Does a field extension equal $F(\alpha) = a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3+\cdots$ Let $F$ be a field, does $F(\alpha)=a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3+\cdots$?
If it does, this was never explicitly stated, and I have gained confusion from this, since in the case that:
$$(*)\quad\Bbb Q(\sqrt{2})=\{a+b\sqrt2:a,b\in \Bbb Q\}$$
I never thought about the fact that $a_0+a_1\sqrt{2}+2a_2+2\sqrt{2}a_3+4a_4+\cdots$ falls into the form written in $(*)$, so I don't really know what this is meant to look like :\

If so, then what does $F(\alpha,\beta)$ look like?
 A: First, $\mathbb{F}(\alpha)$ is a set. Be careful to distinguish between the elements in the set and the set itself. Second, the element $a_o+2a_2+4a_4+ \cdots = c_1 \in \mathbb{Q}$ since the sum of rational numbers is rational. Likewise, the terms with $\sqrt{2}$ also group together to give a single term.
A: If $\alpha$ is an element of a field containing $F$, the definition of $F(\alpha)$ is this:
$$F(\alpha)=\biggl\{\dfrac{P(\alpha)}{Q(\alpha)}\;\bigg\vert\; P(X), Q(X)\in F[X]\enspace\text{and}\enspace Q(\alpha)\neq 0\biggr\}.$$
If $\alpha$  is transcendental over $F$, there is nothing more to say. If  $\alpha$  is algebraic over $F$, say $\alpha$ satisfies an lagebraic equation of degree $d$, an easy induction shows the polynomials $P$ and $Q$ may be taken with degree $<n$.
For two elements, the definition is similar, but with polynomials in two indeterminates:
$$F(\alpha, \beta)=\biggl\{\dfrac{P(\alpha, \beta)}{Q(\alpha, \beta)}\;\bigg\vert\; P(X, Y), Q(X, Y)\in F[X, Y]\enspace\text{and}\enspace Q(\alpha, \beta)\neq 0\biggr\}.$$
