Suppose we have two fields $E$ and $F$ such that $F\subset E$. Then we call $E$ an extension of $F$.
By $"\subset"$ I understand the set-theoretical inclusion.
But couple days ago when I was reading the following theorem: If $k$ is a field and $p(X)\in k[X]$ is irreducible polynomial then there exists an extension $E$ of $k$ such that $E$ contains a root of $p(x)$.
And the field $E= k[X]/(p(X))$ but it does not contain $k$ as a subset. I know that we can form embedding (injective homomorphism) of $k$ into $E$ OR identify $k$ with subfield of $E$. But anyway to be honest we have no $k\subset E$, right?
I would be very grateful can anyone in detail explain this moment to me, please! Because I have very big difficulties and I've asked many people to explain this to me but maybe I am dumb and cannot understand this properly. Maybe somebody from MSE can do this.
In reality a field extension $F$ of $K$ is just a given (injective) homomorphism of fields $K \to F$. Thus you can identify $K$ with a subfield of $F$, but for $F$ to be an extension of $K$, $K$ need not necessarily be defined as a subset of $F$, you just have to find an embedding of $K$ into $F$.
Therefore, if $K[X]$ is the ring of polynomials and $P$ is an irreducible polynomial, then $F:= K[X]/(P)$ is a field, and you can find an obvious embedding of $K$ into $F$ by sending an element $x \in K$ to the constant polynomial $x$ modulo $P$. Therefore $F$ is an extension of $K$.
I hope this answers your question.