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 [x]

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


(b)


(c)


Let Tn End(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 Tn


Continuing the pattern above, the characteristic polynomial for Tn 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 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 tj 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 ddx-


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 Endgrp(p-m) is naturally isomorphic to ∕pm


Since the ring p-mis cyclically generated by p-m, each endomorphism is defined by it’s mapping of p-m. Since each element of p-mis an integer multiple of p-m then let’s denote each element of Endgrp(p-m) by
φ (p-m) := np--m
 n

Given this notation, because each φnm are ring homomorphisms, we are immediately afforded both φnφm = φnm and φn + φm = φn+m.

With this, we will define ϕ : End(p-m) ∕pmby ϕ(φn) = n. Thus using the ring homomorphic properties of φn and φm outlined above and the additive/multiplicative operations on ∕pm, 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(φn) = 0, then φn(p-m) = 0p-m = 0, and so φn = φ0. With this we have the injectivity of ϕ. Now because End(p-m) and ∕pmhave the same cardinality, then ϕ is bijective. Hence we have that End(p-m) is isomorphic to ∕pm.

(b)


For referential reasons, we will number the property each element of 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 (xn)n1,(yn)n1 p their sum is in p because of the closure of addition on ∕pnand 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 ∕pnfor 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 ∕pn.

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 ∕pn.

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 ∕pn.

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

(b2)
Let πn : p ∕pnbe the n-th component projection map. For m ∕pn, define the sequence (xn)1 by xn := m mod pnfor each n 1. Then we will have that πn((xn)1) = m(modpn) = m. Hence the map πn is surjective.

Through use of the additive and multiplicative definitions on both p and ∕pn, 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 πn is a ring homomorphism in addition to being surjective.

(b3)
Let (xm) Kerπn. Then xn 0 mod pn implying that xn 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 xm is a multiple of pn for m n. Likewise, equation 4.2 gives us that xn xk mod pk for all k < n, so since xn is a multiple of pn it is inherently a multiple of pk for k < n. Thus we have that each xk 0 mod pk which also implies that

xk ≡ 0 mod pn
(4.4)

Hence the fact that xn 0 mod pn combined with equations 4.3 and 4.4 implies that (xn) pn p. So we have that Kerπn pn p.

Now if (xn) pnp, then xn would be a multiple of pn, i.e. xn 0 mod pn. So the image of (xn) under πn will therefore be 0 ∕pn. Hence pnp Kerπn.

With the above two results we conclude that Kerπn = pnp.

(c) Extra Credit


(d) Extra Credit


(e) Extra Credit


(f) Extra Credit


(g) Extra Credit


5 Extra Credit