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x3 + x + 1 and f(x) = x + 1, the first step of the Euclidean algorithm gives
p(x) = (x2 +x)f(x)+1. Thus p(x)-(x2 +x)f(x) = 1, and so reducing modulo
p(x) gives [-x2 - x][f(x)] = [1], and thus [x + 1]-1 = [-x2 - x] = [x2 + x].
We next give an alternate solution, which uses the identity [x3] = [x + 1] to
solve a system of equations. We need to solve [1] = [x + 1][ax2 + bx + c] or
[1] = [ax3 + bx2 + cx + ax2 + bx + c]
= [ax3 + (a + b)x2 + (b + c)x + c]
= [a(x + 1) + (a + b)x2 + (b + c)x + c]
= [(a + b)x2 + (a + b + c)x + (a + c)] ,
so we need a + b a" 0 (mod 2), a + b + c a" 0 (mod 2), and a + c a" 1 (mod 2).
This gives c a" 0 (mod 2), and therefore a a" 1 (mod 2), and then b a"
1 (mod 2). Again, we see that [x + 1]-1 = [x2 + x].
13. Find the multiplicative inverse of [x2 + x + 1]
(a) in Q[x]/ x3 - 2 ;
Solution: Using the Euclidean algorithm, we have
x3 - 2 = (x2 + x + 1)(x - 1) + (-1), and so [x2 + x + 1]-1 = [x - 1].
This can also be done by solving a system of 3 equations in 3 unknowns.
(b) in Z3[x]/ x3 + 2x2 + x + 1 .
Solution: Using the Euclidean algorithm, we have
x3 + 2x2 + x + 1 = (x + 1)(x2 + x + 1) + (-x) and
x2 + x + 1 = (-x - 1)(-x) + 1. Then a substitution gives us
1 = (x2 + x + 1) + (x + 1)(-x)
= (x2 + x + 1) + (x + 1)((x3 + 2x2 + x + 1) - (x + 1)(x2 + x + 1))
= (-x2 - 2x)(x2 + x + 1) + (x + 1)(x3 + x2 + 2x + 1) .
Thus [x2 + x + 1]-1 = [-x2 - 2x] = [2x2 + x]. This can be checked by finding
[x2 + x + 1][2x2 + x], using the identities [x3] = [x2 - x - 1] and [x4] = [x - 1].
CHAPTER 4 SOLUTIONS 91
This can also be done by solving a system of equations, or, since the set is
finite, by taking successive powers of [x2 + x + 1]. The latter method isn t
really practical, since the multiplicative group has order 26, and this element
turns out to have order 13.
14. In Z5[x]/ x3 + x + 1 , find [x]-1 and [x + 1]-1, and use your answers to find
[x2 + x]-1.
Solution: Using the division algorithm, we obtain x3 + x + 1 = x(x2 + 1) + 1,
and so [x][x2 + 1] = [-1]. Thus [x]-1 = [-x2 - 1].
Next, we have x3 +x+1 = (x+1)(x2-x+2)-1, and so [x+1]-1 = [x2-x+2].
Finally, we have
[x2 + x]-1 = [x]-1[x + 1]-1 = [-x2 - 1][x2 - x + 2]
= [-x4 + x3 - 2x2 - x2 + x - 2] .
Using the identities [x3] = [-x - 1] and [x4] = [-x2 - x], this reduces to
[x2 + x]-1 = [x2 + x - x - 1 - 3x2 + x - 2]
= [-2x2 + x - 3] = [3x2 + x + 2] .
15. Factor x4 + x + 1 over Z2[x]/ x4 + x + 1 .
Solution: There are 4 roots of x4 + x + 1 in the given field, given by the
cosets corresponding to x, x2, x + 1, x2 + 1. This can be shown by using the
multiplication table, with the elements in the form 10, 100, 11, and 101, or
by computing with polynomials, using the fact that (a + b)2 = a2 + b2 since
2ab = 0. We have x4 + x + 1 a" 0,
(x2)4 + (x2) + 1 = (x4)2 + x2 + 1 a" (x + 1)2 + x2 + 1 a" x2 + 1 + x2 + 1 a" 0,
(x + 1)4 + (x + 1) + 1 a" x4 + 1 + x a" x + 1 + 1 + x a" 0, and
(x2 +1)4 +(x2 +1)+1 a" (x4)2 +1+x2 a" (x+1)2 +1+x2 a" x2 +1+1+x2 a" 0.
Thus x4 + x + 1 factors as a product of 4 linear terms.
92 CHAPTER 4 SOLUTIONS
Chapter 5
Commutative Rings
SOLUTIONS TO THE REVIEW PROBLEMS
1. Let R be the ring with 8 elements consisting of all 3 × 3 matrices with entries
in Z2 which have the following form:
îø ùø
a 0 0
ðø ûø
0 a 0
b c a
You may assume that the standard laws for addition and multiplication of
matrices are valid.
(a) Show that R is a commutative ring (you only need to check closure and
commutativity of multiplication).
Solution: It is clear that the set is closed under addition, and the following
computation checks closure under multiplication.
îø ùø îø ùø îø ùø
a 0 0 x 0 0 ax 0 0
ðø ûø ðø ûø ðø ûø
0 a 0 0 x 0 = 0 ax 0
b c a y z x bx + ay cx + az ax
Because of the symmetry a ”! x, b ”! y, c ”! z, the above computation also
checks commutativity.
(b) Find all units of R, and all nilpotent elements of R.
Solution: Four of the matrices in R have 1 s on the diagonal, and these are
invertible since their determinant is nonzero. Squaring each of the other four
matrices gives the zero matrix, and so they are nilpotent.
(c) Find all idempotent elements of R.
93
94 CHAPTER 5 SOLUTIONS
Solution: By part (b), an element in R is either a unit or nilpotent. The only
unit that is idempotent is the identity matrix (in a group, the only idempotent
element is the identity) and the only nilpotent element that is also idempotent
is the zero matrix.
2. Let R be the ring Z2[x]/ x2 + 1 . Show that although R has 4 elements, it [ Pobierz całość w formacie PDF ]

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