Thread: Automorphisms of finite field of degree 16

since we have a finite field, its multiplicative group is cyclic. what are the automorphisms of Z₁₅? (hint: there are 8 of them).

if you call the generator x+1 + (x⁴ + x + 1), b, then these automorphisms are of the form b→bⁿ, for some n. which n are possible?

EDIT: not all automorphisms of the multiplicative group, will yield field automorphisms. we know that the order of the galois group is the degree of our field of 16 elements over Z₂, namely 4.

so only 4 of the 8 multiplicative automorphisms are additive. can you say which ones (think: frobenius maps)?

that is: for which n do we have (a₁ + a₂b + a₃b² + a₄b³)ⁿ = (a₁)ⁿ + (a₂b)ⁿ + (a₃b²)ⁿ = (a₄b³)ⁿ?

Ah! Yeah, I forgot that the multiplicative group is cyclic...oops, hehehe.
Okay, so for the first part, the automorphisms of the multiplicative group would be all n such that gcd(n,15) = 1, so that we have n = 1,2,4,7,8,11,13,14.

Now, for the second part, you mentioned that the Galois group will be of order 2. That seems to imply that the automorphisms will fix 0 and 1 (elements of Z₂). Will that always be the case in general for finite fields? (i.e. the automorphisms of a finite field of order pⁿ will fix the elements of Z_p).

Continuing, I'm not too familiar with frobenius maps, and solving for the n that satisfy that equality seems tedious ... so, couldn't we take advantage of Lagrange's theorem and simply look for the automorphisms whose order divides 4? So this would give us n = 1,2,4,8. Is there anything wrong with what I did? Would there be cases where it would be difficult to apply my method? Would your way actually be easier/quicker? (could you show me how you would go about doing your way).

Thanks!

no the galois group is of order 4. the galois group has to fix 0 and 1, that goes without saying, since 0 and 1 MUST be preserved by any field automorphism (and thus 1+1+1+..+1 as well). the field generated by 1 is often called the prime subfield of a given field K (and is isomorphic to either Z_p, or Q).

an automorphism of a field K must preserve the prime subfield (think about why this HAS to be true).

for a prime p, the frobenius map is simply: a→a^p. in an extension of Z_p, this is an automorphsim (why...does the binomial theorem ring a bell?).

your answer IS correct....n = 1,2,4 and 8 all yield automorphisms, and we know we have 4, so that's all of them.

recall that Aut(Z₁₅) ≅ U(15) ≅ U(3) x U(5). clearly {1,2,4,8} is the cyclic subgroup of U(15) isomorphic to U(5). so to describe U(15) completely, we need to find an element of order 2 not in {1,2,4,8}. any one would do, but 7 is the smallest, so we can use that: U(15) = {1,7} x {1,2,4,8} (here, we have an INTERNAL direct product).

so to show that b→b⁷ isn't additive, find two elements x,y of F₁₆ for which (x+y)⁷ ≠ x⁷ + y⁷ (i suggest using b and b+1).

b→b⁷ isn't additive then neither is b→b¹⁴, which is the composition of b→b² and b→b⁷,

or b→b¹³, which is the composition of b→b⁴ and b→b⁷,

or b→b¹¹, which is the composition of b→b⁸ and b→b⁷.

(do you see how the group structure of U(15) plays out in the maps here?)