$F[x_1,...,x_n]$, $F(x_1,...,x_n)$, $F(\alpha_1, ..., \alpha_n)$ and $F[\alpha_1, ..., \alpha_n]$ I am having a hard time understanding the difference between $F[x_1,...,x_n]$ and $F[\alpha_1, ..., \alpha_n]$ (let $F$ be a subfield of a field extension $K$). If I understand this correctly, the first is the field $F$ with $n$ variables while the latter is the same field evaluated at $\alpha_1, ..., \alpha_n$.
Then $F(\alpha_1, ..., \alpha_n)$ is also defined as:
$$F(\alpha_1, ..., \alpha_n) = \bigg\{ \frac{\alpha}{\beta} | \alpha,\beta \in F[\alpha_1, ..., \alpha_n], \beta \neq0 \bigg\} $$
Then there is a theorem that states $F(\alpha_1, ..., \alpha_n)$ is the smallest subfield of the field $K$ that contains $F$ and $\alpha_1, ..., \alpha_n$. Why $F[\alpha_1, ..., \alpha_n]$ is not the smallest subfield with those properties?
In genenal, what are the difference between $F[x_1,...,x_n]$, $F(x_1,...,x_n)$, $F[\alpha_1, ..., \alpha_n]$ and $F(\alpha_1, ..., \alpha_n)$? Some concrete examples that shows the differences would be much appreciated. This question helped a bit, but here I am asking for a comparison between all these four fields.
 A: If F is a field, $F[x_1,...,x_n]$ is the ring of polynomials over F (with coefficients in F that is) in the n variables $x_1,...,x_n$, while $F(x_1,...,x_n)$ is defined just like the definition you gave for $F(\alpha_1,...,\alpha_n)$ but with $F[x_1,...,x_n]$ instead of $F[\alpha_1,...,\alpha_n]$. Since $F[\alpha_1,...,\alpha_n]$ is just the ring of polynomials over $\alpha_1,...,\alpha_n$, notice how $F(\alpha_1,...,\alpha_n)$ is just the smallest field containging F and $\alpha_1,...,\alpha_n$ (it is the closure under multiplicative inverses for nonzero elements), while $F[\alpha_1,...,\alpha_n]$ is the smallest ring containing F and $\alpha_1,...,\alpha_n$ (it need not be a field because it might not be closed under multiplicative inverses, e.g. $\frac{1}{x}\notin \mathbb{R}[x]$). But also, sometimes $F[\alpha_1,...,\alpha_n]$ coincides with $F(\alpha_1,...,\alpha_n)$. For instance, consider the ring $\mathbb{Q}(i)$, it can be shown that this is actually the same as $\mathbb{Q}[i]$ (you can take an arbitrary fraction in $\mathbb{Q}(i)$ and multiply by the conjugate of the denominator to show you really just have an element of $\mathbb{Q}[i]$). In particular:
$\mathbb{Q}(i) = \{x+yi\mid x,y\in \mathbb{Q}\}$
replace the above equality with $\sqrt{2}$ and it is still true. Note that always:
$F[x]\subseteq F(x)$,
so $F[x] = F(x)$ iff $F[x]$ is a field (same with $F(x_1,...,x_n)$ and $F[x_1,...,x_n]$).
The other glaring difference between $F[\alpha_1,...,\alpha_n]$ and $F[x_1,...,x_n]$ is that the latter is a polynomial ring in some n variables, and the former is over these elements $\alpha_i$ which are presumed to be in an extension field of F (and not formal indeterminates) (e.g. they could be roots of polynomials over F). Also, the latter (polynomial ring in n variables) will always be a transcendental extension (since the variables won't be roots of nonzero polynomials) while the former might be algebraic or even just F possibly (e.g. $\mathbb{Q}[\sqrt{2}]$ is algebraic over $\mathbb{Q}$, $\mathbb{Q}[7] = \mathbb{Q}$, while $\mathbb{Q}(\pi)$ is transcendental over $\mathbb{Q}$, but $F(x,y)$ is transcendental over F and even over $F(y)$, which is itself transcendental over F).
