<p>Let $K=\frac{\mathbb{Q}[x]}{&lt;f(x)&gt;}$ where $f(x)$ is irreducible over $\mathbb{Q}$ and has even degree. I want to find $K_2$ such that $ \mathbb{Q} \subseteq K_2\subseteq K$ and $[K_2:\mathbb{Q}]=2$.</p>

<p>If K is a <em>cyclic Galois extension</em> over $\mathbb{Q}$ then discriminant of $f(x)$ solves the problem.</p>

<p>But what if $K$ is not a cyclic galois extension?</p>

<p>Or even particular when $K$ is not even a Galois extension.  </p>

CC BY-SA 3.0
2016-08-23T00:12:57.817Z
do you know or can calculate the galois group of (the splitting field of) f(x)?
ricardorr
CC BY-SA 3.0
2016-08-23T03:11:12.720Z
If degree of splitting field is small "poly(deg ($f$))" than we can calculate the galois group of the splitting field, but its degree two subfield might not be a subfield of $K$ as well.
xyz
CC BY-SA 3.0
2016-08-27T05:11:05.110Z
How does the discriminant of $f$ solve the problem when the extension is Galois? $\mathbb{Q}(\sqrt{\Delta_f})$ works when the discriminant isn't a square in $\mathbb{Q}$, but that needn't always be the case. In fact, if $K/\mathbb{Q}$ is a Galois extension with Galois group $A_n$ ($n\ge 4$), there are no index-2 subgroups, so $K_2$ won't even exist. Is it just the case that there is no $K_2$ when the discriminant is a square?
Andrew Morgan
CC BY-SA 3.0
2016-08-27T05:28:10.887Z
To answer my own question at the end of my previous comment: Consider $f(x) = x^4 + 1$. It's irreducible with roots $\pm\sqrt{\pm i}$. It has discriminant 256, which is a square, but its splitting field contains the degree-2 field $\mathbb{Q}(i)$.
Andrew Morgan
CC BY-SA 3.0
2016-08-27T05:36:56.090Z
@AndrewMorgan yes I also observed that ! Actually Discriminant will always work when the galois group is cyclic. Also note that I want a quadratic subfield of K and not splitting field of $f(x)$
xyz
CC BY-SA 3.0
2016-08-27T06:35:11.430Z
For Galois extensions: It's known that the discriminant is a square iff the Galois group is a sub(-permutation-)group of $A_n$ (where $n$ is the degree of $f$). Hence if the Galois group isn't a subgroup of $A_n$, the discriminant method works. This generalizes the case of cyclic Galois extensions: Galois groups act transitively on the roots, so a cyclic Galois group has to be generated by an $n$-cycle; however, when $n$ is even, $n$-cycles are odd permutations, and hence aren't in $A_n$.
Andrew Morgan
<algebra><polynomials><comp-number-theory>
Finding degree two subfield
xyz
# Finding degree two subfield

(Asked by: username_0 on 2016-08-22)

Let $K=\frac{\mathbb{Q}[x]}{}$ where $f(x)$ is irreducible over $\mathbb{Q}$ and has even degree. I want to find $K\_2$ such that $ \mathbb{Q} \subseteq K\_2\subseteq K$ and $[K\_2:\mathbb{Q}]=2$.

If K is a *cyclic Galois extension* over $\mathbb{Q}$ then discriminant of $f(x)$ solves the problem.

But what if $K$ is not a cyclic galois extension?

Or even particular when $K$ is not even a Galois extension.

**Tags:** algebra, polynomials, comp-number-theory
> **Comments:**
> **username_1**: do you know or can calculate the galois group of (the splitting field of) f(x)?  
> **username_0**: If degree of splitting field is small "poly(deg ($f$))" than we can calculate the galois group of the splitting field, but its degree two subfield might not be a subfield of $K$ as well.  
> **username_2**: How does the discriminant of $f$ solve the problem when the extension is Galois? $\mathbb{Q}(\sqrt{\Delta_f})$ works when the discriminant isn't a square in $\mathbb{Q}$, but that needn't always be the case. In fact, if $K/\mathbb{Q}$ is a Galois extension with Galois group $A_n$ ($n\ge 4$), there are no index-2 subgroups, so $K_2$ won't even exist. Is it just the case that there is no $K_2$ when the discriminant is a square?  
> **username_2**: To answer my own question at the end of my previous comment: Consider $f(x) = x^4 + 1$. It's irreducible with roots $\pm\sqrt{\pm i}$. It has discriminant 256, which is a square, but its splitting field contains the degree-2 field $\mathbb{Q}(i)$.  
> **username_0**: @AndrewMorgan yes I also observed that ! Actually Discriminant will always work when the galois group is cyclic. Also note that I want a quadratic subfield of K and not splitting field of $f(x)$  
> **username_2**: For Galois extensions: It's known that the discriminant is a square iff the Galois group is a sub(-permutation-)group of $A_n$ (where $n$ is the degree of $f$). Hence if the Galois group isn't a subgroup of $A_n$, the discriminant method works. This generalizes the case of cyclic Galois extensions: Galois groups act transitively on the roots, so a cyclic Galois group has to be generated by an $n$-cycle; however, when $n$ is even, $n$-cycles are odd permutations, and hence aren't in $A_n$.  

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