Finite-Dimensional Integral Domain over Field is Field
Theorem
Let $k$ be a field.
Let $R$ be a $k$-algebra of finite dimension which is an integral domain.
Then $R$ is a field.
Proof
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We will prove this using the Rank - Nullity Theorem applied to $A$ viewing it as a finite dimensional vector space over $K$. Pick a non-zero element $y \in A$ and consider the $K$ - linear transformation $$\begin{eqnarray*} T&\colon&A \longrightarrow A \\ && x\mapsto yx.\end{eqnarray*} $$
Now the kernel of this linear transformation is trivial because $A$ is an integral domain, and hence by rank-nullity we have that
$$\begin{eqnarray*} \dim_K A &=& \dim \ker T + \dim \operatorname{im} T \\ &=& 0 + \dim \operatorname{im} T\\ &=& \dim \operatorname{im} T \end{eqnarray*}$$
from which it follows that $T$ is surjective. In particular there exists $z \in A$ such that $zy = 1$, so that since $y$ was arbitrary we have shown that every element in $A$ is invertible, i.e. $A$ is a field.
- Edit:** I should say it is important to include the hypothesis that $A$ is an integral domain otherwise the result is not necessarily true. Consider the $\Bbb{C}$ - algebra
$$\Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C} $$
as a finite dimensional vector space over $\Bbb{C}$. This is not an integral domain because $(1,0,0) \cdot (0,1,0) = (0,0,0)$. Now we also have
$$\dim_{\Bbb{C}} \big( \Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C} \big) = 3 < \infty$$
but $\Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C}$ cannot be a field because elements like $(1,0,0)$ don't have an inverse.
If we also drop the hypothesis that $\dim_K A < \infty$ the result is not true as well. For example consider the polynomial ring $\Bbb{C}[T]$ in the indeterminate $T$. This is an integral domain because $\Bbb{C}$ is. Then if we view this as a vector space over $\Bbb{C}$ it is infinite dimensional, but it is already a well known fact that $\Bbb{C}[T]$ is not a field.
$\blacksquare$
Also see
Sources
user38268 (https://math.stackexchange.com/users/38268/user38268), If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field, URL (version: 2012-05-23): https://math.stackexchange.com/q/146859