k-algebra of dimension 3 over an algebraically closed field I am trying resolve this problem, and I would like to ask for your help:

$k$ is a algebraically closed field.
  Let $R$ be a (commutative) $k$-algebra with $\dim_kR=3$. Then $R$ is isomorphic to $k[x]/\langle x^3\rangle$ or $k[x,y]/\langle x^2,xy,y^2\rangle$.

So, I though the following:
Since $R$ is a $k$-vector space, due to the Proposition 6.10 of Atiyah-Macdonald's book, then it satisfies the ascending chain condition, the descending chain condition, and also the length of $R$ is $3$.
Then $$\cdots\subset \mathfrak{m}^3\subset \mathfrak{m}^2\subset\mathfrak{m}\subset R$$
ends in $\mathfrak{m}^3=0$.
We also notice that $R$ is artinian, and by the theorem 3.8 of A-M, we can suppose that $R$ is a local ring.
Seeing R as a $k$-vector space, we can see $R=\langle 1,\alpha,\beta\rangle$, then its maximal ideal $\mathfrak{m}$ will have $\dim_k(\mathfrak{m})=1 \text{ or } 2$.
I got a suggestion about analyzing the relations of
$$\alpha^2=a_1+a_2\alpha+a_3\beta$$
$$\beta^2=b_1+b_2\alpha+b_3\beta$$
$$\alpha\beta=c_1+c_2\alpha+c_3\beta$$
to give conditions on the coefficients and find an onto homomorphism $k[x,y]\rightarrow R$ which will allow us to conclude what we need to prove, but I still can't solve it.
Would anybody give some help or a reference to solve this question?
 A: Either suggestion will work, but some of the ring-theoretical constructions are a bit slicker and take less time. 
Decompose $R$ as a ring direct product of local rings and then analyze those cases. We'll do $R$ is a local ring here, but the cases where $R=R_1\times R_2$ or $R=R_1\times R_2\times R_3$ are easier since any local ring which is $2$-dimensional as a $k$-vectors space is isomorphic to $k[x]/(x)^2$ and any local ring which is $1$-dimensional as a $k$-vector space is just $k$.
Firstly, note that the exact sequence of $k$-modules $0\to \mathfrak{m}\to R\to k\to 0$ shows that $\mathfrak{m}$ is two-dimensional. Since this is a map of vector spaces, it is split, which means we have that $R=k\oplus \mathfrak{m}$ as vector spaces. This means we can pick a basis $1,\alpha,\beta$ for $R$ such that $\alpha,\beta$ are in the maximal ideal. This means that $a_1=b_1=c_1=0$ in your relations, as the product of two elements in the maximal ideal is again in the maximal ideal. Now we need some casework depending on whether $\mathfrak{m}^2=0$ or not.
If $\mathfrak{m}^2=0$, pick two linearly independent elements $\alpha,\beta\in\mathfrak{m}$ and define $k[x,y]\to R$ by $x\mapsto \alpha, y\mapsto\beta$. Note that the map is surjective and the kernel is precisely $(x^2,xy,y^2)$ and by the first isomorphism theorem, you are done.
If $\mathfrak{m}^2\neq 0$, we may pick one basis element in $\mathfrak{m}\setminus\mathfrak{m}^2$. Without loss of generality, let it be $\alpha$. I claim the map of rings $k[x]\to R$ by $x\mapsto \alpha$ is surjective with kernel $(x^3)$, which will finish this by the first isomorphism theorem. Clearly $(x^3)\subset \ker$, so the map factors as $k[x]\to k[x]/(x^3)\to R$. Now $k[x]/(x^3)$ and $R$ are both $k$-vector spaces of dimension 3, so it suffices to show that the map is injective, or equivalently takes a basis for $k[x]/(x^3)$ to a basis of $R$. Since $1\mapsto 1$, $x\mapsto \alpha$, and $x^2\mapsto \alpha^2\in \mathfrak{m}^2$ it suffices to show that $\alpha^2$ is linearly independent from $1,\alpha$. But this is clear, as otherwise $\alpha\notin\mathfrak{m}\setminus\mathfrak{m}^2$.
If one were instead to attempt to analyze the relations, you would be able to follow much the same process. Selecting $\alpha,\beta\in\mathfrak{m}$ is equivalent to altering them by constants (send $\alpha\mapsto \alpha+\sqrt{-a_1}$, similarly for $\beta$) to eliminate $a_1,b_1,c_1$. By selecting one of $\alpha,\beta$ to lie in $\mathfrak{m}\setminus\mathfrak{m}^2$, one can eliminate one of $a_2,b_2$, and then after playing around a bit more, one is well on the way to exhibiting the relations $\alpha^3=0$, $\beta=\alpha^2$ or $\alpha^2=\beta^2=\alpha\beta=0$.
