1. ## Permuting roots....

Hi guys,

This question is about Galois Theory.

One of the theorems in my books states that "Any K-automorphism of the field extension L/K must keep the elements of K invariant & must permute the roots of the minimal polynomial."

I was wondering, do the K-automorphisms permutes the roots of ALL polynomials, or just the minimal polynomial?

Many thanks in advance. :-) x

2. Originally Posted by AAM
Hi guys,

This question is about Galois Theory.

One of the theorems in my books states that "Any K-automorphism of the field extension L/K must keep the elements of K invariant & must permute the roots of the minimal polynomial."

I was wondering, do the K-automorphisms permutes the roots of ALL polynomials, or just the minimal polynomial?

Many thanks in advance. :-) x
All polynomials with coefficients in K, yes. If p(x)= $ax^n+ bx^{n-1}+ \cdot\cdot\cdot+ yx+ z$, then $f(p(x))= f(a)f(x)^n+ f(b)f(x)^{n-1}+ \cdot\cdot\cdot+ f(y)f(x)+ f(z)$, because f is an automorphism, $= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z$ because f "fixes" members of K.

IF x is a root of p(x) then $p(x)= 0$ so $f(p(x))= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z= 0$. that is, if x is a zero of a polynomial with coefficients in K, f(x) is a zero of that same polynomial: f permutes the zeros.

3. Originally Posted by HallsofIvy
All polynomials with coefficients in K, yes. If p(x)= $ax^n+ bx^{n-1}+ \cdot\cdot\cdot+ yx+ z$, then $f(p(x))= f(a)f(x)^n+ f(b)f(x)^{n-1}+ \cdot\cdot\cdot+ f(y)f(x)+ f(z)$, because f is an automorphism, $= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z$ because f "fixes" members of K.

IF x is a root of p(x) then $p(x)= 0$ so $f(p(x))= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z= 0$. that is, if x is a zero of a polynomial with coefficients in K, f(x) is a zero of that same polynomial: f permutes the zeros.
You still need to show more. Let $X$ be the set of zeros of $f(x)\in K[x]$ in some splitting field $L$. Notice that $|X|<\infty$ and that if $\sigma \in \text{Gal}(L/K)$ then $\sigma: X\to X$ as you have shown. However, $\sigma$ is injective (for it is an automorphism) and so $\sigma: X\to X$ is a bijection. Thus, $\sigma$ is a permutation of $X$ - the set of zeros of $f(x)$.

Originally Posted by AAM
I was wondering, do the K-automorphisms permutes the roots of ALL polynomials, or just the minimal polynomial?
In the proof we do not use the fact that the polynomial is irreducible. Thus, this is true for all non-constant polynomials.

4. Thank you! :-D