# $K \le B\le F$, If $F$ Galois over $K$ $\Rightarrow$ $F$ a Galois over $B$

Let $F$ a Galois over $K$, and let $B$ be a subfield of $F$ such that : $K \le B\le F$ $\Rightarrow$ $F$ a Galois over $B$

PROOF: $F$ is a splitting field of $f \in K[x]$ separable over $K$. $F=K(a_1,...,a_n)$ and $a_i$ are the roots of $f$ in $F$. Then $F=B(a_1,...,a_n)$ and $F$ is a splitting field of $f$ over $F$. I have to show that $f$ is separable over B, how?

Can somebody help me with this proof?

EDIT:

From my notes: "If $p$ is an irreducible factor of $f$ in $K$ then $p$ is product of irreducible factors over $B$" why?

• Hint: The roots of $f$ in $F$ don't change, when you view $f$ as coming from $B[x]$ as opposed to $K[x]$. – Jyrki Lahtonen Jul 27 '14 at 10:12
• From my notes: "If $p$ is an irreducible factor of $f$ in $K$ then $p$ is product of irreducible factors over $B$" why? – user158013 Jul 27 '14 at 10:20

A polynomial is separable if and only if $\gcd(f,f')=1$. Hence, separability does not change if you replace the base field from $K$ to $B$.
Let $\alpha \in K$, and let $f(x)=Irr(\alpha,K).$ Then $a(x)f(x)+b(x)f'(x)=1$ in $K[x]$, i.e. $f$ has no multiple roots.
Now, let $g(x)=Irr(\alpha,B).$ Clearly $g(x)|f(x)$ in $B[x]$ (as $f(\alpha)=0$ and $f\in K[x]\subset B[x]$ and $g$ is the polynomial of minimum degree in $B[x]$ with this property), so if $f$ has no multiple roots, nor has $g$.
Your question simply follows from the fact that $B[x]$ is a UFD.