I'm working out of Hungerford's (undergrad) algebra book. Extension fields are defined in a discussion on the structure of $F[x]/(p(x))$, where $p(x)$ is prime in $F[x]$:
"Let $F$ be a field and $p(x)$ an irreducible polynomial in $F[x]$. Let $K$ denote the field of congruence classes $F[x]/(p(x))$. Then $F$ is a subfield of the field $K$ (equivalently: $K$ is an extension field of $F$)."
This concept is introduced to show that irreducible polynomials $f(x)$ in $F[x]$ have roots when considered as polynomials over $K$.
My question is, how can I understand how the original field "behaves" inside of the extension field? Vaguely speaking, does $K$ contain a copy (or an isomorphic copy) of $F$ plus some other "stuff?" Must every property satisfied by $F$ (commutative, inverse, etc.) carry over to the copy of $F$ in $K$?
Finally, is it possible for there to exist an element $u$ in $F$ and an element $w$ in $K$ (and not in the part that's isomorphic to $F$) s.t. $uw = 1_F$?
Thanks in advance!
There is a canonical projection $\pi: F[x] \to F[x]/(p(x))$ and when restricted to $F$, this gives a homomorphism $F \to K$. Note that it preserves the $1 \in F$ so it is a nonzero map of fields. Therefore, it must be injective. This shows we can identify $F$ with the subfield $\pi (F) \subseteq K$ which will be an isomorphic copy of $F$ contained in $K$. The properties you mention, such as commutativity and inverse are always preserved under isomorphisms.