Outline of Galois theory

See this and [xournal 121]. Also [Stillwell 1994]

Given a polynomial $p (x) \in Q (x)$ we can consider its splitting field $K$ and its Galois group $G a l (K | Q)$ .

We say that a field extension of $Q$ , i.e. a pair of fields $Q \subseteq E$ , is cyclotomic if $E = Q (α)$ where $α$ is such that

α^{m} - k = 0

for $m \in N$ and $k \in Q$ .

We say that a field extension $E$ of $Q$ is radical if there is a finite sequence of cyclotomic extension

Q \subseteq Q (α_{1}) \subseteq Q (α_{1}, α_{2}) \subseteq Q (α_{1} α_{2}, α_{3}) \subseteq \dots \subseteq E

Following [Stilwell 1994] pages 4-5, we can add intermediate steps in this sequence in such a way that in each each step the $m_{i}$ is prime and every field in the sequence does not contain an $m_{i}$ th root of unity not in the previous step unless $α_{i}$ is itself one of them.

If the polynomial can be solved by radicals (i.e. we can work out the roots from the coefficients of $p$ and the operations $+, -, \cdot, /, \sqrt[n]{}$ ), then the splitting field is a radical extension.

We will show that the Galois group of a radical extension is a solvable group. But the Galois group of the equation

x^{5} - 2 x + 1

is $S_{5}$ , which is non solvable. Therefore this equation cannot be solved by radicals.

It only remains to see that the Galois group of a radical field extension is solvable. But this can be deduce from the following

Proposition

([Stillwell 1994] theorem 2)
Consider the fields $B \subseteq B (α) \subseteq E$ where $B (α)$ is a cyclotomic extension of $B$ . Assume also that, if $p$ is such that $α^{p} \in B$ , then $B (α)$ does not contain a $p$ th root of unity not in $B$ (unless $α$ itself is one). Then $G a l (E | B (α))$ is a normal subgroup of $G a l (E | B)$ and

G a l (E | B) / G a l (E | B (α))

is abelian.

Proof and visualization:

([Stillwell 1994] theorem 2)
![[galois theory.pdf]]

Essence of Galois theory

The essence, then, in Galois theory is what follows. The Galois group of an irreducible polynomial $p (x)$ acts on the space of their roots shuffling them, not in a transitive way. Every time we "compute" a radical we are adding a new number to the ground field. The important radicals (for the solvability) are those that let us decompose the polynomial in new irreducible factors.
Example: Consider $x^{3} - 1$ . If we introduce the number $e^{\frac{2 π}{3} i}$ in the ground field we can factor

x^{3} - 1 = (x - 1) (x - e^{\frac{2 π}{3} i}) (x - e^{\frac{4 π}{3} i})

When we add such a radical, we obtain a new Galois group with a new ground field, which is a subgroup of the previous one that shuffle the roots inside every irreducible polynomial. We can take quotients, both in the space of roots and in the original Galois group and get a group action of the quotient space on the space of orbits, because these irreducible polynomials are on the same footing (this implies the subgroup is normal). If the roots of the polynomial $p (x)$ cannot be separate this way, it is not solvable!

Galois theorem?

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