Math 342
Homework Assignment #6 (Due Friday, April 8, 5PM in my office)
1. Which of the following polynomials in F [x] are irreducible?
(a) f (x) = x2 + x + 1, F = Z2
(b) f (x) = x2 + x + 1, F = GF (4)
(c) f (x) = x4 + x + 1, F = Z2
(d) f (x) = ax3 + bx2 + ax + 1, F = GF (4)
solution:
a. The polynomial x2 + x + 1 has no solutions in Z2 . Since it is of degree 2, it is
irreducible over Z2 .
b. a2 + a + 1 = b + a + 1 = 1 + 1 = 0. Similarly, b2 + b + 1 = 0. So, a and b are roots in
GF (4). So, f (x) = (x − a)(x − b) = (x + a)(x + b) and thus is reducible over GF (4).
c. One check that the polynomial x4 + x + 1 has no roots in Z2 . Thus, the only possible
factors are the quadratics x2 , x2 + x, x2 + 1 and x2 + x + 1. However, we can eliminate
the first two, since they would imply that 0 is a root. So, we need only check if x2 + 1
and x2 + x, x2 + 1 divide x4 + x + 1. Apply the division algorithm
g(x) = h(x)q(x) + r(x)
to see if there is a remainder, which yields:
x4 + x + 1 = (x2 + 1)(x2 + 1) + x
and
x4 + x + 1 = (x2 + x + 1)(x2 + x) + 1
So, x4 + x + 1 is irreducible.
d. Check that this polynomial has no roots in GF (4). Since it is a cubic, it is irreducible.
2. Find the gcd of 2x5 + x4 + 2x3 + x2 + 1 and 2x4 + x3 + x + 1 in Z3 [x]
solution:
Apply the Euclidean Algorithm.
2x5 + x4 + 2x3 + x2 + 1 = (2x4 + x3 + x + 1)(x) + (2x3 + 2x + 1)
2x4 + x3 + x + 1 = (2x3 + 2x + 1)(x + 2) + (x2 + 2x + 2)
2x3 + 2x + 1 = (x2 + 2x + 2)(2x + 2)
So, up to a constant multiple, the gcd is the last nonzero remainder, which is x2 +2x+2.
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3. Show that for any finite field GF (q) and any nonzero element α ∈ GF (q), αq−1 = 1.
solution:The proof is almost identical to the proof of Fermat’s little Theorem.
Write GF (q) \ {0} = {a1 , a2 , . . . , aq−1 }. Since GF (q) is a field, the product of two
nonzero elements is nonzero. It follows that the elements {αa1 , αa2 , . . . , αaq−1 } are
distinct and thus exhaust all of GF (q). Thus,
a1 a2 · · · aq−1 = (a1 α)(a2 α) · · · (aq−1 α) = (a1 a2 · · · aq−1 )αq−1
Since a1 a2 · · · aq−1 6= 0, it has an inverse and thus αq−1 = 1.
Note: it follows that for all α ∈ GF (q), αq = α.
4. A primitive element of a finite field GF (q) is an element α ∈ GF (q) such that
{α, α2 , . . . , αq−1 } = GF (q) \ {0}. It is a fact that every finite field has a primitive
element.
Find all primitive elements of each of the fields Z3 , GF (4) and Z7 .
solution:
Clearly, 0 is never a primitive element, and 1 is primitive element of GF (q) iff q = 2.
Z3 : {2, 22 } = {1, 2} = Z3 \ {0}. So, 2 is the only primitive element of Z3 .
GF (4) : {a, a2 , a3 } = {a, b, 1} = GF (4) \ {0}, ; {b, b2 , b3 } = {b, a, 1} = F3 \ {0}. So a
and b are the primitive elements of GF (4).
Z7 : 2 is not primitive since 23 = 1; then 43 = (22 )3 = (23 )2 = 1, and so 4 is not
primitive; 6 is not primitive since 62 = 1;
{3, 32 , 33 , 34 , 35 , 36 } = {3, 2, 6, 4, 5, 1} = Z7 \ {0}
{5, 52 , 53 , 54 , 55 , 56 } = {5, 4, 6, 2, 3, 1} = Z7 \ {0}
So, 3 and 5 are the only primitive elements of Z7 .
5. Recall from class that the polynomials f (x) = x3 + x + 1 and g(x) = x3 + x2 + 1 are
both irreducible over Z2 . So, both Rf := Z2 [x]/f (x) and Rg := Z2 [x]/g(x) are fields.
(a) Find |Rf | and |Rg |.
(b) Write out the addition and multiplication tables for both fields Rf and Rg .
(c) Show that these two fields are isomorphic. That is, find a bijection k : Rf → Rg
such that for all α, β ∈ Rf such that k(α + β) = k(α) + k(β) and k(α · β) =
k(α) · k(β).
solution:
a. From class we know |Rf | = q deg(f ) = 23 = 8. Similarly, |Rg | = q deg(f ) = 23 = 8.
b. Rf = Z2 [x]/f (x):
2
+
0
1
x
x2
1+x
1 + x2
x + x2
1 + x + x2
0
0
1
x
x2
1+x
1 + x2
x + x2
1 + x + x2
·
0
1
x
x2
1+x
1 + x2
x + x2
1 + x + x2
0
0
0
0
0
0
0
0
0
1
1
0
1+x
1 + x2
x
x2
1 + x + x2
x + x2
1
0
1
x
x2
1+x
1 + x2
x + x2
1 + x + x2
x
x
1+x
0
x + x2
1
1 + x + x2
x2
1 + x2
x
0
x
x2
1+x
x + x2
1
1 + x + x2
1 + x2
x2
x2
1 + x2
x + x2
0
1 + x + x2
1
x
1+x
x2
0
x2
1+x
x + x2
1 + x + x2
x
1 + x2
1
1+x
1+x
x
1
1 + x + x2
0
x + x2
1 + x2
x2
1+x
0
1+x
x + x2
1 + x + x2
1 + x2
x2
1
x
1 + x2
1 + x2
x2
1 + x + x2
1
x + x2
0
1+x
x
1 + x2
0
1 + x2
1
x
x2
1 + x + x2
1+x
x + x2
x + x2
x + x2
1 + x + x2
x2
x
1 + x2
1+x
0
1
1 + x + x2
1 + x + x2
x + x2
1 + x2
1+x
x2
x
1
0
x + x2
0
x + x2
1 + x + x2
1 + x2
1
1+x
x
x2
1 + x + x2
0
1 + x + x2
1 + x2
1
x
x + x2
x2
1+x
1 + x2
x + x2
0
0
1 + x2
x + x2
x2 + x + 1 1
1+x
x
x
1 + x + x2
2
x+x
x2
2
x
1+x
1
1 + x2
1 + x + x2
0
1 + x + x2
1+x
x + x2
x2
1
1 + x2
x
Rg = Z2 [x]/g(x): the addition table is the same as above.
·
0
1
x
x2
1+x
1 + x2
x + x2
1 + x + x2
0
0
0
0
0
0
0
0
0
1
0
1
x
x2
1+x
1 + x2
x + x2
1 + x + x2
x
0
x
x2
1 + x2
x + x2
1 + x + x2
1
1+x
x2
0
x2
1 + x2
1 + x + x2
1
1+x
x
x + x2
1+x
0
1+x
x + x2
1
1 + x2
x
1 + x + x2
x2
Note: The tables are symmetric since addition and multiplication are commutative. This reduces
the effort needed to compute the tables. Also, as discussed in class, the multiplication table can be
simplified by representing the nonzero elements as powers of a primitive element.
c. x is a primitive element of Rf ; to see this, use the 2nd column of the multiplication table to find:
{x, x2 , x3 , x4 , x5 , x6 , x7 } = {x, x2 , 1 + x, x + x2 , 1 + x + x2 , 1 + x2 , 1} = Rf \ {0}
x is also a primitive element of Rg ; so a bijection between powers of x in Rf to corresponding powers
of x in Rg will preserve multiplication, but it does not preserve addition.
However, y = x3 = x2 + 1 is a primitive element of Rg : use the 6th column of the multiplication table
to find:
{y, y 2 , y 3 , y 4 , y 5 , y 6 , y 7 } = {1 + x2 , x + x2 , x2 , 1 + x, x2 , 1 + x + x2 , 1} = Rg \ {0}
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And the bijection between powers of x in Rf and corresponding powers of y in Rg will preserve
multiplication and addition: in terms of x:
k(x) = 1+x2 , k(x2 ) = x+x2 , k(1+x) = x2 , k(x+x2 ) = 1+x, k(1+x+x2 ) = x, k(1+x2 ) = 1+x+x2 , k(1) = 1
To verify that this bijection actually preserves multiplication: transform the addition table above for
Rf by applying k to each element of the table. This yields:
+
0
1
1 + x2
x + x2
x2
1 + x + x2
1+x
x
0
0
1
1 + x2
x + x2
x2
1 + x + x2
1+x
x
1
1
0
x2
1 + x + x2
1 + x2
x + x2
x
1+x
1 + x2
1 + x2
x2
0
1+x
1
x
x + x2
1 + x + x2
x + x2
x + x2
1 + x + x2
1+x
0
x
1
1 + x2
x2
x2
x2
1 + x2
1
x
0
1+x
1 + x + x2
x + x2
1 + x + x2
1 + x + x2
x + x2
x
1
1+x
0
x2
1 + x2
1+x
1+x
x
x + x2
1 + x2
1 + x + x2
x2
0
1
which is exactly the addition table for Rg (which is the same as the addition table for Rf ) except
that the rows and columns have been permuted.
6. Find all primitive elements of GF (8). Use the representation: GF (8) = Z2 [x]/(x3 + x + 1).
solution:
Using the multiplication table from problem 5, we get that for each α ∈ GF (8) \ {0, 1}, the sequence
of powers {α, . . . , α7 } is
α
x
x2
1+x
1 + x2
x + x2
1 + x + x2
α2
x2
x + x2
1 + x2
1 + x + x2
x
1+x
α3
1+x
1 + x2
x2
x + x2
1 + x + x2
x
α4
x + x2
x
1 + x + x2
1+x
x2
1 + x2
α5
1 + x + x2
1+x
x
x2
1 + x2
x + x2
α6
1 + x2
1+x+x∗2
x + x2
x
1+x
x2
α7
1
1
1
1
1
1
Since each row exhausts GF (8) \ {0, 1}, all elements of GF (8) \ {0, 1} are primitive.
Alternative proof: Suppose that some α ∈ GF (8) \ {0, 1} were not primitive. Then for some 1 ≤ i <
j ≤ 7, αi = αj and so αj−i = 1.
Then H = {α, α2 , . . . , αj−i = 1} is a subgroup of G = GF (8) \ {0}. Thus |H| = j − i divides |G| = 7,
which is prime. So, j − i = 1 or 7. But j − i 6= 1 since α 6= 1 and j − i 6= 7 by the conditions on j and
i above. This is a contradiction.
7. Let {a1 , . . . , a7 } denote the 7 non-zero elements of GF (8). Let C be the linear code over GF (8)
defined by the parity check matrix


1 1 1 1 1 1 1
H =  a1 a2 a3 a4 a5 a6 a7 
a21 a22 a23 a24 a25 a26 a27
(a) Find the dimension and minimum distance of C.
(b) How many cosets does C have?
(c) Find 51 coset leaders and their corresponding syndromes.
4
x
x
1+x
1 + x + x2
x2
x + x2
1 + x2
1
0
a. This is a Reed-Solomon code. Recall from class that such a code is an [n, n − r, r + 1]-linear code,
where n is the length and r is the number of rows. Recall that this follows from the Vandermonde
structure. So, in this case, the code is a [7, 4, 4]-linear code. In particular, its dimension is 4 and its
minimum distance is 4.
b. Recall that the number of cosets for a k-dimensional linear code of length n over GF (q) is q n /q k =
q n−k which in this case is 83 = 512.
c. Since d(C) = 4 ≥ 3, it can correct a single error and thus every vector of weight at most 1 must be a
coset leader (and they are all in different cosets). Since the code is of length 7 and there are 7 non-zero
symbols, there are 72 = 49 vectors over F8 of weight 1. Together with the zero vector, this accounts
for 50 coset leaders. The syndromes of these cosets are the vectors of length 3 which are either the
zero vector or a non-zero scalar multiple of a single column. More specifically, for b ∈ F8 \ {0}, let
ei,b be the vector with entry b in position i an all other entries 0. Then ei,b is a coset leader with
syndrome [b, bai , ba2i ].
Among the remaining syndromes are all vectors of length three that have at least one zero entry (and
not all zeros). The vector [1, 1, 0, 0, 0, 0, 0] has syndrome x = [1 + 1, a1 + a2 , a21 + a22 ]. Each element
of F8 is its own additive inverse (the main diagonal of the addition table is all zeros). Thus, the first
entry is 0 and the second entry is nonzero (otherwise, a2 would be the additive inverse of a1 ). Thus,
[1, 1, 0, 0, 0, 0, 0] can be chosen as a coset leader and has syndrome [0, a1 + a2 , a21 + a22 ].
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