How does a field extension maintain the field structure, and are all field extensions fields, or only algebraic extensions? Suppose we adjoin a symbol $k$ to the field $F_p$, as in $F_p(k)$. What is an intuitive understanding of the structure of this field and its elements? Since multiplication needs to be closed, all new polynomials are part of this field. But also multiplicative inverses need to be included, does that mean all rational functions over $F_p$ should also be included? What about additive inverses? Do you have a mental model of this structure that gives you a sense of what the elements look like and how they preserve the structure, or is that something you can only reasonably inspect case-by-case?
Let's consider $F_2(k)$. If my understanding above is correct, all the following are new elements added to $F_2$: 
$$0,1,k,k+1,k^2,k^2+1,k^2+k+1$$
The list goes on forever, which doesn't seem right. Perhaps the structure is not well-defined if $k$ is just some arbitrary symbol, and it needs to be a root of a polynomial that doesn't have a root in $F_2$. This would mean I can only make $F_p(k)$ a field if it is an algebraic extension of $F_2$. Is that so?
If it is, then say I choose $p(x)=x^2+x+1$ in $F_2$, and I add the root $k$. How do I define $k^n$ for any power $n$, and how do I define $k+k+k$. In other words, how do I force the field structure? $k$ being the root of $p(x)$ should somehow help me fill in the multiplication and addition tables, but I don't see how.
 A: What we have are two closely related concepts - a field extension $F(k)$ and a ring extension $F[k]$. The difference? In the former, we adjoin rational functions of $k$ as well, so $F(k)$ is the field of quotients of $F[k]$.
The two are equal if and only if $k$ is algebraic over $F$.
So, your example, adjoining a transcendental element $k$ to $F_2$? Done as a ring extension $F_2[k]$, we get the polynomial ring in one variable over $F_2$, with such elements as $k$, $k^2$, $k^3+k+1$, and so on - $2^m$ new elements of each degree $m>0$. Done as a field extension, we get all of their quotients as well, for such elements as $\frac{(k+1)^2(k^2+k+1)}{k(k^3+k+1)}$.
If instead your $k$ is a root of the irreducible polynomial $k^2+k+1$? Then that ring extension collapses - everything of degree $2$ or higher can be reduced to an equivalent element of degree $1$ or less, by taking it mod $k^2+k+1$. For example, $k^3+k+1=(k+1)(k^2+k+1)+k\equiv k$. Moreover, everything that's not zero is invertible; we can apply the Euclidean algorithm to find the GCD of any polynomial and $k^2+k+1$, and write it as a linear combination (with polynomial coefficients) of those two. Since $k^2+k+1$ is invertible, that GCD is either $1$ or $k^2+k+1$ - and writing $1$ as a linear combination constructs the inverse mod $k+2+k+1$. That makes it a field.
Since there are only four elements, we can write out everything cleanly. We have $x^2\equiv x^2-(x^2+x+1)\equiv x+1$, and $x^3\equiv x(x+1)=x^2+x\equiv 1$. Of the four elements $0,1,x,x+1$, we've got them all accounted for, and the multiplicative group is simply cyclic of order $3$.
And we did all this with just the elements from the ring extension. There's no need to bring in the rational functions when everything's invertible already.
