# What is the automorphism group of the field K=Q(21/4)?

• wattsup03
In summary, the automorphism group Aut(K/Q) is the set of all homomorphisms from K to itself that are the identity on Q, which is the same as the set of all automorphisms from K to itself. Since a homomorphism from a field to itself is an automorphism, and there are four homomorphisms from K to C, we know that there are four automorphisms in Aut(K/Q). These automorphisms are determined by their values on the fourth roots of 2, 1, and 2^(1/2), and must take conjugate pairs to conjugate pairs. Thus, the automorphism group Aut(K/Q) is a group of four elements, determined by the
wattsup03

## 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.

$(2^{1/4}^2= 2^{1/2}$ and $(2^{1/4}^3=2^{3/4}$ are irrational but $(2^{1/4})^4= 2$ is rational so any number in [/itex]Q(2^{1/4})[/itex] is of the form $a+ b2^{1/4}+ c2^{1/2}+ d2^{3/4}$ for rational numbers a, b, c, d. For any automorphism, f, $f(a+ b2^{1/4}+ c2^{1/2}+ d2^{3/4})= a + bf(2^{1/4})+ cf(2^{1/2})+ df(2^{3/4})$ so the possible values of f depend entirely upon the possible values of $f(2^{1/4})$, $f(2^{1/2})$ and $f(2^{3/4})$.

Further, $(f(2^{1/4})^4= f(2)$ is rational so $f(2^{1/4}$ must be a fourth root of 2. Also, $f(2^{1/2})^2= f(2)$ so $f(2^{1/2}) must be [itex]2^{1/2} (or [itex]-2^{1/2}$ which is just a rational number times $2^{1/2}. Each f permutes the fourth roots of 1 while fixing [itex]2^{1/2}$.

it is somewhat misleading to say the values of f depend on the values of f(21/4), f(21/2) and f(23/4).

f is a field automorphism, so (for example) f(23/4) = f((21/4)3)= (f(21/4)3,

so f ONLY depends on the value of f(21/4).

while it is true that f permutes the roots of x4-2, it is not true that any such permutation yields an "f". f takes conjugate pairs to conjugate pairs, as well (the square of a fourth root of 2 must be a square root of 2).

so if one regards the 4 roots of x4-2 as α1234, where:

αj = αe(j-1)πi/4

then (α2 α4) yield a member of Aut(K/Q), but (α2 α3) does not.

## 1. What is the definition of a morphism in Galois Theory?

A morphism in Galois Theory is a function between two fields that preserves the structure of the fields. It is a homomorphism that maps elements of one field to elements of another field while maintaining the field operations of addition, multiplication, and inverses.

## 2. How are morphisms related to Galois extensions?

Morphisms play a crucial role in Galois extensions. They are used to determine whether a field extension is Galois or not. If a morphism between two fields preserves the structure of the fields, then the field extension is considered Galois.

## 3. Can there be multiple morphisms between two fields in Galois Theory?

Yes, there can be multiple morphisms between two fields in Galois Theory. However, there is always a unique morphism called the identity morphism, which maps elements to themselves. Other morphisms may exist depending on the specific fields and their properties.

## 4. How are morphisms used in proving the Fundamental Theorem of Galois Theory?

In the proof of the Fundamental Theorem of Galois Theory, morphisms are used to show that there is a one-to-one correspondence between the intermediate fields of a Galois extension and the subgroups of its Galois group. This correspondence is essential in understanding the structure of Galois extensions.

## 5. Can morphisms be used to determine the solvability of polynomials in Galois Theory?

Yes, morphisms can be used to determine the solvability of polynomials in Galois Theory. This is because the solvability of a polynomial is related to the structure of its Galois group, which can be studied using morphisms. In particular, if a polynomial is solvable by radicals, then its Galois group must be solvable, and this can be determined using morphisms.

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