Proving $R$ is a division ring Theorem says that 
Suppose that $R$ is a domain that is an algebra over a subfield $k$. Assume that $R$ is finite dimensional $k$-vector space. Prove that $R$ is a division ring. 
I suppose should start by saying $R=\{x_1,x_2,...,x_n\}$ and I can assume $1$ is in $R$ from the definition of a ring given in the course. 
Do I need to use the fact that $R$ is a vector space?
As I seriously don't know where to start. I'm looking at the axioms of $k$-vector space and $k$-algebra and can't seem to get this. Vector spaces are my weakness. 
 A: Every k-algebra is necessarily a vector space, so that doesn't in itself tell you anything. What you need is that it is finite-dimensional when viewed as a vector space.
Under no circumstances should you say $R=\{x_1,x_2,\ldots,x_n\}$ -- nothing guarantees you that $R$ has a finite number of elements. That's something quite different from having finite dimension. Having finite dimensional means that $R$ is isomorphic as a vector space to $k^n$ for some $n$. So another way to pose the problem would be:

Let $k$ be a field, and assume that some binary operation $*$ on $k^n$ is given that makes $k^n$ into a $k$-algebra. Suppose also that $(k^n,+,*)$ viewed as a ring is an integrity domain. Show $k^n$ is actually a division ring.

Hint: For $a\in k^n\setminus\{0\}$, consider the mapping $T: b\in k^n\mapsto a*b$. This is a linear operator on $k^n$ when $k^n$ is considered a vector space (show this). What more can you say about $T$ given the assumptions?
A: Yes, you need to use the fact that $R$ is a vector space. In particular, you need to use the fact that $R$ is a finite dimensional vector space. Given an element $a \in R-${$0$} you can define a $k$-linear map $f_a: A \to A$ by left multiplication i.e. $f_a(b) = ab$. The fact that $A$ is a domain tells you that ker($f$) = {$0$} and so by the rank-nullity theorem, $f$ is surjective meaning $a$ has a right inverse. Similarly, one can show that $a$ has a left inverse and that the two must be equal.    
A: Here's how I would do it:  we need to show every $\alpha \in R \setminus k$ has a multiplicative inverse, that is, 
$\exists \beta \in R \mid \beta \alpha = \alpha \beta = 1; \tag 1$
we note in passing that we must have $\beta \in R \setminus k$; otherwise, $\alpha = \beta^{-1} \in k$.  In any event, the hypothesis that $R$ is finite-dimensional over $k$ implies that the sequence of elements of $R$
$\{\alpha^i  \mid i \in \Bbb N \} = \{ 1, \alpha, \alpha^2, \ldots, \alpha^m, \ldots, \} \tag 2$
must at some point become linearly dependent; that is, for some $n \in \Bbb N$ there are $p_0, p_1, p_2, \ldots, p_n \in k$ with
$\sum_0^n p_i \alpha^i = 0; \tag 3$
we set
$k[x] \ni p(x) = \sum_0^n p_i x^i; \tag 4$
then 
$p(\alpha) = 0. \tag 5$
We can in fact take $\deg p(x)$ to be minimal; that is, we can assume that there is no $q(x) \in k[x]$, $\deg q(x) < \deg p(x)$ and $q(\alpha) = 0$.  We also have $n = \deg p(x) \ge 2$, lest $\alpha \in k$.
I claim that for such $p(x)$, $p_0 \ne 0$; otherwise, we have $p_1(x) \in k[x]$ with $p(x) = xp_1(x)$; indeed,
$p_1(x) = \sum_1^n p_i x^{i - 1}; \tag 6$
then
$\alpha p_1(\alpha) = p(\alpha) = 0, \tag 7$
and since $0 \ne \alpha \in R \setminus k$, and $R$ is a domain, we conclude that
$p_1(\alpha) = 0, \tag 8$
which contradicts our assumption that $p(x)$ is of minimal degree.  So, since now we have $p_0 \ne 0$, and
$p(x) = xp_1(x) + p_0, \tag 9$
we see that
$\alpha p_1(\alpha) = -p_0, \tag{10}$
whence
$\alpha(-p_0^{-1}p_1(\alpha)) = (-p_0^{-1}p_1(\alpha)) \alpha = 1; \tag{11}$
thus
$\alpha^{-1} = -p_0^{-1}p_1(\alpha),\tag{12}$
and $R$ is a division ring.
NB:  Note I've used the fact that $R$ is a finite-dimensional vector space over $k$ (see ca. (2)-(3)); I don't see how this can be avoided.
End of Note.
