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Problem 1. (Euclidean division algorithm).
(a) For the set F[x] of polynomials over any field F, show that the distributive law
holds: (f1(x) + f2(x))h(x) = f1(x)h(x) + f2(x)h(x).
Using the associative and commutative laws and the rules of polynomial arithmetic,
we have
and
Finally, (f1i + f2i)hj = f1ihj + f2ihj by the distributive law over F.
(b) Use the distributive law to show that for any given f(x) and g(x) in F[x], there is a
unique q(x) and r(x) with deg r(x) < deg g(x) such that f(x) = q(x)g(x) + r(x).
Suppose that f(x) can be written in two ways:
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where deg r(x) < deg g(x) and deg r’ (x) < deg g(x). Using the distributive law, we have
(q(x) − q’(x))g(x)+(r(x) − r’ (x)) = 0. (1)
If q(x) = q’(x), then (q(x) − q’ (x))g(x) = 0, so (1) implies r(x) = r’(x). If q(x) ≠ q’(x),
then (q(x) − q’ (x))g(x) ≠ 0 and has degree ≥ deg g(x), whereas r(x) − r’ (x) has degree
< deg g(x), so (1) cannot hold. Thus the quotient q(x) and remainder r(x) are unique.
Problem 2. (unique factorization of the integers).
Following the proof of Theorem 7.7, prove unique factorization for the integers Z
We follow the statement and proof of Theorem 7.7, replacing statements about
polynomials by corresponding statements about integers:
Theorem 7.0 (Unique factorization of integers) Every positive integer n ∈ Z with
magnitude |n| ≥ 2 may be written in the form

where each pi, 1 ≤ i ≤ k, is a prime integer. This factorization is unique, up to the order of
the factors.
Proof. If n is a prime, then n = n is the desired unique factorization. If n is not a prime,
then n can be factored into the product ab of two nontrivial factors each less than n, which
in turn can be factored, and so forth. Since magnitudes decrease with each factorization,
this process can only terminate in a prime factorization.
Now we need to prove uniqueness. Thus assume hypothetically that the theorem is false
and let n be the smallest integer that has more than one such factorization,
n = a1 ··· ak = b1 ··· bj ; j, k ≥ 1, (2)
where a1,... , ak and b1,... , bj are prime integers. We will show that this implies an integer
n smaller than n with non-unique factorization, and this contradiction will prove the
theorem. Now a1 cannot appear on the right side of (2), else it could be factored out for
an immediate contradiction. Similarly, b1 cannot appear on the left.

Without loss of generality, assume b1 ≤ a1. By the Euclidean division algorithm, a1 = qb1
+ r. Since a1 is prime, r ≠0 and 0 < r < b1 ≤ a1. Now r has a prime factorization r = r1 ··· rn,
where 1 is not a divisor of any of the ri, since it has greater magnitude. Substituting into
(2), we have
(qb1 + r1 ··· rn)a2 ··· ak = b1 ··· bj ,
or, defining n’ = r1 ··· rna2 ··· ak and rearranging terms,
n’= r1 ··· rna2 ··· ak = b1 (b2 ··· bj − qa2 ··· ak).
Now n’ is positive, because it is a product of positive integers; it is less than n, since r <
a1; and it has two different factorizations, with b1 a factor in one but not a divisor of any
of the factors in the other; contradiction.
Problem 3. (finding irreducible polynomials).
(a) Find all prime polynomials in F2[x] of degrees 4 and 5. [Hint: There are three prime
polynomials in F2[x] of degree 4 and six of degree 5.]

We can immediately eliminate all polynomials which have the degree-1 factor x (i.e.,
whose constant term is 0) or the degree-1 factor x + 1 (i.e., which have an even number
of nonzero coefficients). This eliminates ¾ of the candidate polynomials. We then need to
sieve out only multiples of the degree-2 prime polynomial x2 +x +1 and the two degree-3
prime polynomials, x3 + x + 1 and its reverse, x3 + x2 + 1.
This leaves four degree-4 polynomials. One of these is (x2 + x + 1)2 = x4 + x2 + 1. The
remaining three are prime:
x4 + x + 1, x4 + x3 + 1, x4 + x3 + x2 + x + 1.
Similarly, this leaves eight degree-5 polynomials. Two of these are multiples of the
degree2 prime polynomial with one of the two degree-3 prime polynomials, namely (x2 +
x + 1) (x3 + x + 1) = x5 + x4 + 1 and its reverse, x + x + 1. The remaining six are prime:
x 5+ x2 + 1, x5 + x3 + x2 + x + 1, x5 + x4 + x2 + x + 1,
and their reverses x5 + x3 + 1, x5+ x4 + x3 + x2 + 1, x5 + x4 + x3 + x + 1.

(b) Show that x16 + x factors into the product of the prime polynomials whose degrees
divide 4, and x32 + x factors into the product of the prime polynomials whose degrees
divide 5.
The prime polynomials whose degrees divide 4 are x, x + 1, x2 + x + 1 and the three
degree-4 prime polynomials above. Straightforward polynomial multiplication shows
that
x(x + 1)(x2 + x + 1)(x4 + x3 + 1)(x4 + x3 + x2 + x + 1)(x4 + x + 1) = x16 + x.
(Note that (x2+x+1) (x4+x3+1) = x10+x5+1 and (x+1)(x4+x3+x2+x+1) = x5 + 1.)
Similarly, the prime polynomials whose degrees divide 5 are x, x + 1 and the six
degree-5 prime polynomials above. Again, straightforward polynomial multiplication
shows that their product is x32 + x.
Problem 4. (The nonzero elements of Fg(x) form an abelian group under
multiplication).

Let g(x) be a prime polynomial of degree m, and r(x), s(x), t(x) polynomials in Fg(x). (a)
Prove the distributive law, i.e., (r(x) + s(x)) ∗ t(x) = r(x) ∗ t(x) + s(x) ∗ t(x). [Hint:
Express each product as a remainder using the Euclidean division algorithm.]
By the distributive law for ordinary polynomials, we have
(r(x) + s(x))t(x) = r(x)t(x) + s(x)t(x).
Following the hint, write r(x)t(x) = q1(x)g(x) + r1(x), s(x)t(x) = q2(x)g(x) + r2(x), and (r(x)
+ s(x))t(x) = q3(x)g(x) + r3(x), where deg ri(x) < deg g(x) for i = 1, 2, 3. Then r(x) ∗ t(x) =
r1(x), s(x) ∗ t(x) = r2(x), and (r(x) + s(x)) ∗ t(x) = r3(x). Now from the equation above,
q3(x)g(x) + r3(x) = q1(x)g(x) + r1(x) + q2(x)g(x) + r2(x).
which implies
0=(q3(x) − q1(x) − q2(x))g(x)+(r3(x) − r1(x) − r2(x))
Since 0 = 0q(x) + 0 and such a decomposition is unique, we have r3(x) = r1(x) + r2(x). (b)
For r(x) ≠ 0, show that r(x) ∗ s(x) ≠ r(x) ∗ t(x) if s(x) ≠ t(x).

The equation r(x) ∗ s(x) = r(x) ∗ t(x) implies r(x) ∗ (s(x) − t(x)) = 0; but since g(x) is
irreducible, this implies either r(x)=0 or s(x) = t(x).
(c) For r(x) ≠ 0, show that as s(x) runs through all nonzero polynomials in Fg(x), the
product r(x) ∗ s(x) also runs through all nonzero polynomials in Fg(x).
By part (b), the products r(x) ∗ s(x) are all nonzero and are all distinct as as s(x) runs m
through the |F|m − 1 nonzero polynomials in Fg(x), so they must be all of the |F|m − 1
nonzero polynomials in Fg(x).
(d) Show from this that r(x) ≠ 0 has a mod-g(x) multiplicative inverse in Fg(x); i.e., that
r(x) ∗ s(x)=1 for some s(x) ∈ Fg(x).
By part (c), the products r(x) ∗ s(x) include every nonzero polynomial in Fg(x), including
1. Therefore, given r(x) ≠ 0 ∈ Fg(x), there exists a unique s(x) ≠ 0 ∈ Fg(x) such that r(x) ∗
s(x) = 1; i.e., such that s(x) is the multiplicative inverse of r(x) in Fg(x).
Since the multiplication operation ∗ is associative and commutative and has identity 1, it
follows that the nonzero elements of Fg(x) form an abelian group under multiplication.

Problem 5. (Construction of F32)
(a) Using an irreducible polynomial of degree 5 (see Problem 5.3), construct a finite
field F32 with 32 elements.
We can construct F32 using any of the 6 irreducible polynomials of degree 5 found in
Problem 5.3. Using g(x) = x5 + x2 + 1, the field F32 is defined as the set of all 32 binary
polynomials of degree 4 or less under polynomial arithmetic modulo g(x).
(b) Show that addition in F32 can be performed by vector addition of 5-tuples over F2.
The sum of two polynomials of degree 4 or less is obtained by a componentwise sum of
their coefficients, whether modulo g(x) or not.
(c) Find a primitive element α ∈ F32. Express every nonzero element of F32 as a distinct
power of α. Show how to perform multiplication and division of nonzero elements in F32
using this “log table.”

The set F*32 of nonzero elements of F32 is the set of roots of the equation x31 = 1 in F32;
i.e., every β ∈ F∗
32 satisfies β31 = 1, so the multiplicative order of every element 32 must
divide 31, which is prime. There is one element of multiplicative order 1, namely 1. The
remaining 30 elements must therefore have multiplicative order 31; i.e., there are 30 ∗
primitive elements in F*32. Therefore α = x must be primitive. We compute its powers,
reducing x5 to x2 + 1 as necessary:
α = x,
α2 = x2 ,
α3 = x3,
α4 = x4 ,
α5 = x2 + 1,
α6 = x3 + x,
α7 = x4+ x2,
α8 = x3 + x2 + 1,
α9 = x4 + x3 + x,
α10 = x4 + 1,
α 11= x2 + x + 1,
α12 = x3 + x2 + x,
α13 = x4 + x3+ x2 ,

α14 = x4 + x3 + x2 + 1,
α15 = x4 +x3 + x2 +x + 1,
α16 = x4+x3+x+1,
α17 = x4+x+1,
α18 = x+1,
α19 = x2+x,
α20 = x3+x2,
α21 = x4+x3,
α22 = x4+x2+1
α23 = x3+x2+x+1
α24 = x4+x3+x2+x,
α25 = x4+x2+x+1,
α26 = x4+x2+x+1
α27 = x3+x+1,
α28 = x4+x2+x
α30 = x4+x,
α31 = 1.
The product of αi and αj is αi+j . The quotient of αi divided by αj is αi−j . In both cases the
exponents are computed modulo 31, since α31 = 1.

(d) Discuss the rules for multiplication and division in F32 when one of the field elements
involved is the zero element, 0 ∈ F32.
The product of 0 with any field element is 0. Division by 0 is not defined; i.e., it is illegal
(as with the real or complex field).
Problem 6. (Second nonzero weight of an MDS code)
Show that the number of codewords of weight d + 1 in an (n, k, d) linear MDS code over
Fq is
where the first term in parentheses represents the number of codewords with weight ≥ d
in any subset of d + 1 coordinates, and the second term represents the number of
codewords with weight equal to d.

Consider any subset of d + 1 = n − k + 2 coordinates. Take two of these coordinates and
combine them with the remaining k − 2 coordinates to form an information set. Fix the
components in the k − 2 coordinates to zero, and let the remaining two coordinates run
freely through Fq. These q2 information set combinations must correspond to q2
codewords. (In fact, we may view this subset of codewords as a shortened (d + 1, 2, d)
MDS code.)
One of these codewords must be the all-zero codeword, since the code is linear. The
remaining q2 − 1 codewords must have weight d or d + 1. Since there are q − 1
codewords of weight d with support in any subset of d coordinate positions, the
number of codewords of weight d whose support is in any subset of d + 1 coordinate
positions is (d+1
d) (q-1) (the number of codewords of weight d in a (d + 1, 2, d) MDS
code). So the number of codewords of weight d + 1 in any d + 1 coordinate positions
is
Since there are (n
d+1) distinct subsets of d + 1 coordinate positions, the given expression
for Nd+1 follows.

Note that
This implies that if n>d, then d ≤ q, since otherwise Nd+1 would become negative. In
other words, there exists no (n, k, d = n − k + 1) MDS code over Fq with q < d < n. For
example, there exist no binary MDS codes other than the (n, n, 1),(n, n − 1, 2) and (n, 1,
n) codes (and (n, 0,∞), if you like). More generally, when n ≥ q + 2, there exist forbidden
values of d, namely q + 1 ≤ d ≤ n − 1.
Similarly, by considering shortened codes of lengths d + 2, d + 3,... , n, we can compute
Nd+2, Nd+3,... , Nn.
Problem 7. (Nd and Nd+1 for certain MDS codes)
(a) Compute the number of codewords of weights 2 and 3 in an (n, n − 1, 2) SPC code
over F2.
We n have N2 = (q − 1) (n
2) = (n
2) ; i.e., there is a weight-2 codeword for every
coordinate pair. Then N3 = (q − d)(q − 1) (n
3) = 0. This is consistent with the definition
of an SPC code as the set of all even-weight n-tuples.

(b) Compute the number of codewords of weights 2 and 3 in an (n, n − 1, 2) linear code
over F3.
Here we have N2 = (q − 1) (n
2) = 2(n
2) ; i.e., there are two weight-2 codewords for every
coordinate pair. Then N3 = (q − d)(q − 1) (n
3) =2(n
3).
For example, the (3, 2, 2) RS code over F3 has generators (111, 012) and codewords
{000, 111, 222, 012, 120, 201, 021, 102, 210}, with N2 = 6 and N3 = 2. Thus in general a
zero-sum code over a field larger than F2 has odd-weight codewords.
(c) Compute the number of codewords of weights 3 and 4 in a (4, 2, 3) linear code over
F3.
Here we have N3 = (q − 1) (4
3) = 2(4) = 8, so all non-zero codewords have weight 3.
(Verification: N4 = 0 because q = d.) The (4, 2, 3) linear code over F3 given in the
introduction of Chapter 8 (or in the next problem) is an example of such a code, called a
“tetracode.”
Problem 8. (“Doubly” extended RS codes)

(a) Consider the following mapping from (Fq)k to (Fq)q+1. Let (f0, f1,... , fk−1) be any k-
tuple over Fq, and define the polynomial f(z) = f0 + f1z + ··· + fk1 zk−1 of degree less than
k. Map (f0, f1,... , fk−1) to the (q + 1)-tuple ({f(βj ), βj ∈ Fq}, fk−1)— i.e., to the
RS codeword corresponding to f(z), plus an additional component equal to fk−1.
Show that the qk (q + 1)-tuples generated by this mapping as the polynomial f(z) ranges
over all qk polynomials over Fq of degree less than k form a linear (n = q + 1, k, d = n −
k + 1) MDS code over Fq. [Hint: f(z) has degree less than k − 1 if and only if k−1 = 0.]
The code evidently has length n = q + 1. It is linear because the sum of codewords
corresponding to f(z) and f’ (z) is the codeword corresponding to f(z) + f’ (z), another
polynomial of degree less than k. Its dimension is k because no polynomial other than
the zero polynomial maps to the zero (q + 1)-tuple.
To prove that the minimum weight of any nonzero codeword is d = n − k + 1, use the
hint and consider the two possible cases for fk−1:
• If fk−1 ≠ 0, then deg f(z) = k − 1. By the fundamental theorem of algebra, the RS
codeword corresponding to f(z) has at most k − 1 zeroes. Moreover, the fk−1 component
is nonzero. Thus the number of nonzero components in the code (q + 1)- tuple is at least
q − (k − 1) + 1 = n − k + 1.

If fk−1 = 0 and f(z) ≠ 0, then deg f(x) ≤ k − 2. By the fundamental theorem of algebra, the
RS codeword corresponding to f(z) has at most k − 2 zeroes, so the number of nonzero
components in the code (q+1)-tuple is at least q−(k−2) = n−k+1.
(b) Construct a (4, 2, 3) linear code over F3. Verify that all nonzero words have weight 3.
The generators of an extended RS (4, 2, 3) “tetracode” over F3 are (1110, 0121), and the
code is {0000, 1110, 2220, 0121, 1201, 2011, 0212, 1022, 2102}, with N3 = 8 and N4 = 0
(as shown in Problem 5.7(c); compare the zero-sum (3, 2, 2) code of Problem 5.7(b)).

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