Let $k$ a field of characteristic different from 2.

a) Show that for all extension $K/k$ of degree $2$, there is a $a\in k$ s.t. $K=k(\alpha)$ and $\alpha^2=a$.

b) Show that all extension of $k$ of degree $2$ is galoisienne and that it's Galois group is isomorphic to $\mathbb Z/2\mathbb Z$.

My attempts

a) done.

b) Let $K/k$ an extension of degree $2$. By a) there is a $a\in k$ s.t. $K=k(\alpha)$ and $a=\alpha^2$. Let $X^2-a\in k[X]$. Since $\alpha\notin k$, this polynomial is irreducible. My teacher told me that the fact that the characteristic is not $2$, the polynomial is separable.

Q1) What is the correlation between the fact that the characteristic is not $2$ and the fact that the polynomial is separable ?

Therefore $k(\alpha)$ is the splitting field of $X^2-a$ and thus the extension is Galoisienne. The Galois group is of order $2$, what prove that it's isomorphic to $\mathbb Z/2\mathbb Z$.

Q2) I have the impression that I can do this for any extension of degree $n$. Why is it specific of the extension of degree $2$ ?

  • $\begingroup$ What is your definition of "separable"? What equivalent conditions for separability do you know? $\endgroup$ Nov 29, 2015 at 21:14
  • 1
    $\begingroup$ Re Q2: Can do what for arbitrary $n$? Show that $K=k(\alpha)$ for some $\alpha$? That we can achieve $\alpha^n\in k$? That all extensions of degree $n$ are Galois? That the Galois group is $\Bbb Z/n\Bbb Z$? $\endgroup$ Nov 29, 2015 at 21:17
  • 1
    $\begingroup$ Part a) is taught in high school and called 'Completing the square'. $\endgroup$
    – MooS
    Nov 29, 2015 at 21:19

2 Answers 2

  1. In characteristic $2$, $\alpha$ is a double root of $X^2-a$ cause $(X-\alpha)^2=X^2-\color{red}2\alpha X+\alpha^2=X^2-\alpha^2=X^2-a$. In characteristic $\ne 2$, the polynomial $f(X)=X^2-a$ has formal derivative $f'(X)=2X$, which has no root in common with $f$ (as $f'(\alpha)=2\alpha\ne0$), hence $f$ has no multiple roots. (The fact that $f'$ is identically zero in characteristic $2$ is an alternate proof that $f$ is not separable in characteristic $2$)

  2. There is (up to isomorphism) only one group of order $2$. One element is neutral and the other must be its own inverse.

  • $\begingroup$ How do you answer part a). specifically, why only if characteristic $\ne$ 2 is an extension of degree 2 a splitting field of a polynomial of the form $x^2-a$ = 0. I'm confused about what happens in the case of characteristic 2. $\endgroup$ Nov 29, 2015 at 22:58

Part (a) is not at all trivial and is very very special to degree $2$. You want to prove this using the quadratic formula and the fact that the characteristic is not $2$. Most extensions of degree $n$ for $n>2$ cannot be obtained by adjoining a root of $x^n-a$ for any $a$. It is also not necessarily true that adjoining one such root adjoins all of them (and thus gives a Galois extension); you need to know that $k$ contains a primitive $n$th root of unity. This is always true for $n=2$, since a primitive square root of unity is just $-1$.

As for question 1, an irreducible polynomial $p(x)$ is separable iff $p'(x)\neq 0$. If $p(x)=x^2-a$, then $p'(x)=2x$ is nonzero iff the characteristic is not $2$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .