Understanding Galois theory

[dhaya777]

I'm trying to understand galois theory.

I came across this text online that I find useful as it provides examples

Galois theory text ( this takes you to a Google results page, please click the first result, sorry couldn't attach the file too big and couldn't link it any other way.)

I hope this link works. See pages 158-161

In summary...

f(x)=x³​-2, has roots a,b and c.

The Galois resolvent and its conjugates are formed thus..

t₀=a+2b+3c, t₁=c+2a+3b....t₅=a+2c+3b


F(X)=(X-t₀)(X-t₅)

This will lead to a polynomial in X with coefficients in a,b,c that are symmetrical, hence they can be represented with just one of the roots say "a".

so F(X,a)=X²+3aX+3a² and f(x)=x³-2

The text then makes the claim (indirectly) that "a" is a common factor of F(X) and f(x), because (x-a) divides
both F(X,a) and f(x).

This is where my confusion starts,

How can "a" be a factor of F(X,a)? I plug "a" into F(X,a) and get 7a² which cannot equal to 0.

"a" can be factor of f(x) as it is a root of f(x).

Then it replaces "a" with Y and x with t₀.

It goes on to claim that the g.c.d of F(t₀,Y) and f(Y) is

d(Y)=1/3(2t_0²Y+t_0³-6)

I just can't figure out how they got to that.

I tried dividing f(x) by F(X,a) but I run into trouble with unwieldy terms.

I am truly at my wits end, any guidance would be much appreciated.

Thank you for your time.

 

[Limax]

The text then makes the claim (indirectly) that "a" is a common factor of F(X) and f(x), because (x-a) divides
both F(X,a) and f(x).

I've been following the text to see what the problem is. What is asserted (indirectly) is that \(\displaystyle a\) is a root of \(\displaystyle F(t_0,X)\), not \(\displaystyle F(X,a)\). Replacing \(\displaystyle X\) by \(\displaystyle t_0\) and \(\displaystyle a\) by \(\displaystyle X\) gives

\(\displaystyle F(t_0,X)\ =\ t_0^2+3t_0X+3X^2\)​

\(\displaystyle \therefore\ F(t_0,a)\ =\ t_0^2+3t_0a+3a^2\)

Recalling that \(\displaystyle a,b,c\) are roots of the equation \(\displaystyle x^3-2=0\), we have \(\displaystyle a+b+c=0\), \(\displaystyle ab+bc+ca=0\) and \(\displaystyle abc=2=b^3\ \implies\ ac=b^2\). Hence \(\displaystyle t_0=a+2b+3c=-a+c+2(a+b+c)=-a+c\) and so

\(\displaystyle t_0^2+3t_0a+3a^2\)​

\(\displaystyle =\ (-a+c)^2+3a(-a+c)+3a^2\)

\(\displaystyle =\ a^2+b^2+c^2\)

\(\displaystyle =\ (a+b+c)^2-2(ab+bc+ca)\)

\(\displaystyle =\ 0\)

 

[Limax]

If you're still interested in learning Galois theory, I suggest this very readable book.

I'm currently reading it on my own, and I've reached Chapter 11, on field automorphisms.