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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.

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Just try some examples. For example try the $\mathbb{Q}$-vector space $\{a+b\sqrt{2}: a,b\in\mathbb{Q}\}$. Why is this a division ring (actually a field, but in any case, you just need to prove existence of inverses)? –  Alex B. Oct 13 '11 at 0:09
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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?

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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.

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