TITLE: Artin's Algebra Exercise: Characterize all rings $R$ which contains $\mathbb C$ and has dimension $2$ as a vector space over $\mathbb C$.
QUESTION [5 upvotes]: Here is a problem from Artin's Algebra text:

Characterize all rings $R$ which contains $\mathbb C$ and has dimension $2$ as a vector space over $\mathbb C$.

Here is my idea: Suppose $\{1, \alpha\}$ is a basis of $R$ as a $\mathbb C$-vectorspace, then as $\alpha^2 \in R$ so there exist $a,b \in \mathbb C$ such that $\alpha^2=a\alpha + b$. Now the whole multiplication is defined, since $(s\alpha + t)(u\alpha+v) = su\alpha^2 + (sv+tu)\alpha + tv = (sua + sv + tu)\alpha + (sub + tv)$, for generic elements $s\alpha + t, u\alpha+v \in R$. Such a ring is clearly commutative. Is this solution correct ? Moreover,

What can be said more about ring $R$ ?
  How can we find all rings which contains $\mathbb C$ has dimensions $n$ as a vector space Over $\mathbb C$ ? Any ideas?

REPLY [4 votes]: Following Jyrki Lahtonen hint, here is my solution:
Let's work with an arbitrary field. Let $k$ be any field. If $R$ is a $2$-dimensional $k$-algebra, it must have a basis $\{ 1, x \}$ where $x$ is not a scalar multiple of $1$; in particular, it must be commutative, since it is generated by $x$. Since it is $2$-dimensional, $x^2 = ax + b$ for some $a, b \in k$, and it follows that $R \cong k[x]/(x^2 - ax - b)$. The isomorphism type of $R$ is now controlled by the possible types of monic quadratic polynomials over $k$. There are three cases:

$x^2 - ax - b$ has two distinct roots in $k$. In this case $R \cong k \times k$ by the Chinese remainder theorem.
$x^2 - ax - b$ is irreducible over $k$. In this case $R$ is a quadratic field extension of $k$.
$x^2 - ax - b$ has two repeated roots in $k$. In this case $R \cong k[x]/x^2$. 

If $k$ is algebraically closed then the second case never occurs and we conclude that there are exactly two isomorphism types of $2$-dimensional $k$-algebras.