I have been thinking that a field $F$ and then adding in an element $\alpha \in E$ ($E$ is some larger field properly containing $F$) was denoted by $F(\alpha)$.
It seems like $F(\alpha)$ is more specifically a field of fractions where each element is a ratio of polynomials (from $F[X]$) that may involve $\alpha$. For example, $\frac{2\alpha^2 + \alpha}{\alpha^3}$ (assuming $1,2 \in F$) could be an element of $F(\alpha)$.
This somehow results in $F(\alpha)$ being the smallest field that contains $F$ and $\alpha$.
Furthermore, there is notation such as $F(\alpha, \beta)$ and $F(\alpha)(\beta)$ that to me, looks to me would result in the exact same thing.
So (assuming I am roughly correct so far), why do we start with a polynomial ring of $F$, then turn it into a field of fractions?
Why does a field of fractions result in the smallest ring containing $F$ and $\alpha$ (or, what is an intuitive explanation as to how I could show it is the smallest)?
What is an example of a result of $F(\alpha, \beta)$ being different from $F(\alpha)(\beta)$?
Best Answer
Let's make it clear. Suppose $E/F$ is a field extension, $S\subseteq E$ is a subset. (might be an infinite subset as well) Then $F(S)$ is the smallest subfield of $E$ which contains the field $F$ and the set $S$. The meaning is smallest by inclusion, i.e if $K$ is a subfield of $E$ which contains $F$ and $S$ then $F(S)\subseteq K$. Now how do we know such smallest field really exists? Because we can take all the subfields of $E$ which contain $F$ and $S$ and intersect them. It is easy to check that the intersection is a subfield which contains $F$ and $S$, and obviously it is the smallest with such property.
Ok, so we know $F(S)$ really exists. However, the method I described (taking the intersection of all such subfields) doesn't give us much information about how the elements of $F(S)$ look like. So we would like to find another way to describe this field. And it really can be described like that:
$K=\{\frac{p(s_1,...,s_n)}{q(t_1,...,t_m)}: n,m\in\mathbb{N}, p\in F[x_1,...,x_n], q\in F[x_1,...,x_m]\, s_1,...,s_n,t_1,...,t_m\in S , q(t_1,...,t_m)\ne 0\}$
It can be checked that this set is a subfield of $E$ which contains $F$ and $S$. Hence $F(S)\subseteq K$. Now why is $K$ exactly $F(S)$? Because any field which contains $F$ and $S$ must be closed under sums, products and fractions of their elements, i.e any such field must contain all the elements from $K$.
What I described is the general case, now $F(\alpha)$ is when we take $S=\{\alpha\}$, a set with one element. And $F(\alpha,\beta)$ is when we take $S=\{\alpha,\beta\}$. Note that $F(\alpha,\beta)$ and $F(\alpha)(\beta)$ is really the same thing. Let's show this.
$F(\alpha)(\beta)$ by definition contains $F(\alpha)$ and the element $\beta$. From here it follows that it contains the field $F$ and the elements $\alpha,\beta$. So $F(\alpha,\beta)\subseteq F(\alpha)(\beta)$. On the other hand, $F(\alpha,\beta)$ is a field which contains the field $F$ and the element $\alpha$, hence $F(\alpha)\subseteq F(\alpha,\beta)$. But it also contains the element $\beta$, so $F(\alpha)(\beta)\subseteq F(\alpha,\beta)$.