Question

D PROVE THAT N-M.

Answer 1

(1) Let
K DO
be the splitting field, and let $\alpha\in K$ be a root of . Since
G = Gal(K : Q)
is abelian, every subgroup of it is normal. By fundamental theorem of Galois theory, every subfield of
K DO
is a normal extension of $\mathbb Q$. In particular, the extension
Q(a) Q
is normal (and separable too because we are in characteristic zero). Since
Q(a) Q
is normal and
Q(a)
contains one root of the irreducible polynomial $f\in \mathbb Q[x]$ , it must contain all the roots of $f\in \mathbb Q[x]$ . But then,
Q(a)
is the splitting field of $f\in \mathbb Q[x]$ . This implies

$M=|G|=[\mathbb Q(\alpha):\mathbb Q]=[\mathbb Q[\alpha]:\mathbb Q]=\deg f=N$

(2) Consider the polynomial in $\mathbb Q[x]$. We claim that is irreducible in $\mathbb Q[x]$. This is equivalent to

f(x + 1) = (x + 1)4+1=14 +42.3 + 62.2 + 4x + 2

being irreducible in$\mathbb Q[x]$, and this irreducibility follows from Eisenstein.

Note that the zeros of are as follows: since

$\begin{align*}f(x)&=x^4+1\\ &=x^4-i^2\\ &=(x^2-i)(x^2+i)\\ &=\begin{pmatrix}x^2-\begin{pmatrix}{\frac{1+i}{\sqrt 2}}\end{pmatrix}^2\end{pmatrix}\begin{pmatrix}x^2+\begin{pmatrix}{\frac{1+i}{\sqrt 2}}\end{pmatrix}^2\end{pmatrix}\\ &=\begin{pmatrix}x-{\frac{1+i}{\sqrt 2}}\end{pmatrix}\begin{pmatrix}x+{\frac{1+i}{\sqrt 2}}\end{pmatrix}\begin{pmatrix}x^2-i^2\begin{pmatrix}{\frac{1+i}{\sqrt 2}}\end{pmatrix}^2\end{pmatrix}\\ &=\begin{pmatrix}x-{\frac{1+i}{\sqrt 2}}\end{pmatrix}\begin{pmatrix}x+{\frac{1+i}{\sqrt 2}}\end{pmatrix}\begin{pmatrix}x+{\frac{1-i}{\sqrt 2}}\end{pmatrix}\begin{pmatrix}x-{\frac{1-i}{\sqrt 2}}\end{pmatrix}\end{align*}$

the zeros of are

$\begin{align*}{\frac{1+i}{\sqrt 2}},-{\frac{1+i}{\sqrt 2}},{\frac{1-i}{\sqrt 2}},-{\frac{1-i}{\sqrt 2}}\end{align*}$

The splitting field of thus contains

iv2 iv 2. = 2.

But all the roots above lies in $\begin{align*}\mathbb Q(i,\sqrt 2)\end{align*}$ , which must thus be the splitting field. Therefore, the Galois group of is

$\begin{align*}\mbox{Gal}(\mathbb Q(i,\sqrt 2):\mathbb Q)\cong \mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\end{align*}$

Thus, the Galois group of irreducible $\begin{align*}f\in \mathbb Q[x]\end{align*}$ is abelian does not necessarily imply it is cyclic.