How do we know that $F[\alpha]$ is a ring? Note that this question is not about the proposition $\alpha$ is algebraic over $F \iff F[\alpha] = F(\alpha)$
I've got the following note in my textbook (given without proof, suggesting the statement is rather trivial), but I've been having some difficulty with it:
Given field $F$, extension field $L/F$ and $\alpha\in L$ then $F[\alpha] = \{a_0 + \alpha a_1 + \cdots + \alpha^m a_m:a_1,\ldots,a_m \in F\}$ is the smallest subring of $L$ containing $\alpha$.
If this is indeed a ring, I agree that it is the smallest, but how do we know that this is indeed a ring? If we have some finite ring $R$ with $m$ elements, then $\alpha^{m+1} \not \in F[\alpha]$ (or so it seems to me), and thus $F[\alpha]$ isn't closed.
Thanks
 A: There is an implicit quantification over $m$ here; it might be clearer to write $$F[\alpha]=\{a_0+\alpha a_1+\alpha^2a_2+...+\alpha^ma_m: \color{red}{m\in\mathbb{N}}, a_1,...,a_m\in F\}.$$ There are three key points here:

*

*$m$ is not fixed, and elements of $F[\alpha]$ are allowed to be "arbitrarily (finitely) long."


*There is no requirement that the $a_i$s be distinct.


*Different sequences may name the same element of $L$ (and so of $F[\alpha]$); more abstractly, the obvious homomorphism $F[X]\rightarrow F[\alpha]$ need not be injective.
This should eliminate your concerns about running out of elements if $F$ (or even $L$) is finite. For example, if $F=\mathbb{F}_2$, $L=\mathbb{F}_4$ (where for prime $p$ $\mathbb{F}_{p^n}$ is the unique-up-to-isomorphism finite field of cardinality $p^n$ - note that $\mathbb{F}_2$ sits inside $\mathbb{F}_4$ in a unique way), and $\alpha\in L\setminus F$, one of the elements of $F[\alpha]$ is the sum $$\gamma:=1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6+\alpha^7.$$ This element $\gamma$ is defined by a sequence of length longer than even the size of $L$, but that's not a problem; it still is an element of $F[\alpha]$.
