# The Galois group of a polynomial over a field and over some extension (updated)

Let $f(x)\in F[ x]$ and $K/F$ be a field extension. Show that the Galois group of $f(x)$ over $K$ is isomorphic to a subgroup of the Galois group of $f(x)$ over $F$.

Let $E$ be the splitting field of $f(x)$. Then we need to construct a monomorphism $Gal(E/K)\to Gal(E/F)$. But every automorphism $E\to E$ that fixes $K$ also fixes $F$ so $Gal(E/K)\subseteq Gal(E/F)$. Then trivially the inclusion is a monomorphism.

I think this is just too easy. In fact, my question is if we can indeed asume that if $E$ is the splitting field of $f\in F[x]$ then it is also the splitting field of $f(x)\in K[x]$?

Updated.

It seems they are not the same. Let $E$ be the splitting field of $f(x)$ over $F$ and $E'$ be the splitting field of $f(x)$ over $K$. We need to show there is a monomorphism $Gal(E'/K)\to Gal(E/F)$.

However, since $F\subseteq K\subseteq E'$ and $f(x)$ can be descomposed as linear factors in $E'$ then $E\subseteq E'$.

If $\sigma\in Gal(E'/K)$ then clearly $\sigma\restriction_E\in Gal(E/F)$. My question is if there is any way to prove $\sigma\mapsto\sigma\restriction_E$ is injective?

• THis is confusing: what's the relation between $\;F, K, E\;$ ? what is contained in what? – DonAntonio May 15 '16 at 19:49
• Sorry @Joanpemo. It was $K/F$ then $F\subseteq K$. I edited it. – Talexius May 15 '16 at 19:54
• Ok, so $\;F\subset K\subset E\;$ , and this last is a (the) splitting field of $\;f\;$ over $\;F\;$, am I right? Of course, it is also a splitting field for $\;f\;$ over $\;K\;$ as it is the minimal overfield of $\;F\;$ in which $\;f\;$ splits in linear factors. – DonAntonio May 15 '16 at 19:57
• I think that in the statement of the problem you are not supposed to assume that $K$ lies within any particular splitting field of $f$ over $F$. This means that you need to keep track of the distinction between the splitting field for $f$ over $F$ and the splitting field for $f$ over $K$. – Rolf Hoyer May 15 '16 at 19:59
• But how do you know $K\subseteq E$? We only know $K$ is a field extension of $E$. How can we say $K\subseteq E$? @Joanpemo – Talexius May 15 '16 at 20:02

You say that if $\sigma\in \text{Gal}(E'/K)$, then 'clearly' $\sigma$ restricts to an element of $\text{Gal}(E/F)$. Note that this requires demonstrating that $\sigma(E) = E$. This follows since $E$ is a splitting field over $F$. Namely, $\sigma$ fixes $F$ and thus permutes the roots of $f(x)$. This is enough to show that $\sigma(E) = E$.
To show that the defined restriction map $\text{Gal}(E'/K) \to \text{Gal}(E/F)$ is injective, you need to demonstrate that if $\sigma$ acts trivially on $E$, then it acts trivially on $E'$ too. Note that $\sigma$ restricting to the identity on $E$ implies that $\sigma$ acts trivially on the roots of $f(x)$. This is enough to show that $\sigma$ acts trivially on all of $E'$, since $E'$ is a splitting field for $f(x)$ over $K$ and thus is generated over $K$ by roots of $f(x)$.