My question concerns Kronecker's theorem and Extension fields.
Let $F$ be a field and let $f (x)$ be a nonconstant polynomial in $F[x]$. Then there is an extension field $E$ of $F$ in which $f(x)$ has a zero.
The proof of Kronecker's theorem is given as follows:
Since $F[x]$ is a unique factorization domain, $f (x)$ has an irreducible factor, say, $p(x)$. Clearly, it suffices to construct an extension field $E$ of $F$ in which $p(x)$ has a zero. Our candidate for $E$ is $F[x]/ \langle p(x) \rangle$. Due to a previous theorem, this is a field. Also, since the mapping of $\phi: F \rightarrow E$ given by $\phi(a) = a + \langle p(x) \rangle$ is one-to-one and preserves both operations, $E$ has a subfield isomorphic to $F$. We may think of $E$ as containing $F$ if we simply identify the coset $a + \langle p(x) \rangle$ with its unique coset representative a that belongs to $F$.
Then to show $p(x)$ has a zero in $E$, we substitute in $x + \langle p(x) \rangle$ into $p(x)$ and get $p(x) + \langle p(x) \rangle$ = $0 + \langle p(x) \rangle$.
My question is: This does not show that $F[x] / \langle p(x) \rangle$ is an extension field of $F$, only that there's a homomorphic image of $F$ that's a subfield of $E$. Because by the definition of an extension field:
A field $E$ is an extension field of a field $F$ if $F \subseteq E$ and the operations of $F$ are those of $E$ restricted to $F$.
But $F$ is not a subset of $F[x] / \langle p(x) \rangle$ because elements of $F$ are just the elements while elements of $F[x] / \langle p(x) \rangle$ are of the form $g(x) + \langle p(x) \rangle$, which are cosets. Can someone explain to me what I'm missing here?