Finite extension of $K$,a finite extension of $\mathbb{Q}_p$, that is Galois? Suppose I have $K$ a finite extension of $\mathbb{Q}_p$. Can I always find 
$K'$ which is a finite extension of $K$ and is Galois extension over $\mathbb{Q}_p$? or does $K'$ have to be infinite in some cases?
I would greatly appreciate any comments. Thank you!
 A: If $L/K$ is any finite, separable extension of fields, the primitive element theorem states that $L=K(\alpha)$ for some $\alpha\in L$. 
Let $f$ be the minimal polynomial of $\alpha$ over $K$. Then its splitting field is the field you’re after. 
This gives an explicit construction of the Galois closure of a separable extension $L/K$, namely a field $K’$ containing $L$ such that $K’/K$ is Galois, and such that $K’$ is minimal amongst all such extensions. 
A: Let $K/F$ be any finite field extension (we don't need to assume separability or use the primitive element theorem). Fix an algebraic closure $\overline{F}$ of $F$. It is well-known that the number of $K$-linear field homomorphisms $K \to \overline{F}$ is finite. Let $\sigma_1, \dots \sigma_n$ be all such homomorphisms.
Then consider the composite field $\sigma_1(K) \cdot \sigma_2(K) \cdot \ldots \cdot\sigma_n(K):=K'$ inside $\overline{F}$. Since composites of finite extensions are finite, this is a finite extension of $F$.
I claim that $F$ is a normal extension. If $f$ is an irreducible polynomial over $K$ that has a root $\alpha$ in $K'$, then for any other root $\beta$ of $f$, there is a $F$ automorphism $\tau$ of $\overline{F}$ with $\tau(\alpha)=\beta$.
But since $\sigma_1, \dots, \sigma_n$ is a complete list of $F$-embeddings $K \to \overline{F}$ and for every $i$, we have that $\tau \circ \sigma_i$ is also a $f$-embedding $K \to \overline{F}$, we see that $\tau \circ \sigma_1, \dots \tau \circ \sigma_n$ is just a permutation of the list $\sigma_1, \dots, \sigma_n$. From this, we see that $$\tau(K')=\tau(\sigma_1(K) \cdot \ldots \cdot\sigma_n(K))=\tau(\sigma_1(K)) \cdot \ldots \cdot \tau(\sigma_n(K))=\sigma_1(K) \cdot \ldots \cdot\sigma_n(K)=K'$$ so $K'$ is invariant under $\tau$ (some authors take a condition like this as the definition of normality). In particular, since $\alpha \in K'$, $\beta=\tau(\alpha) \in \tau(K')=K'$, so $K'$ is normal as claimed.
If we started with a separable extension $K/F$, then because the separability does not depend on the embedding and the composite of separable extension is again separable, $K'/F$ is Galois.
