Automorphisms in the Galois Group of f(x) over K Permute the Roots of f

Automorphisms in the Galois Group of f(x) over K Permute the Roots of f

Recall from the Galois Groups page that if $K$ is a field and $F$ is an extension field of $K$ then the Galois group of $F$ over $K$ is defined as:

(1)
\begin{align} \quad \mathrm{Gal} (F/K) = \{ \phi \in \mathrm{Aut}(F) : \phi(k) = k, \: \forall k \in K \} \end{align}

Furthermore, if $K$ is a field, $f(x) \in K[x]$, and $F$ is the splitting field of $f(x)$ over $K$ then the Galois group of $f(x)$ over $K$ is $\mathrm{Gal} (F/K)$.

We will now look at a very important result which says that if $K$ is a field and $f(x) \in K[x]$ then every automorphism in the Galois group of $f(x)$ over $K$ permutes the roots of $f(x)$.

Theorem 1: Let $K$ be a field and let $F$ be an extension field of $K$. Let $f(x) \in K[x]$. Then every automorphism in $\mathrm{Gal} (F/K)$ permutes the roots of $f(x)$ in $F$.

In particular, if $K$ is a field and $f(x) \in K[x]$ then every automorphism in the Galois group of $f(x)$ over $K$ permutes the roots of $f(x)$ (since here, the field extension $F$ is the splitting field of $f(x)$ over $K$).

  • Proof: Let $f(x) \in K[x]$ and write:
(2)
\begin{align} \quad f(x) = a_0 + a_1x + ... + a_nx^n \end{align}
  • Where $a_0, a_1, ..., a_n \in K$. Let $u \in F$ be a root of $f$. Then $f(u) = 0$. Let $\phi \in \mathrm{Gal} (F/K)$. Then:
(3)
\begin{align} \quad 0 &= \phi(f(u)) \\ &= \phi(a_0 + a_1u + ... + a_nu^n) \\ &= \phi(a_0) + \phi(a_1u) + ... + \phi(a_nu^n) \\ &= \phi(a_0) + \phi(a_1)\phi(u) + ... + \phi(a_n)\phi(u^n) \\ \end{align}
  • Since $a_0, a_1, ..., a_n \in K$ and $\phi \in \mathrm{Gal} (F/K)$ we have that $\phi(a_0) = a_0$, $\phi(a_1) = a_1$, …, $\phi(a_n) = a_n$, and so:
(4)
\begin{align} \quad 0 = a_0 + a_1\phi(u) + ... + a_n \phi(u)^n = f(\phi(u)) \end{align}
  • Therefore $\phi(u)$ is a root of $f$. Observe that a polynomial $f$ can have only finitely many roots in $F$ and since $\phi$ is an automorphism, it is bijective. Therefore $\phi$ permutes the roots of $f$ in $F$. $\blacksquare$
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