Determine Aut(K/Q): Calculating Automorphism Group of K=Q(21/4)

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SUMMARY

The discussion focuses on determining the automorphism group Aut(K/Q) for the field K = Q(21/4). It is established that there are four homomorphisms from K to C, which correspond to the images of the fourth root of 2, specifically 4√2, -4√2, i*4√2, and -i*4√2. The automorphisms are defined as isomorphisms from the field K to itself that fix the base field Q. The key takeaway is that Aut(K/Q) consists of these four mappings, which are completely determined by the image of 4√2.

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Homework Statement



Let K = Q(21/4)

Determine the automorphism group Aut(K/Q)


Homework Equations



An automorphism is an isomorphism from a Field to itself

Aut(K/Q) is the group of Automorphisms from k/Q to K/Q

Definition: A K-Homomorphism from L/K to L'/K is a homomorphism L---> L' that is the identity on K

The Attempt at a Solution



I am completely at a loss really. I have calculated there are four homomorphisms from K to C and think from there if I know how many are K-homomorphisms then that'll be the number of automorphisms, because a homomorphism from a field to itself is an automorphism (Please correct me if I'm wrong on this). Then that'll give me the set of Automorphisms.

My problem is that I don't know how to go from the number of homomorphisms to the actual homomorphisms. I think it has a relation to the roots of 2(1 /4) in C (which I have calculated to be 2(1 /4), - 2(1 /4), i*2(1 /4), -i2(1 /4) )

Please help, this lack of understanding is preventing me from moving forward with other questions and my notes from lectures completely gloss over how to do this.
 
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one problem that I see is that not every homomorphism is an automorphism. Consider the identity mapping. this is clearly a homomorphism but it is not bijective.
Also you said there are four homomorphisms from K to C. But what is C? is C just Aut(K/Q)?
Also I just want to clarify something: Is this what you mean by K:

K= \{ a+b \sqrt[4]{2} :a,b \in \mathbb{Q} \}
 
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no C is not Aut(K/Q).

Aut(K/Q) is the group of all automorphisms of K that fix Q.

note K is NOT {a+b(4√2)}, this set is not even closed under multiplication, K is more precisely the set:

{a + b(4√2) + c√2 + d(4√8)}

which as a vector space is isomorphic to Q4.

what IS clear, is that any element of Aut(K/Q) sends 4√2 to another 4th root of two, and that any such automorphism is completely determined by the image of 4√2.

there are exactly 4 of these in C:

4√2, -4√2, (4√2)i and -(4√2)i.

if we agree to set α = 4√2, of the maps:

α→α
α→-α
α→iα
α→-iα

how many of these are in Aut(K/Q)?

(the complex numbers play only an indrect role, here, as a field contaning the algebraic closure of Q(α). we could have just used this field (call it E) instead, but it doesn't lend itself to a nice easy description like C does. the important fact is that there is SOME subfield of C that is the algebraic closure of Q(α), as Q(α) is a subfield of R, and C is the algebraic closure of R (a fact called the fundamental theorem of algebra). note we don't even need all of C, just A, where A is the set of all (real) algebraic numbers).
 

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