How to test if a finite extension contains any zero of an irreducible polynomial? So specifically, if $E/F$ is a finite extension of degree $n$ and $f\in F[x]$ if an irreducible of degree dividing $n$, then is it always true that all zeros of $f$ falls into $E$ ?
If $E=GF(q^n)$ is finite, then $F(a)=GF(q)(a)\cong GF(q^k), k|n, f(a)=0$ but how does it implies that $F(a)\subseteq E$ ? I mean, just because it is isomorphic to a subfield of $E$ doesn't mean that it is a subfield of $E$, right ?
 A: In field theory, or more generally in abstract algebra, we are interested in algebraic structures up to isomorphism. So yes technically $F(a)$ may not exactly be a subfield of $E$, but it is isomophic to a subfield of $E$ so we say that $F(a) \subseteq E$. Only the field structure is interesting in that case.
Your question is true for finite field, as there is only one field of order $q^n$ up to isomorphism and $\mathbb{F}_{q^k} \subseteq \mathbb{F}_{q^n}$ if and only if $k \mid n$. In your example, if $f$ is irreducible of degree $k$, all the zeroes of $f$ lies in $\mathbb{F}_{q^k} \subseteq \mathbb{F}_{q^n}$. 
In general, this is completely false. For example $[\mathbb{Q}[i] \ : \ \mathbb{Q}]=2$ and $f(x)=x^2-2$ is irreducible over $\mathbb{Q}$ but $\mathbb{Q}[i]$ contains no roots of $f$. The real problem here is that there is infinitely many field extensions of degree $n$ whereas there is only one degree $n$ extension of $\mathbb{F}_q$.
A: It wasn't clear to me from the question that our OP Taylor Huang's main interest lay with finite fields, so I cooked up this counterexample:
I don't think this is true in general:
Take $F = \Bbb Q$ and $E = \Bbb Q(\sqrt[3] 2)$; then 
$[E:F] = [\Bbb Q(\sqrt[3]2): \Bbb Q] =3; \tag 1$
now take
$f(x) = x^3 - 2 \in \Bbb Q[x]; \tag 2$
then $f(x)$ is irreducible over $\Bbb Q$, $\deg f(x) \mid [E:F]$, but $f(x)$ has precisely one zero in $E = \Bbb Q(\sqrt[3] 2)$.
