What does it mean for an element to generate a field as an F-algebra? What does it mean for an element "x" to generate a field E as an F-algebra, where F is a field? I am struggling to find a clear definition, and as a result I am confused as to the equivalence of this statement and the statement "x is algebraic over F".
I think if I can get a clear definition I will be able to understand the equivalence.
This is the claim in the notes that I'm struggling with. 
Proposition 26 Let E/F be a field extension such that E = F(x) for some
element x ∈ E (meaning that E is the smallest field containing F and x).
The following are equivalent
(1) E/F is a finite extension.
(2) x is algebraic over F.
(3) E is generated by x as an F-algebra.
(4) E is finitely generated as an F-algebra
Now that I have thought a bit more about the problem and read the answers I will try to provide a proof of (3) implies (2).
(3) implies $f(x) = x$ is in E, which is a field, so has an inverse $g(x)$. Then $xg(x)-1 = 0$.
We can write $g(t) = c_o + c_1t +...+ c_rt^r $, and if $tg(t) = 1$ in F[t] then we must have $tg(t) = c_ot + c_1t^2 +...+ c_rt^{r+1}= 1 $, a contradiction. Hence $tg(t)-1 \neq 0$, and $xg(x) - 1 = 0$, that is $x$ is algebraic over F.
 A: Generating as an $F$-algebra means you can get everything using polynomials with coefficients in $F$: use sums and products and multiplication by elements of $F$. Generating as a field over $F$ means you are allowed to use division also. For example, $i$ generates $\mathbf C$ as an $\mathbf R$-algebra and $\sqrt[3]{2}$ generates $\mathbf Q(\sqrt[3]{2})$ as a $\mathbf Q$-algebra since $\mathbf C = \mathbf R[i] = \mathbf R + \mathbf R{i}$ and $\mathbf Q(\sqrt[3]{2}) = \mathbf Q[\sqrt[3]{2}] = \mathbf Q + \mathbf Q\sqrt[3]{2} + \mathbf Q\sqrt[3]{2}^2$. But $x$ does not generate $\mathbf R(x)$ as an $\mathbf R$-algebra, since all you get as an $\mathbf R$-algebra with the indeterminate $x$ is $\mathbf R[x]$. The ring $\mathbf R[x]$ is not a field: it does not contain an inverse of $x$, or more generally an inverse of any nonconstant polynomial.
An example with explicit numbers: since $\pi$ is transcendental, the field $\mathbf Q(\pi)$ is not generated by $\pi$ as a $\mathbf Q$-algebra, since $\mathbf Q(\pi) \not= \mathbf Q[\pi]$. If $E/F$ is a field extension and $\alpha \in E$ then $\alpha$ algebraic over $F$ if and only if $F(\alpha) = F[\alpha]$. That last equation is what saying $F(\alpha)$ is generated by $\alpha$ as an $F$-algebra means. Also $\alpha$ is algebraic over $F$ if and only if the field $F(\alpha)$ is finite-dimensional as an $F$-vector space. 
