Math 502: Abstract Algebra

Homework 8

Lawrence Tyler Rush

<me@tylerlogic.com>

January 5, 2014

http://coursework.tylerlogic.com/courses/upenn/math502/homework08

1

Let p(x) = x³ - x - 1 ∈ ℚ[x]

(a) Extra Credit: Show that p(x) is irreducible in ℚ[x]

(b)

(c)

Let T_n ∈ End_ℚ(V _n) be defined by

Tn(f (x) +p(x)nℚ[x]) = x⋅f (x) +p(x)nℚ[x] ∀ f(x) ∈ ℚ[x]

For n = 1 The images of the basis elements in part (b) are

T(1+ p(x)ℚ[x]) = x+ p(x)ℚ[x] T(x+ p(x)ℚ[x]) = x2 + p(x)ℚ [x] 2 3 T(x + p(x)ℚ[x]) = x + p(x)ℚ [x] = (x + 1)+ p(x )ℚ [x]

and so the matrix representation is

( ) ( 1 ) 1 1 1

For n = 2 Since

2 6 4 3 2 p(x) = x - 2x - 2x +x + 2x+ 1

then the images of the basis elements in part (b) are

T (1 + p(x)2ℚ [x]) = x+ p(x)2ℚ[x] T (x + p(x)2ℚ [x]) = x2 + p(x)2ℚ [x ] 2 2 3 2 T (x + p(x) ℚ [x]) = x + p(x)ℚ [x ] T(x3 - x - 1 + p(x)2ℚ [x]) = (x4 - x2 - x)+ p(x)2ℚ[x] T(x(x3 - x - 1)+ p(x)2ℚ [x]) = (x5 - x3 - x2)+ p(x)2ℚ [x ] T (x2(x3 - x - 1)+ p(x)2ℚ [x]) = (x6 - x4 - x3)+ p(x)2ℚ [x ] = (x4 + x3 - x2 - 2x - 1) + p(x)ℚ [x]

which results in the following matrix representation

( 1| ) | 1 1| | || 1 | || || -------1|------1-|| |( |1 1 |) | 1

with vertical and horizontal lines to better see the nicities of the matrix.

(d)

For n = 1 The characteristic polynomial for the matrix

( ) ( 1 ) 1 1 1

from above is

( ( 1 ) ( 1) ) det( λ( 1 ) - ( 1 1) ) = λ3 - λ- 1 1 1

According to part (a), this polynomial is irreducible, so because the minimal polynomail divides the characteristic polynomial, this polynomial is also the minimal polynomial.

For n = 2 The characteristic polynomial for the matrix

( 1 ) || 1 1 || || 1 || || 1 1 || ( 1 1 ) 1

from above is

( ( ) ( )) 1 1 || || 1 || || 1 1 |||| det|| λ|| 1 || - || 1 |||| || || 1 || || 1 1 |||| ( ( 1 ) ( 1 1 )) 1 1

which is

2 2 2 λ(λ (λ(λ(λ - 1)- 1))- (- 1)(- 1)(3λ(λ3 - 1)- 1))(- 1)(3- 1)(- 1)(λ(λ3- 1)- 1) λ (λ - λ - 1)- λ(λ - λ- 1)- (λ - λ - 1) (λ3 - λ - 1)(λ3 - λ - 1)

Again because neither of the two factors of the above product are reducible, then the minimal polynomial is simply λ³ - λ - 1

(e) Extra Credit: Minimal and Characteristic polynomial for T_n

Continuing the pattern above, the characteristic polynomial for T_n will be

n 3 n p(x) = (x - x - 1)

and the minimal polynomial will be

p(x) = x3 - x - 1

2

(a)

(b) Extra Credit

3

For each n ∈ ℕ define an F-linear operator, ∂^[n], on F[x] by

f(x+ t) = ∑ ∂[n](f)⋅tn n≥0

for all f(x) ∈ F[x]. So for an arbitrary m-degree polynomial f(x) ∈ F[x] defined as

∑m i aix i=0

we have, through use of the binomial formula, that

∑m i m∑ ∑i (i) i-j j ∑m ∑i (i) i- jj f(x+ t) = ai(x+ t) = ai j x t = ai j x t i=0 i=0 j=0 i=0j=0

Rearranging the indexing variables, we can morph the right-hand side of the above equation into

∑m ∑m (i) ai xi-jtj j=0i=j j

which in turn allows us to move the t^j outside the inner summation to obtain

( ( ) ) ∑m ( m∑ i i-j) j f(x+ t) = ai j x t j=0 i=j

which finally allows us to clearly see the coefficients of f(x + t) and therefore the formulation of ∂^[j](f) to be

∑m (i) ∂ [j](f) = ai j xi- j i=j

(3.1)

(a) Show that ∂^[1] is given by the standard formula for

Letting f(x) ∈ F[x] be a polynomial of degree m, then the formula in equation 3.1, we have

( ) [1] m∑ i i-1 ∑m i-1 ∂ (f) = ai 1 x = aiix i=1 i=1

which is exactly the formula for f′(x).

(b) Show that n! ⋅ ∂^[n](f) yields the “n-th derivative of f”

Letting f(x) ∈ F[x] be a polynomial of degree m, then the formula in equation 3.1, we have

m∑ ( i) n!⋅∂[n](f) = n! ai n xi-n i=n m∑ ---i!--- i- n = n! ain!(i- n)!x m i=n = ∑ a --i!--xi-n i=n i(i - n )! m∑ = aii(i- 1)(i- 2)⋅⋅⋅(i- (n + 1))xi-n i=n

which is exactly the formula for fⁿ(x).

(c) Extra Credit

4

(a) Show that End_grp(p^-mℤ∕ℤ) is naturally isomorphic to ℤ∕p^mℤ

Since the ring p^-mℤ∕ℤ is cyclically generated by p^-m, each endomorphism is defined by it’s mapping of p^-m. Since each element of p^-mℤ∕ℤ is an integer multiple of p^-m then let’s denote each element of End_grp(p^-mℤ∕ℤ) by

φ (p-m) := np--m n

Given this notation, because each φ_n,φ_m are ring homomorphisms, we are immediately afforded both φ_nφ_m = φ_nm and φ_n + φ_m = φ_n+m.

With this, we will define ϕ : End(p^-mℤ∕ℤ) → ℤ∕p^mℤ by ϕ(φ_n) = n. Thus using the ring homomorphic properties of φ_n and φ_m outlined above and the additive/multiplicative operations on ∕p^mℤ, we have

ϕ(φnφm ) = ϕ(φnm) = nm- --- = nm = ϕ(φn)ϕ(φm)

and

ϕ(φ + φ ) = ϕ(φ ) n m ---n+m = n-+m = nm = ϕ(φn)ϕ(φm)

and so ϕ is a ring homomorphism.

Now if ϕ(φ_n) = 0, then φ_n(p^-m) = 0p^-m = 0, and so φ_n = φ₀. With this we have the injectivity of ϕ. Now because End(p^-mℤ∕ℤ) and ℤ∕p^mℤ have the same cardinality, then ϕ is bijective. Hence we have that End(p^-mℤ∕ℤ) is isomorphic to ℤ∕p^mℤ.

(b)

For referential reasons, we will number the property each element of ℤ_p has

n am ≡ an(modp ℤ ) ∀ m ≥ n

(4.2)

(b1)

There exists a zero element The sequence of all zeros, denote it by (0), will be the zero element since:

(0)+ (xn)n∈ℕ = (0+ xn)n∈ℕ = (xn + 0)n∈ℕ = (xn)n∈ℕ + (0) ≥1 ≥1 ≥1 ≥1

Addition is closed For each (x_n)_{n∈ℕ_≥1},(y_n)_{n∈ℕ_≥1} ∈ ℤ_p their sum is in ℤ_p because of the closure of addition on ℤ∕pⁿℤ and because

n n n xm + ym ≡ (xn mod p ℤ )+ (yn mod p ℤ) ≡ (xn + yn) mod p ℤ

for all m ≥ n.

Additive inverses The sequence of negatives of the elements of a sequence is the additive inverse since

(xn)n∈ℕ + (- xn)n∈ℕ = (xn - xn)n∈ℕ = (0) = (- xn + xn)n∈ℕ = (- xn)n∈ℕ + (xn)n∈ℕ ≥1 ≥1 ≥1 ≥1 ≥1 ≥1

Addition is commutative by the following

(xn)n∈ℕ≥1 + (yn)n∈ℕ≥1 = (xn + yn)n∈ℕ≥1 = (yn + xn)n∈ℕ≥1 = (yn)n∈ℕ≥1 + (xn)n∈ℕ≥1

which is due to the commutative addition of ℤ∕pⁿℤ for each n.

There exists a 1 element which is the sequence of all ones, which we will denote by (1). It is the multiplicative identity by

(1)(xn)n∈ℕ≥1 = (1xn)n∈ℕ≥1 = (xn1)n∈ℕ≥1 = (xn)n∈ℕ≥1(1)

Multiplication is closed since

xmym ≡ (xn mod pnℤ)(yn mod pnℤ) ≡ (xnyn) mod pnℤ

for all m ≥ n

Multiplication is associative by the following

( ) (xn)n∈ℕ≥1(yn)n∈ℕ≥1 (zn)n∈ℕ≥1 = (xnyn)n∈ℕ≥1(zn)n∈ℕ≥1 = ((xnyn)zn)n∈ℕ≥1 = (xn(ynzn))n∈ℕ≥1 = (x ) (y z ) n n∈ℕ≥1( n nn∈ℕ≥1 ) = (xn)n∈ℕ≥1 (yn)n∈ℕ≥1(zn)n∈ℕ≥1

where we make use of associativity on ℤ∕pⁿℤ.

Multiplication distributes over addition by the following

( ) (xn)n∈ℕ≥1 (yn)n∈ℕ≥1 + (zn)n∈ℕ≥1 = (xn)n∈ℕ≥1(yn + zn)n∈ℕ≥1 = (xn(yn + zn))n∈ℕ ≥1 = (xnyn + xnzn)n∈ℕ≥1 = (x(nyn)n∈ℕ≥1 + (xnzn))n∈ℕ(≥1 ) = (xn)n∈ℕ≥1(yn)n∈ℕ≥1 + (xn)n∈ℕ≥1(zn)n∈ℕ≥1

where we make use of the distributive law on ℤ∕pⁿℤ.

Multiplication is commutative by the following

(xn)n∈ℕ≥1(yn)n∈ℕ≥1 = (xnyn)n∈ℕ≥1 = (ynxn)n∈ℕ≥1 = (yn)n∈ℕ≥1(xn)n∈ℕ≥1

in which we make use of the commutative property of multiplication on ℤ∕pⁿℤ.

Finally, given all the above properties, we have that ℤ_p is a commutative ring.

(b2)

Let π_n : ℤ_p → ℤ∕pⁿℤ be the n-th component projection map. For m ∈ ℤ∕pⁿℤ, define the sequence (x_n)_{ℕ_≥1} by x_n := m mod pⁿℤ for each n ∈ ℕ_≥1. Then we will have that π_n((x_n)_{ℕ_≥1}) = m(modpⁿℤ) = m. Hence the map π_n is surjective.

Through use of the additive and multiplicative definitions on both ℤ_p and ℤ∕pⁿℤ, we obtain

------- --- -- πn((xn)ℕ≥1 + (yn)ℕ≥1) = πn((xn + yn)ℕ≥1) = xn + yn = xn + yn = πn((xn)ℕ≥1)+ πn((yn)ℕ≥1)

and

----- ----- πn((xn )ℕ≥1(yn)ℕ≥1) = πn((xnyn)ℕ≥1) = xnyn = xn(yn) = πn((xn)ℕ≥1)πn((yn)ℕ≥1)

which reveals that π_n is a ring homomorphism in addition to being surjective.

(b3)

Let (x_m) ∈ Kerπ_n. Then x_n ≡ 0 mod pⁿ implying that x_n is a multiple of pⁿ. Furthermore given equation 4.2 we have that

x ≡ 0 mod pn m

(4.3)

for all m ≥ n. Hence each x_m is a multiple of pⁿ for m ≥ n. Likewise, equation 4.2 gives us that x_n ≡ x_k mod p^k for all k < n, so since x_n is a multiple of pⁿ it is inherently a multiple of p^k for k < n. Thus we have that each x_k ≡ 0 mod p^k which also implies that

xk ≡ 0 mod pn

(4.4)

Hence the fact that x_n ≡ 0 mod pⁿ combined with equations 4.3 and 4.4 implies that (x_n) ∈ pⁿ ⋅ ℤ_p. So we have that Kerπ_n ⊂ pⁿ ⋅ ℤ_p.

Now if (x_n) ∈ pⁿℤ_p, then x_n would be a multiple of pⁿ, i.e. x_n ≡ 0 mod pⁿ. So the image of (x_n) under π_n will therefore be 0 ∈ ℤ∕pⁿℤ. Hence pⁿℤ_p ⊂ Kerπ_n.