Download Assignment 4 – Solutions

LINNÆUS UNIVERSITY
Department of Computer Science,
Physics, and Mathematics
Per-Anders Svensson
Assignment 4 – Solutions
Exercise 1. Suppose the polynomial f (x) = x4 + mx + p ∈ Q[x], where p 6= 3
is a prime and m any integer, has no zeros in Q. Show that f (x) is irreducible
over Q.
Solution. Suppose f (x) is reducible over Q. Since f (x) has no zeros in Q, it
must then be that f (x) is a product of two quadratic polynomials
f (x) = (x2 + ax + b)(x2 + cx + d),
where we may assume that a, b, c, d ∈ Z. We obtain
x4 + mx + p = x4 + (a + c)x3 + (b + d + ac)x2 + (ad + bc)x + bd,
and by identifying the coefficients, we get

a+c=0



b + d + ac = 0

ad + bc = m



bd = p.
According to the last equation, bd = p. Since p is a prime, we then must have
b = 1, d = p or b = −1, d = −p (and we could also shift the roles of b and d,
but due to symmetry, it is enough to to consider the two cases mentioned).
If b = 1 and d = p, then


a + c = 0
1 + p + ac = 0


ap + c = m.
Plugging a = −c into the second equation yields
1 + p − c2 = 0 ⇐⇒ p = c2 − 1 = (c + 1)(c − 1).
But p is a prime, so one of the factors in the right hand side must be equal to 1.
This in turn implies that c = 2, and thereby p = c2 − 1 = 22 − 1 = 3. But p 6= 3
according to the premises.
We then look at the case when b = −1 and d = −p. Then


a + c = 0
−1 − p + ac = 0


−ap − c = m.
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Combining the first and second equation yields
p = ac − 1 = −c2 − 1 < 0,
which is not possible (since a prime by definition is positive). This proves that
f (x) cannot be reducible.
Exercise 2. Find all irreducible polynomials of degree 3 in Z2 [x].
Solution. Let ax3 + bx2 + cx + d ∈ Z2 [x] be a polynomial of degree 3. Then we
must have a = 1. For this polynomial to be irreducible we must also have d = 1
since otherwise we will have a polynomial x3 + bx2 + cx = x(x2 + bx + c) that
can be factored an therefore reducible.
So we thus have a polynomial f (x) = x3 + bx2 + cx + 1, where b, c ∈ Z2 [x]
must be chosen so that this polynomial has no zero in Z2 [x]. Hence we may not
have f (0) = 0 or f (1) = 0. Now f (0) = 1 no matter how b and c is chosen, and
f (1) = b + c. Hence we have to choose b and c so that b + c = 1, and there are
two such possibilities: b = 1, c = 0 or b = 0, c = 1. We conclude that there are
two irreducible polynomials of degree 3 in Z2 [x], namely
f1 (x) = x3 + x2 + 1
f2 (x) = x3 + x + 1.
Exercise 3. Let f (x) ∈ Z2 [x] be one of those polynomials you found in the previous Exercise. Write down all elements in the quotient ring F = Z2 [x]/ hf (x)i
and a Cayley table for the multiplication on F .
Solution. We let f (x) = x3 + x + 1. The factor ring F = Z2 [x]/ hf (x)i can be
seen as the set of all possible remainders that can be obtained, if any polynomial
in Z3 [x] is divided by f (x), according to the Division Algorithm. Now since
deg f (x) = 3, the degree of such a remainder can be at most 2, and since its
coefficients belongs to Z2 , there will be 23 = 8 such remainders, namely
0
x2
1
x2 + 1
x
x2 + x
x+1
x2 + x + 1
The binary operations on F are addition and multiplication modulo f (x). Now
x3 + x + 1 ≡ 0 (mod f (x)), which also can be written as
x3 ≡ −x − 1 ≡ x + 1
(mod f (x)).
This means that we may multiply the elements in F as if the were ”ordinary”
polynomials i Z2 [x], but every occurrence of x3 is to be replaced by x + 1. For
instance x2 (x + 1) = 1 in F , since
x2 (x + 1) ≡ x3 + x ≡ (x + 1) + x ≡ 1 (mod f (x)).
Similarly
(x2 + 1)(x2 + x) = x4 + x3 + x2 + x = x3 (x + 1) + x2 + x
= (x + 1)(x + 1) + x2 + x = x2 + 1 + x2 + x = x + 1.
Keeping this in mind, a Cayley table for multiplication will be readily constructed.
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Exercise 4. Study the electronic circuit:
x1
x2
x3
x1
x2
x1
x′2
x′3
x3
x′3
x1
x′2
(a) Represent this electronic circuit by means of a Boolean mapping.
(b) Find an as simple circuit as possible, that is equivalent to the circuit above.
Solution. (a) The circuit is a serial coupling of four parallel coupled circuits,
and can be described by means of the Boolean mapping
f (x1 , x2 , x3 ) = (x1 ∨ x2 ∨ x3 ) ∧ (x2 ∨ x1 ∨ x′3 ) ∧ (x3 ∨ x′2 ∨ x1 ) ∧ (x1 ∨ x′3 ∨ x′2 ).
(b) By using the fact that xi ∧ x′i = 0 for each i and the distributive laws (which
hold in any Boolean algebra), we obtain
f (x1 , x2 , x3 ) = [(x1 ∨ x2 ) ∨ (x3 ∧ x′3 )] ∧ [(x1 ∨ x′2 ) ∨ (x3 ∧ x′3 )]
= [(x1 ∨ x2 ) ∨ 0] ∧ [(x1 ∨ x′2 ) ∨ 0] =
= (x1 ∨ x2 ) ∧ (x1 ∨ x′2 ) = x1 ∨ (x2 ∧ x′2 ) = x1 ∨ 0 = x1 .
Hence the given circuit is equivalent to a circuit that contains one relay only,
namely x1 .
Exercise 5. Let L be a distributive lattice. Show that
(a ∨ b) ∧ c a ∨ (b ∧ c)
for all a, b, c ∈ L.
Solution. In any lattice x y is equivalent to x ∨ y = y. With x = (a ∨ b) ∧ c
and y = a ∨ (b ∧ c) we thus must show that
[(a ∨ b) ∧ c] ∨ [a ∨ (b ∧ c)] = a ∨ (b ∧ c).
We have been told that L is distributive, whence the distributive laws holds
in L. Using them, we obtain
[(a ∨ b) ∧ c] ∨ [a ∨ (b ∧ c)] = [(a ∨ b) ∧ c] ∨ [(a ∨ b) ∧ (a ∨ c)]
= (a ∨ b) ∧ [c ∨ (a ∨ c)]
= (a ∨ b) ∧ (a ∨ c)
= a ∨ (b ∧ c),
as required. Note that we above also have used the fact that ∨ is associative
and commutative, and that all elements in L are idempotent with respect to ∨
(i.e. that c ∨ c = c).
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