Every $2$-dimensional commutative $k$-algebra with only one prime ideal is isomorphic to $k[x]/(x^2)$

This is about the case when $dim_k A=3$.

Consider the $k$-algebras $A_1=k[x]/(x^3)$ and $A_2=k[x,y]/(x^2,xy,y^2)$. As vector spaces, $A_1$ is generated by $1,x,x^2$ and $A_2$ is generated by $1,x,y$ so they're both $3-$dimensional. The prime ideals of $A_1$ and $A_2$ are $(x)/(x^3)$ and $(x,y)/(x^2,xy,y^2)$ respectively (since $text{rad}(x^2,xy,y^2)=(x,y)$, a maximal ideal).

I claim these two algebras are nonisomorphic. To prove it I'll just prove their ideal lattices are nonisomorphic.

First, the ideals of $A_1$ are in a $1-1$ lattice-preserving correspondence (which also preserves prime ideals) with the ideals $(f)$ of $k[x]$ containing $(x^3)$. The generator $f$ of any such ideal must divide $x^3$, so, up to a scalar the choices are $fin {1,x,x^2,x^3}$. This means that the ideal lattice of $A_1$ is just a chain $$begin{matrix} A_1\ uparrow\ (x)/(x^3)\ uparrow\ (x^2)/(x^3)\ uparrow\ 0 end{matrix}$$

Similarly, the ideals of $A_2$ are in a $1-1$ lattice-preserving correspondence (which also preserves prime ideals) with the ideals of $k[x,y]$ containing $I=(x^2,xy,y^2)$. The primary decomposition of $I$ is $$I=(y^2,x)cap (x^2,y).$$ Since $ynotin (y^2,x)$ and $xnotin (x^2,y)$ those ideals form an antichain, which wouldn't be possible if the ideal lattice of $A_2$ were a chain. This is, the ideal lattice of $A_2$ contains the following sublattice: $$begin{matrix} & &(x,y)/I& & \ &nearrow & & nwarrow&\ (x^2,y)/I & & & & (x,y^2)/I\ & nwarrow & & nearrow&\ & &0 & & end{matrix}$$

There's an easier way to approach this. Suppose $A$ is a 2-dimensional $k$-algebra with exactly one prime ideal. There are two possibilities:

1. $0$ is prime (hence $A$ is a domain)
2. The unique maximal ideal $m$ of $A$ is not equal to $0$ (hence $dim_k m = 1$)

In case 1, we have that $A$ is a domain. Therefore, for all $a in A setminus {0}$, $x mapsto ax : A to A$ is an injective $k$-linear map. Since $dim_k A$ is finite, this implies that $x mapsto ax$ is bijective, so $a$ is a unit in $A$. Thus, $A$ is a field, so $A$ is an extension of $k$ of degree $2$. This is impossible because $k$ is algebraically closed!

Since case 1 was impossible, we better find some way to prove that $A$ is isomorphic to $k[x]/(x^2)$ in case 2. In other words, we want to build an isomorphism $k[x]/(x^2) to A$, so in particular we need to define a morphism $k[x]/(x^2) to A$. The "only" way to do this is to use the universal properties of $k[x]$ and quotient rings: given an element $a in A$ such that $a^2 = 0$, $[p] mapsto p(a) : k[x]/(x^2) to A$ is a well-defined $k$-algebra homomorphism, and every $k$-algebra homomorphism $k[x]/(x^2) to A$ can be produced uniquely in this way.

If this is going to be possible, and we do end up proving that $A cong k[x]/(x^2)$ in case 2, it must be the case that the maximal ideal of $A$ corresponds to $(x)$ via this isomorphism. Thus, the easy way to produce an element of $A$ which squares to $0$ should be to pick an element of the maximal ideal $m$! Is there a best element to choose? Well, $0$ is definitely the worst element to choose, since then the induced map $k[x]/(x^2) to A$ is $p mapsto p(0)$, which can't be an isomorphism because its image is $1$-dimensional! However, since in case 2 we have that $dim_k m = 1$, we see that there is a unique choice of nonzero element of $m$, up to scaling (the scaling doesn't matter because $p mapsto p(a)$ and $p mapsto p(xa)$ have the same range for any $x in k setminus {0}$ and $a in A$ – you should absolutely check this if it's not 100% clear why).

So, let's make that choice: let $a in m setminus {0}$ be arbitrary and define $f : k[x]/(x^2) to A$ by $f(p) = p(a)$ (check that this is a well-defined $k$-algebra homomorphism if you're unfamiliar with the aforementioned universal properties). Now we want to show that $f$ is either injective or surjective ($k$-linearity will take care of the rest). Either is easy, but surjectivity is easiest. Since $0 neq 1 notin m$, $0 neq a in m$, and $dim_k A = 2$, we must have that $A = operatorname{span}_k{1,a}$. At the same time, we have $1, a in operatorname{img}(f)$, since $f(1) = 1$ and and $f(x) = a$. $f$ is $k$-linear, so we conclude that $f$ is surjective, and we are done.