Math 593 Assignment # 5 Solutions

Bryden R. Cais

October 23, 2002

1. The k[x] module

M = \frac{k[x]}{(x^{3} + x + 1)} ⊕ \frac{k[x]}{x^{2}}

is given to us in primary decomposition form. As such, we can read off the rational canonical form of the matrix T: it has two blocks corresponding to the two summands, and each block is the companion matrix of the polynomial in the “denominator” of the summand. That is, we have

T = (\begin{matrix}0 & 0 & −1 & 0 & 0 \\ 1 & 0 & −1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{matrix})

up to a change of basis of V = k^{5}.

What is going on is simple. Let V = κ^{5}. Then we know that V has standard basis e_{i} for 1 ≤ i ≤ 5. Observe that M has basis

f_{1} = (1, 0), f_{2} = (x, 0), f_{3} = (x^{2}, 0), f_{4} = (0, 1), f_{5} = (0, x).

We then have an isomorphism (because we are sending a basis to a basis)

M ⟶ V

given by

f_{i} ↦ e_{i}.

To determine T we simply need to see what multiplication by x does to the basis f_{i}. We have

xf_{1} = (x, 0) = f_{2}
xf_{2} = (x^{2}, 0) = f_{3}
xf_{3} = (−x − 1, 0) = −f_{2} − f_{1}
xf_{4} = (0, x) = f_{5}
xf_{5} = (0, x^{2}) = 0.

It follows that multiplication by x on M has the matrix T above.

2. (a) Let ρ be a representation of G on V. We make V into a k[G] module by defining

gv := ρ(g)v

for any g ∈ G and v ∈ V and extending by linearity to all of k[G]. Clearly V is an abelian group under +. It is easy to verify that

i. 1v = ρ(1)v = v since ρ(1) is the identity matrix (since ρ is a homomorphism). ii. (gh)v = ρ(gh)v = ρ(g)ρ(h)v = ρ(g)(hv) = g(hv), since ρ is a homomorphism. iii. (g + h)v = (ρ(g) + ρ(h))v = ρ(g)v + ρ(h)v = gv + hv iv. g(v + w) = ρ(g)(v + w) = ρ(g)v + ρ(g)w = gv + gw, since ρ(g) is a linear transformation. Hence, we have made V into a k[G] module.

(b) Now suppose that V has the structure of a k[G] module where scalars λ ∈ k[G] act by scalar (vector space) multiplication. Define ρ : G → GL(V) by

ρ(g)v := gv

for any g ∈ G, v ∈ V. We must check that ρ defines a homomorphism ρ : G → GL(V). Observe that

ρ(gh)v := (gh)v = g(hv) = g(ρ(h)v) = ρ(g)ρ(h)v

since V is a k[G] module. Moreover, since

ρ(1g)v = ρ(1)ρ(g)v = ρ(g)ρ(1)v = ρ(g)v

for all v ∈ V, we have ρ(1) = 1. It follows that ρ is a homomorphism. The usual axioms for a k[G] module imply that ρ(g) is a linear transformation of V. Since

ρ(g)ρ(g^{−1}) = ρ(1) = 1

we see that ρ(g) is an invertible linear transformation. It follows that ρ is a representation of G on V. The fact that V ≃ V_{ρ} as k[G] modules follows from our definition in part a).

3. Let k be a field and f, g ∈ k[x] nonzero polynomials. For any p, q ∈ k[x], we let p̄_{q} denote the image of p in k[x]/(q) under the natural quotient map π_{q}. Let

R = Hom_{k[x]} (k[x]/f, k[x]/g) ,

and suppose that ϕ ∈ R. Let ϕ̃ be any lift of ϕ to an element of Hom_{k[x]} (k[x], k[x]); that is, ϕ̃ is any k[x] module homomorphism making the following diagram commute:

k[x] ⟶^{ϕ̃} k[x]
π_{f} ↓  ↓^{π_{g}}
k[x]/(f) ⟶_{ϕ} k[x]/(g)

Then we have

ϕ̃(f) = fϕ̃(1),

but since

ϕ ∘ π_{f}(f) = ϕ(¯0_{f}) = ¯0_{g},

we must have

π_{g}(fϕ̃(1)) = ¯0_{g},

or what is the same, g|fϕ̃(1). Since k is a field, k[x] is a PID (therefore also a UFD), so putting

(g, f) = (h)

we see that

g_{1} = g/h

divides ϕ̃(1). Now define the map

Π : k[x] ⟶ R

by

p ↦ ϕ_{p}

where ϕ_{p} is defined by

ϕ_{p}(¯1_{f}) = p̄_{g}ḡ_{1}_{g}.

(Realize that an element of R is completely determined by where it sends ¯1_{f}). It is not difficult to check that Π is a k[x] module homomorphism. Indeed,

(a)

Π(p + q)(¯1_{f}) = ϕ_{p+q}(¯1_{f})
= (p̄_{g} + q̄_{g})ḡ_{1}_{g}
= p̄_{g}ḡ_{1}_{g} + q̄_{g}ḡ_{1}_{g}
= (Π(p) + Π(q))(¯1_{f}),

so that Π(p + q) = Π(p) + Π(q).

(b)

Π(pq)(¯1_{f}) = ϕ_{pq}(¯1_{f})
= p̄_{g}q̄_{g}ḡ_{1}_{g}
= p̄_{g}Π(q)(¯1_{f})

so that Π(pq) = pΠ(q) (modulo g, of course, but that is all we care about).

That Π is surjective follows from our opening remarks. Now Π(p) = 0 if and only if Π(p)(¯1_{f}) = ¯0_{g}, that is, if and only if g|pg_{1}, or equivalently, if and only if h|p. Therefore, ker Π = (h) and we have

k[x]/(h) ≃ Hom_{k[x]} (k[x]/(f), k[x]/(g)) .

4. (a) Let V = ℝ^{2} and T_{θ} : V → V be rotation by θ. Then V_{θ} is irreducible if and only if θ ∉ πℤ. First suppose that that θ ∉ πℤ. Then sin(θ) ≠ 0 so that for any nonzero v ∈ V, the magnitude of v × T_{θ}v is not zero; that is, v, T_{θ}v form the sides of a parallelogram in ℝ^{2} with nonzero area. As such, v, T_{θ}v span ℝ^{2}. (There are lots of ways of saying this; the idea is that as long as θ is not a multiple of π, the vectors v, T_{θ}v are nor multiples of each other). Let W ⊂ V be a T_{θ} invariant nontrivial subspace and let w ∈ W be nonzero. Then T_{θ}w ∈ W so that W contains a basis for V and hence W = V so that V is irreducible. Conversely, if θ = nπ for some n ∈ ℤ then T_{θ}v = ±v for any v ∈ V; as such V_{θ} has the nontrivial proper submodules M_{v} = {λv : λ ∈ ℝ} for all v ≠ 0 in V and is thus reducible.

(b) A nontrivial submodule of V_{ρ} is a nontrivial subspace W ⊂ V that is preserver by the action of G. Therefore, V is irreducible if and only if for every W ⊂ V, there exists some g ∈ G such that ρ(g)W ⊈ W. Equivalently, V is irreducible if and only if the G orbit of any nonzero vector v ∈ V spans V (if the orbit doesn’t span V then it spans a proper, nontrivial subspace W, which is then preserved by the action of G).

(c) V is reducible as a k[S_{3}] module. To see this, let W ⊂ V be the subspace given by

W = {(a, a, a) : a ∈ k}.

Clearly this is a nontrivial proper subspace of V. Moreover, W is S_{3} invariant since S_{3} acts by permuting the components of any vector v ∈ V. It follows that W is a nontrivial, proper k[S_{3}] submodule of V.

5. (a) Suppose that M is an irreducible R module. Let ϕ ∈ End_{R}(M) and suppose that ϕ ≠ 0. Then since ker ϕ ⊂ M is a submodule, ker ϕ = 0 or M; since ϕ ≠ 0, ϕ is then injective. Hence ϕ is an isomorphism and therefore invertible; that is, End_{R}(M) is a division ring. Notice that we really do need to show that ϕ is both injective and surjective: if M were finite, one implies the other, but in general injectivity of a map ϕ : M → M does not imply surjectivity. For example, the map ϕ : ℤ → ℤ given by ϕ(n) = 2n is certainly injective, but definitely not surjective.

(b) There are many ways of seeing this. The intuition is that T acts just like i = \sqrt{−1} and gives R^{2} complex structure. Moreover, since x acts by T, ℝ[x] is like ℂ. It is clear that the ℂ linear endomorphisms of ℂ is isomorphic to ℂ since any such endomorphism is completely determined by where the identity is sent, and it can be sent anywhere. To make this intuition into a proof, we proceed as follows: We take the standard basis e_{1}, e_{2} for V. In this basis, any endomorphism ov V is given as a 2 × 2 matrix with entries in ℝ. Suppose φ ∈ End_{ℝ[x]}(V_{T}) and let

(\begin{matrix}a & b \\ c & d \end{matrix})

be the matrix of φ. Since φ must commute with T, and since the matrix of T is given by

(\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix}) ,

a short calculation shows that we must have

(\begin{matrix}−c & −d \\ a & b \end{matrix}) = (\begin{matrix}b & −a \\ d & −c \end{matrix}) ,

that is, b = −c, a = d. It follows that every φ ∈ End_{ℝ[x]}(V_{T}) has the form

a + bT

for some a, b ∈ ℝ. Therefore, we define

ϕ : ℂ ⟶ End_{ℝ[x]}(V_{T})

by

ϕ(a + bi) = a + bT.

By our remarks above, ϕ is surjective. Injectivity is also clear. That ϕ is a homomorphism follows from a short calculation:

ϕ((a + bi)(c + di)) = ϕ(ac − bd + (ad + bc)i) = (ac − bd) + (ad + bc)T,

on the other hand,

ϕ(a + bi)ϕ(c + di) = (a + bT)(c + dT) = ac + adT + bcT + bdT^{2} = (ac − bd) + (ad + bc)T

since T^{2} = −1. It is obvious that ϕ preserves addition and that ϕ(1) = 1. Since ϕ is a bijective ring homomorphism, it is an isomorphism, and we are done.