TITLE: $K$ is a splitting field $\iff$ any irreducible polynomial with a root in $K$ splits completely over $K$.
QUESTION [9 upvotes]: Let $K/F$ be a finite extension. I want to show that $K$ is a splitting field over $F$ $\iff$ any irreducible polynomial $p(x)\in F[x]$ that has a root in $K$ splits completely over $K$.

I haven't even been able to show this in a single direction. My efforts have been to show that if $K$ is the splitting field for some set of polynomials $\{f_1,\ldots,f_r\}\subset F[x]$, than any irreducible polynomial $p(x)\in F[x]$ that has a root in $K$ must split completely. Let $p(\alpha)=0$ with $\alpha\in K$. Then we can factor $p(x)$ over $K$: $p(x)=(x-\alpha)p'(x)$.
From here I am actually stuck. It seems showing $p'(x)$ must have a root in $K$ would be useful but isn't really sufficient to show that it splits completely. My progress on that front is as follows. Letting $\beta_1$ and $\beta_2$ be roots of some irreducible (over $F$) factor of one of the $f_i$, we have an automorphism $\phi$ of $K$ obtained by extending the isomorphism $F(\beta_1)\to F(\beta_2)$ to $K$, which in turn extends the identity map on $F$. This fixes $F$ so fixes $p(x)$. Then we have $\phi((x-\alpha)p'(x))=(x-\phi(\alpha))\phi(p'(x))=p(x)$. Thus, if $\phi(\alpha)\neq\alpha$, then we have a new root for $p(x)$ over $K$. But I don't see why one of these automorphisms necessarily doesn't fix $\alpha$, or how we can show $p(x)$ necessarily splits completely this way.
The book (Dummit and Foote) asks this question before introducing any Galois theory, so I would like to solve it without any of those tools. Thank you.

REPLY [4 votes]: I have found a direct way proving this:
Suppose $K$ is the splitting field of $g(x)\in F[x]$ and $p(x)\in F[x]$ is irreducible over $F$. Moreover, $p(x)$ has a root $\alpha$ in $K$, we want to show that $p(x)$ splits in $K[x]$.
If $p(x)$ is linear, then we are done.
Otherwise, suppose $p(x)=(x-\alpha)(x-\beta)\tilde p(x)$ over the splitting field $\mathcal F$ of $p(x)$, where $\alpha\in K$, $\tilde p(x)\in\mathcal F[x]$ and $\beta\in\mathcal F\setminus K$ is taken to be any root of $p(x)$ that is not in $K$ and we want to get a contradiction.
Note that there is a natural isomorphism between $F(\alpha)$ and $F(\beta)$, since $p(x)$ is irreducible. Therefore, we can extend this isomorphism naturally to an isomorphism between the splitting field of $g(x)$ over $F(\alpha)$ and the splitting field of $g(x)$ over $F(\beta)$. Since $\alpha\in K$ and we conclude that the splitting field of $g(x)$ over $F(\alpha)$ is $K$. If we let $K'$ denote the splitting field of $g(x)$ over $F(\beta)$, then we have $K\cong K'$ and of course $[K':F]=[K:F]$ for their extension degrees.
Note that $K'$ can be viewed as adjoining $\beta$ to $K$ and $\beta\in\mathcal F\setminus K$ implies that $$[K': F]=[K': K][K: F]>[K:F]$$ which is a contradiction.
Thus, no root of $p(x)$ can be taken from $\mathcal F\setminus K$ which implies $p(x)$ splits completely over $K[x]$.

The other direction is simple, please refer to k.stm's answer for "(3)⇒ (1)".