Let $k$ be a field and let $A$ be a non-unital, commutative algebra over $k$. Suppose that the only ideals of $A$ are $A$ itself and $\{0\}$. The task is to show that $A$ is isomorphic to the one-dimensional $k$-vector space with zero multiplication.
Of course, $A^2$ is an ideal of $A$, so we have to show that $A^2 = \{0\}$ ruling out the case where $A^2 = A$.
Any hints appreciated.
If $A^2\neq \{0\}$, there must be an $a\neq 0$ such that $(a)^2=(a)=A$.
We can adjoin an identity to $A$ in such a way that $A$ is an ideal of the unital ring $A'$. Now that we have a ring with identity, we can pull a trick to show that $A$ has identity.
In $A'$, $A$ is a finitely generated idempotent ideal. By Nakayama's lemma, there exists an $x\in A$ such that $(1−x)A=\{0\}$. This means that $x$ is an identity for $A$, but that was precluded in the hypotheses. Thus it's not possible for $A^2=A$.