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Math Help - Automorphisms of finite field of degree 16

  1. #1
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    Automorphisms of finite field of degree 16

    Hello, this was a problem on an exam. The problem was to generate a field of order 16, and list the elements as powers of a generator (i.e. find an element that would generate the field), and then to list the automorphisms of that field.

    I got the first parts by noting that 16 = 2^4, so I took quotient field of the polynomial ring Z_2 \[ x \] over x^4+x+1 (an irreducible polynomial in Z_2 \[ x \] ). The generator would be a polynomial that is relatively prime to x^4+x+1 ( I can't remember what exactly I used, but I think it was x+1). I'm okay with that part.

    I had problems getting the automorphisms of the field. There's the obvious identity mapping, but I'm not sure about the others. After doing some reading, I know that if we treat each element as a function f(x), then the map that sends f(x) to f(x+1) will be another automorphism. I believe these will be all the automorphisms that fixes the constants (not sure).

    In a nutshell, my problems are: 1) Would the automorphisms necessarily be K-automorphisms? (K is some field) If yes, why, and what would K be?; and 2) What are the automorphisms?

    Thanks!
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  2. #2
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    Re: 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?
    Last edited by Deveno; April 17th 2012 at 07:48 AM.
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    Re: Automorphisms of finite field of degree 16

    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 Zp).

    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!
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    Re: Automorphisms of finite field of degree 16

    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 Zp, 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→ap. in an extension of Zp, 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?)
    Thanks from Bingk
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