# Extension field definition vs Kronecker's theorem

Extension field definition:

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 $F$.

Kronecker's theorem:

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.

My confusion comes from the proof of Kronecker's theorem: For a nonconstant polynomial $f(x)$ in $F[x]$, you take one of its irreducible factors $p(x)$ and form the extension field out of $F[x]/ \langle p(x) \rangle$, but how does this satisfy the definition of an extension field?

I don't see how $F \subseteq F[x]/ \langle p(x) \rangle$, because elements of $F$ are just elements of the field, while elements of $F[x]/ \langle p(x) \rangle$ are of the form $g(x) + \{h(x)p(x) : h(x) \in F[x]\}$ which is a coset?

For every $c\in F$, let $g(x)$ the constant polynomial $c$ in the expression $g(x) + \{h(x)p(x) : h(x) \in F[x]\}$, this enables to identify $F$ as a subfield of $F[x]/(p(x))$.
• I don't understand how it's a subfield, $c$ is an element of $F$ and $c + \{h(x)p(x) : h(x) \in F[x]\}$ is a coset. How can they any element of $F$ be in $F[x] / \langle p(x) \rangle$? – Oliver G Sep 20 '16 at 14:25
• You can add $(c + \{h(x)p(x) : h(x) \in F[x]\}) + (c' + \{h(x)p(x) : h(x) \in F[x]\})=c+c' + \{h(x)p(x) : h(x) \in F[x]\}$, – Tsemo Aristide Sep 20 '16 at 14:31
• I still don't understand. An element of $F$ is just an element, while an element of $F[x] / \langle p(x) \rangle$ is a coset. How does adding together two cosets show that $F$ is a subset of $F[x] / \langle p(x) \rangle$? – Oliver G Sep 20 '16 at 14:39