# Determining Galois extension based on degree of extension

• PsychonautQQ
In summary, if Char(K) = 0 and [L:K]=2, then L:K is a galois extension because the minimal polynomial of any element in L splits in K[x] and is separable. Additionally, this extension is normal because all roots of the minimal polynomial are in L.
PsychonautQQ

## Homework Statement

If Char(K) = 0 and [L:K]=2, is L:K a galois extension?

## The Attempt at a Solution

My gut is saying yes because if [L:K]=2 then it seems that any polynomial in K[x] with a root in L should split in L[x]. Something about how some hypothetical minimal polynomial of some element m call it m(x) where L[x] is isomorphic to the quotient field K[x]/m(x), then K(u) is isomorphic to L where given the conditions that [L:K] =2 then deg(m(x)) = 2 and so if L has one of the roots of m(x) then that means that m(x) factors linearly because it only has degree 2 and thus all the roots are in L thus the extension is normal. Am I leaving anything important out?

I've thought along the same lines. More formal we can choose an element ##u \in L-K##. Then ##\{1,u\}## is ##K-##linear independent and ##\{1,u,u^2\}## is not. This means we can write ##u^2=\alpha \cdot 1 + \beta \cdot u## which means ##u## is a root of ##x^2- \beta x - \alpha##.
Now we can write down both solutions ##u## and ##v## and see that ##v \in K(u)##, i.e. our minimal polynomial splits and is separable. (Why?)

PsychonautQQ
fresh_42 said:
I've thought along the same lines. More formal we can choose an element ##u \in L-K##. Then ##\{1,u\}## is ##K-##linear independent and ##\{1,u,u^2\}## is not. This means we can write ##u^2=\alpha \cdot 1 + \beta \cdot u## which means ##u## is a root of ##x^2- \beta x - \alpha##.
Now we can write down both solutions ##u## and ##v## and see that ##v \in K(u)##, i.e. our minimal polynomial splits and is separable. (Why?)
Our minimal polynomial is separable because K is given to have Characteristic 0, and our minimal polynomial splits because both of it's roots are in K(u). Thanks!

PsychonautQQ said:
Our minimal polynomial is separable because K is given to have Characteristic 0, and our minimal polynomial splits because both of it's roots are in K(u). Thanks!
One can also see it directly: If ##u## were a double root, then ##u=-\frac{\beta}{2} \in K## which we ruled out. And with ##u=-\frac{\beta}{2}-\sqrt{sth.}\, , \,v=-\frac{\beta}{2}+\sqrt{sth.}## we get ##v=-u-\beta \in K(u)##.

## 1) What is a Galois extension?

A Galois extension is a type of field extension in abstract algebra that has a specific set of properties related to the concept of symmetry. It is named after the French mathematician Évariste Galois and is used in the study of Galois theory.

## 2) How do you determine the degree of a Galois extension?

The degree of a Galois extension is determined by the number of elements in the extension field that are generated by the base field and the root of the polynomial that defines the extension. It is also equal to the dimension of the extension as a vector space over the base field.

## 3) What are some common examples of Galois extensions?

Some common examples of Galois extensions include the splitting fields of polynomials, such as the quadratic, cubic, and quartic extensions. Other examples include the cyclotomic extensions and the algebraic closure of a field.

## 4) How does the degree of a Galois extension affect its properties?

The degree of a Galois extension has a significant impact on its properties. In particular, the degree determines the number of automorphisms of the extension field, which are isomorphisms from the field to itself that preserve the operations. It also plays a role in determining the solvability of certain polynomial equations over the extension field.

## 5) What is the relationship between Galois extensions and Galois groups?

Galois extensions are closely connected to Galois groups, which are groups of automorphisms of the extension field that fix the base field. The structure of the Galois group is determined by the properties of the Galois extension, and it provides a powerful tool for studying the properties of the extension.

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