not Cohen-Macaulay. For instance, the ring of invariants of Z/2 acting on k[x,y] by swapping x and y is k[x+y, xy] which is Cohen-Macaulay. Another: the ring k[x,y,z]/(xy, xz) is not normal. Actually, a standard example is the Segre product of two polynomial rings: R = k[x,y] # k[z,w] = k[xz, xw, yz, yw]/(xz*yw - xw*yz) is normal but not Cohen-Macaulay in characteristic 0? I recall that the coordinate ring of the Segre embedding of P^1 × P^1 is normal and Cohen-Macaulay. Need a concrete non-Cohen-Macaulay normal ring: the ring k[x,y,u,v]/(xu, xv, yu, yv) is not normal. Actually, the r
2. Give an example where the exponential map is not surjective.
For the Lie group GL(2, C), the exponential map is surjective because every invertible matrix has a logarithm. However, for SL(2, R) (real 2×2 matrices with determinant 1), the exponential map is not surjective. For instance, the matrix [[-1,0],[0,-1]] is in SL(2,R) but is not the exponential of any real traceless matrix because its eigenvalues are -1, and the exponential of a real matrix with negative eigenvalues would have positive determinant? Actually, it is not in the image because exp(X) for X in sl(2,R) has positive determinant? Wait, det(exp(X)) = exp(tr(X)) = 1, so exp(X) always has positive determinant? But [[-1,0],[0,-1]] has determinant 1, so it might be in the image? Actually, the exponential of [[0,π],[-π,0]] is [[-1,0],[0,-1]]? Check: exp(π J) where J = [[0,1],[-1,0]] gives cos π I + sin π J = -I. So it is surjective? For SL(2,R), the exponential map is actually surjective? Wait, known fact: For SL(2,R), the exponential map is not surjective; elements with trace less than -2 are not in the image. For example, [[-2,0],[0,-1/2]]? Actually, need a correct example: The mat
3. Find the invariant factors of the group Z/2Z ⊕ Z/4Z ⊕ Z/8Z.
The invariant factors are a sequence where each divides the next. For Z/2Z ⊕ Z/4Z ⊕ Z/8Z, we list the orders: 2,4,8. The largest is 8, then we need a divisor of 8 that also divides something? Actually, we combine to get a single cyclic factor? Wait: The group is not cyclic because gcd(2,4,8)=2, but the direct sum of cyclic groups of orders 2,4,8 has invariant factors 2 and 8? Let's compute: The group is isomorphic to Z/2Z ⊕ Z/8Z? No, because Z/4Z is not a direct summand of Z/8Z? The correct method: For a finite abelian group, the invariant factors are the orders of the cyclic groups in a decomposition where each divides the next. Here, we can rewrite Z/2Z ⊕ Z/4Z ⊕ Z/8Z as Z/2Z ⊕ Z/2Z ⊕ Z/8Z? Actually, Z/4Z is not isomorphic to Z/2Z⊕Z/2Z. The invariant factors are (2,8)? Let's check: The group has exponent 8, and the largest invariant factor is 8. The next invariant factor must divide 8 and be such that the product of all invariant factors equals the order (2*4*8=64). If invariant factors are (2,8), product is 16, not 64. So we need three invariant factors? For a group of order 64, th
4. Give an example showing that the lattice of convex sets in R^2 is not distributive.
In R^2, consider three convex sets: A = the line segment from (0,0) to (1,0), B = the segment from (0,0) to (0,1), and C = the segment from (1,0) to (0,1). Their convex hulls? Actually, take A = {(0,0)}, B = {(1,0)}, C = {(0,1)}. Then A ∧ (B ∨ C) = {0} ∧ convex hull of {(1,0),(0,1)} = {0} ∧ (the segment) = empty? Wait, {0} is not in the convex hull of B and C? The convex hull of B and C is the line segment between (1,0) and (0,1), which does not contain (0,0). So meet is empty. Meanwhile, (A ∧ B) ∨ (A ∧ C) = empty ∨ empty = empty. That doesn't show non-distributivity. Better: take A = triangle with vertices (0,0),(1,0),(0,1), B = segment from (0,0) to (1,0), C = segment from (0,0) to (0,1). Then A ∧ (B ∨ C) = A ∩ convex hull of B∪C = A ∩ triangle = A. (A ∧ B) ∨ (A ∧ C) = B ∨ C = convex hull of B∪C = triangle. So A = triangle, but triangle ≠ triangle? Actually they are equal? Need a proper counterexample. In general, the lattice of convex sets is not distributive; one can find three convex sets where the distributive law fails. For instance, in R, take A = [0,2], B = [1,3], C = [2,4].
5. Give an example where naive truncation changes homology at the truncation point.
Let C be the complex ... → 0 → Z → Z → 0 → ... with differential multiplication by 2, where Z is in degrees 1 and 0. Then H_1(C) = 0 (kernel of multiplication by 2 is 0), H_0(C) = Z/2Z. Naive truncation σ_{≤0}C has only Z in degree 0, so H_0(σ_{≤0}C) = Z, which is different from H_0(C) = Z/2Z. Canonical truncation τ_{≤0}C has ker(d_0)=0 in degree 0, so H_0=0, which is also wrong? Actually τ_{≤0}C has (τ_{≤0}C)_0 = ker(d_0)=0, so H_0=0, but H_0(C)=Z/2Z. Wait, check: For C: d_1: Z→Z, multiplication by 2. Then ker(d_0) is all of Z? No, d_0: Z→0, so ker(d_0)=Z. So τ_{≤0}C has (τ_{≤0}C)_0 = ker(d_0)=Z, and (τ_{≤0}C)_1 = ker(d_1)=0. So H_0 = Z/0 = Z, again wrong. Actually canonical truncation τ_{≤0} should give H_0 = H_0(C) = Z/2Z. Let's compute correctly: For τ_{≤0}C, we take (τ_{≤0}C)_0 = ker(d_0) = Z (since d_0: Z→0), and (τ_{≤0}C)_1 = C_1 = Z, but differential from degree 1 to 0 is the restriction of d_1 to ker(d_1)? No, the definition: (τ_{≤n}C)_i = C_i for i < n, (τ_{≤n}C)_n = ker(d_n), and differentials are induced. So for n=0: (τ_{≤0}C)_0 = ker(d_0)=Z, (τ_{≤0}C)_1 = C_1 = Z, and di
6. Give an example of applying a reflection functor to a representation of a quiver.
Consider the quiver Q: 1 → 2, and let v=2 be a sink. A representation V of Q is (V1, V2; f: V1→V2). The reflected quiver σ_2 Q is 1 ← 2. The reflection functor F_2^+ sends V to (V1', V2'; f') where V1' = ker f, V2' = V1/ker f? Actually, the standard construction is: For a sink v, define (F_v^+ V)_v = direct sum over arrows into v of the source spaces, modulo a certain relation. For this simple quiver, F_2^+ (V) is (V1, V2; f') where V1' = V1, V2' = coker f? Wait, careful: The reflection functor at a sink v: For each arrow a: u→v, we take the vector space at u. Then (F_v^+ V)_v = ⊕_{a: u→v} V_u, and the maps are given by the canonical injections. For the quiver 1→2, with v=2 sink, we have (F_2^+ V)_2 = V1, and the map from (F_2^+ V)_1 to (F_2^+ V)_2 is the identity? Actually, the reflection functor is defined on the category of representations of Q to representations of σ_v Q. For the example, it's easier to look at the inverse functor. But a concrete example: Let V be the representation with V1 = k, V2 = k, f = 1. Then F_2^+ V is the representation of 1←2 with spaces: at vertex 1: k,
7. Explain the Third Isomorphism Theorem for rings with an example.
The Third Isomorphism Theorem: if I and J are ideals of R with I ⊆ J, then J/I is an ideal of R/I, and (R/I)/(J/I) ≅ R/J. For instance, let R=Z, I=6Z, J=12Z. Then R/I = Z/6Z, J/I = 12Z/6Z = {0,6} in Z/6Z, which is an ideal. The quotient (Z/6Z)/({0,6}) has two elements, and it is isomorphic to Z/12Z? Wait, check: R/J = Z/12Z has 12 elements, but (Z/6Z)/({0,6}) has 3 elements? Actually, J=12Z, I=6Z, so J/I = {0,6} in Z/6Z, which has 2 elements, so (Z/6Z)/(J/I) has 3 elements, but Z/12Z has 12 elements — this example is wrong because I⊆J but J/I is not an ideal of R/I? Wait, J/I is an ideal of R/I, and (R/I)/(J/I) ≅ R/J. Here R/J = Z/12Z has 12 elements, but (Z/6Z)/(J/I) has 6/2=3 elements, so this is not an isomorphism. The error: J must contain I, but here 12Z does not contain 6Z? Actually 12Z ⊆ 6Z, so I=6Z, J=12Z, then J⊆I, not I⊆J. So the condition is I⊆J. Correct example: R=Z, I=12Z, J=6Z? No, J must contain I. Let I=12Z, J=6Z? Then J contains I? 6Z contains 12Z, yes. So I=12Z, J=6Z. Then R/I=Z/12Z, J/I=6Z/12Z = {0,6} in Z/12Z, which is an ideal. (Z/12Z)/({0,6}) has 6 elements, and
8. Can two different matrices have the same minimal polynomial? Give an example.
Yes, two different matrices can have the same minimal polynomial. For example, the 2x2 identity matrix and the 2x2 zero matrix have different characteristic polynomials: (λ-1)^2 and λ^2, but their minimal polynomials are λ-1 and λ, respectively, which are different. However, consider the 2x2 matrix with Jordan block [[2,1],[0,2]] and the 2x2 diagonal matrix diag(2,2). Both have minimal polynomial λ-2? No, the Jordan block has minimal polynomial (λ-2)^2, while the diagonal has λ-2. So they differ. Actually, two matrices with the same minimal polynomial: the 3x3 Jordan block for eigenvalue 0 and a 3x3 nilpotent matrix with two Jordan blocks of sizes 2 and 1 both have minimal polynomial λ^2? Let's check: a 3x3 nilpotent matrix with one 2x2 block and one 1x1 block has minimal polynomial λ^2, while a 3x3 Jordan block has minimal polynomial λ^3. So not same. A better example: two different diagonalizable matrices with the same eigenvalues but different multiplicities can have the same minimal polynomial if the distinct eigenvalues are the same. For instance, diag(1,1,2) and diag(1,2,2) bot
9. Give an example of a module that is not injective and explain why.
The ℤ-module ℤ itself is not injective. To see this, consider the inclusion map f: 2ℤ → ℤ (sending 2n to 2n) and the identity map g: 2ℤ → ℤ (sending 2n to 2n). If ℤ were injective, there would exist a homomorphism h: ℤ → ℤ such that h∘f = g. But then h(2)=2, and since h is a ℤ-linear map, h(1) must be an integer a with 2a=2, so a=1. Then h(1)=1, but then h(2)=2, which is fine. However, the map h would have to satisfy h(2n)=2n, but that defines h uniquely as the identity, which is indeed a homomorphism. Wait, that actually works. Let's try a different example: the ℤ-module ℤ/2ℤ is not injective. Consider the inclusion f: 2ℤ → ℤ and the map g: 2ℤ → ℤ/2ℤ sending 2n to n mod 2. This map is well-defined because 2n mod 2 = 0, so g(2n)=0? Actually, g(2n)=n mod 2, which is not zero for odd n. But 2ℤ consists of even numbers, so 2n is even, and n can be odd, so g(2n)=1 mod 2 for odd n. This is a homomorphism. If ℤ/2ℤ were injective, there would exist h: ℤ → ℤ/2ℤ extending g. Then h(1) would be some element a in ℤ/2ℤ. Then h(2)=2a = 0, but g(2)=1 mod 2, contradiction. So ℤ/2ℤ is not injective.
10. What is a 'bent' Boolean function, and why is it useful in cryptography?
A bent function is a Boolean function that is maximally nonlinear; it is as far as possible from any linear function. This property makes it resistant to linear cryptanalysis. Bent functions exist only for even number of variables. For example, the function f(x,y) = x AND y XOR (x XOR 1) AND (y XOR 1) is bent for 2 variables? Actually, bent functions for 2 variables are like XOR? Wait, bent functions for 2 variables are the ones with nonlinearity 1? Let's correct: For 2 variables, the function f(x,y)=x XOR y is linear, not bent. Bent functions for 2 variables: f(x,y)=x AND y is not bent either. Actually, for 2 variables, the maximum nonlinearity is 1, and functions like f(x,y)=x AND y have nonlinearity 1? I need to be accurate. Bent functions exist for n even, n>=2. For n=2, all functions are either linear or affine? Actually there are 16 functions; the bent ones are those with nonlinearity 1? Let's not give a wrong example. Instead, say: A bent function on 4 variables has nonlinearity 6. They are used in stream ciphers and as components of S-boxes.
11. Describe the free Boolean algebra on one generator.
The free Boolean algebra on one generator x has four elements: 0, 1, x, and ¬x. The operations are determined: x ∧ x = x, x ∨ ¬x = 1, etc. It is isomorphic to the power set of a one-element set {a}: map x to {a}, ¬x to ∅^c? Actually, the power set of a one-element set has two elements: ∅ and {a}. But here we have four. Wait, correction: The free Boolean algebra on one generator has four elements: 0, 1, x, ¬x. That's because we also have complements. So it is isomorphic to the direct product of two copies of the two-element algebra? Actually, it's the Boolean algebra of all subsets of a two-element set? No, the free algebra on one generator is the four-element Boolean algebra, which is isomorphic to the power set of a two-element set? Wait, power set of a two-element set has 4 elements: ∅, {a}, {b}, {a,b}. That is indeed a four-element Boolean algebra. So the free algebra on one generator is the four-element Boolean algebra. The generator x corresponds to one of the atoms, say {a}, and ¬x corresponds to the other atom {b}.
12. Find the Levi decomposition of the Lie algebra of upper triangular 2x2 matrices.
The Lie algebra of upper triangular 2x2 matrices has basis e11, e22, e12 with brackets [e11,e12]=e12, [e22,e12]=-e12, [e11,e22]=0. The radical is the set of strictly upper triangular matrices, which is the 1-dimensional ideal spanned by e12. A Levi subalgebra can be the subalgebra of diagonal matrices spanned by e11 and e22, which is abelian and semisimple? Actually diagonal matrices form a 2-dimensional abelian Lie algebra, which is not semisimple because it has a nontrivial abelian ideal. Wait, the Levi decomposition requires the semisimple part to be semisimple. For this algebra, the radical is solvable (it is nilpotent), and a Levi subalgebra is the subalgebra of traceless diagonal matrices? But the algebra of all diagonal matrices is not semisimple. Actually, the Lie algebra of upper triangular matrices is solvable, so its radical is the whole algebra, and the Levi subalgebra is 0. So the Levi decomposition is just the algebra itself as its own radical.