lpetrich
Science Advisor
- 998
- 180
I'll take on problem 1, though I'm only somewhat familiar with its subject matter.
(a) This is presumably for finding the primitive polynomials in GF8. These are cubic polynomials with coefficients in GF2 that cannot be expressed as products of corresponding polynomials for GF4 and GF2. I will use x as an undetermined variable here.
For GF2, the primitive polynomials are x + (0,1) = x, x+1.
For GF4, we consider primitive-polynomial candidates x2 + (0,1)*x + (0,1): x2, x2 + 1 = (x + 1)2, x2 + x = x(x + 1), x2 + x + 1. That last one is the only primitive polynomial for GF4.
For GF8, we consider primitive-polynomial candidates x3 + (0,1)*x2 + (0,1)*x + (0,1): x3, x3 + 1 = (x2 + x + 1)*(x + 1), x3 + x = x * (x + 1)2, x3 + x + 1, x3 + x2 = x2 * (x + 1), x3 + x2 + 1, x3 + x2 + x = x*(x2 + x + 1), x3 + x2 + x + 1 = (x + 1)3.
Thus, GF8 has primitive polynomials x3 + x + 1 and x3 + x2 + 1.
(b) There is a problem here. A basis is easy to define for addition: {1, x, x2} where multiplication uses the remainder from dividing by a primitive polynomial. The additive group is thus (Z2)3. The multiplicative group is, however, Z7, and it omits 0. That group has no nontrivial subgroups, so it's hard to identify a basis for it.
(c) That is a consequence of every finite field GF(pn) being a subfield of an infinite number of finite fields GF(pm*n), each one with a nonzero number of primitive polynomials with coefficients in GF(pn). Since each field's primitive polynomials cannot be be factored into its subfields' ones, each field adds some polynomial roots, and thus, there are an infinite number of such roots.
I will now try to show that every finite field has a nonzero number of primitive polynomials with respect to some subfield. First, itself: for all elements a of F relative to F, (x - a) is primitive. Thus, F has N primitive polynomials. For GF(pm*n) relative to GF(pn), I will call the number N(m). One can count all the possible candidate polynomials for GF(pm*n), and one gets
$$ \sum_{\sum_k k m_k = r} \prod_k P(N(k),m_k) = N^r $$
If N is a prime, then the solution is known:
$$ N(m) = \frac{1}{m} \sum_{d|m} N^{m/d} \mu(d) $$
where the μ is the Moebius mu function, (-1)^(number of distinct primes if square-free), and 0 otherwise. I don't know if that is correct for a power of a prime.
For GF2, the primitive polynomials are x + (0,1) = x, x+1.
For GF4, we consider primitive-polynomial candidates x2 + (0,1)*x + (0,1): x2, x2 + 1 = (x + 1)2, x2 + x = x(x + 1), x2 + x + 1. That last one is the only primitive polynomial for GF4.
For GF8, we consider primitive-polynomial candidates x3 + (0,1)*x2 + (0,1)*x + (0,1): x3, x3 + 1 = (x2 + x + 1)*(x + 1), x3 + x = x * (x + 1)2, x3 + x + 1, x3 + x2 = x2 * (x + 1), x3 + x2 + 1, x3 + x2 + x = x*(x2 + x + 1), x3 + x2 + x + 1 = (x + 1)3.
Thus, GF8 has primitive polynomials x3 + x + 1 and x3 + x2 + 1.
(b) There is a problem here. A basis is easy to define for addition: {1, x, x2} where multiplication uses the remainder from dividing by a primitive polynomial. The additive group is thus (Z2)3. The multiplicative group is, however, Z7, and it omits 0. That group has no nontrivial subgroups, so it's hard to identify a basis for it.
(c) That is a consequence of every finite field GF(pn) being a subfield of an infinite number of finite fields GF(pm*n), each one with a nonzero number of primitive polynomials with coefficients in GF(pn). Since each field's primitive polynomials cannot be be factored into its subfields' ones, each field adds some polynomial roots, and thus, there are an infinite number of such roots.
I will now try to show that every finite field has a nonzero number of primitive polynomials with respect to some subfield. First, itself: for all elements a of F relative to F, (x - a) is primitive. Thus, F has N primitive polynomials. For GF(pm*n) relative to GF(pn), I will call the number N(m). One can count all the possible candidate polynomials for GF(pm*n), and one gets
$$ \sum_{\sum_k k m_k = r} \prod_k P(N(k),m_k) = N^r $$
If N is a prime, then the solution is known:
$$ N(m) = \frac{1}{m} \sum_{d|m} N^{m/d} \mu(d) $$
where the μ is the Moebius mu function, (-1)^(number of distinct primes if square-free), and 0 otherwise. I don't know if that is correct for a power of a prime.