$F[a] \subseteq F(a)?$ I think this is probably an easy question, but I'd just like to check that I'm looking at it the right way.

Let $F$ be a field, and let $f(x) \in F[x]$ have a zero $a$ in some extension field $E$ of $F$. Define $F[a] = \left\{ f(a)\ |\ f(x) \in F[x] \right\}$. Then $F[a]\subseteq F(a)$.

The way I see this is that $F(a)$ contains all elements of the form $c_0 + c_1a + c_2a^2 + \cdots + c_na^n + \cdots$ ($c_i \in F$), hence it contains $F[a]$. Is that the "obvious" reason $F[a]$ is in $F(a)$?
And by the way, is $F[a]$ standard notation for the set just defined?
 A: (1) Yes, you are correct. Note that $F(a)=\{\frac{f(a)}{g(a)}:f,g\in F[x], g(a)\neq 0\}$; in other words, $F(a)$ is the field of fractions of $F[a]$ and therefore certainly contains $F[a]$. 
(2) Yes, the notation $F[a]$ is standard for the set you described.
Exercise 1: Prove that if $a$ is algebraic over $F$, then $F[a]=F(a)$. (Hint: prove first that $\frac{1}{a}\in F[a]$ (if $a\neq 0$) using an algeraic equation of minimal degree of $a$ over $F$.)
Exercise 2: Prove that if $a$ is transcendental over $F$, then $F[a]\neq F(a)$. (Hint: Prove that $F[a]\cong F[x]$ where $F[x]$ denotes the polynomial ring in the variable $x$ over $F$. Note that $F[x]$ is never a field if $F$ is a field.)
A: Yes, these are standard notations: given some fields $E\supseteq F$ and some $a\in E$,
$$F(a)=\left\{\left.\frac{f(a)}{g(a)}\right\vert\,\,\, f,g\in F[x], g(a)\neq0\right\}$$
$$F[a]=\left\{\left.f(a)\,\,\right\vert\,\,\, f\in F[x]\right\}$$
Because any $f(a)=\frac{f(a)}{g(a)}$ where $g=1$, we have that $F[a]\subset F(a)$ (this is just a restatement of your argument). 
A: My preferred definition for $F[a]$ is the smallest ring that contains $F$ and $a$ and for $F(a)$ is the smallest field that contains $F$ and $a$. With these definitions, the inclusion is clear, because every field is a ring. The concrete descriptions in terms of polynomial and rational functions are then theorems of course.
