I am currently working with an $F$-algebra $A$ with basis $\{1,a\}$ and product $a^2=0$ (here $1$ is the identity element). This algebra is commutative and associative and has a nonzero nilpotent element $a$. I wish to apply the Wedderburn-Artin theorem to express $A$ as a direct sum.
First, there are only two algebras of dimension 1 over the base field $F$, those are $F'=Fn, n^2=0$ and $F''=Fe, e^2=e$. Since the subspace generated by $a$ is an ideal of $A$ the quotient $A/\langle a\rangle$ is semissimple and by the mentioned theorem it is a sum of matrix algebras over division rings. That matrix algebra can only be the underlying feld as it would have dimension greater than 1 otherwise, in other words $A/\langle a\rangle\simeq F$. Since $\langle a \rangle \simeq F'$ (as it is nilpotent) we would have $A\simeq F\oplus F'$.
A table may also be constructed by taking a basis $\{f,f'\},f\in F,f'\in F'$ and the products: $ff'=f'f=0, (f')^2=0,f^2=f$. Here a problem arises as this table describes an algebra that is not unital. Clearly something wrong has been done by me at some point and I'm looking for guidance.
Extra: Going the other way around, if I would avoid the use of A-W theorem I could just note that $A\simeq\frac{A}{\langle a\rangle}\oplus \langle a\rangle$ and check what can each factor be isomorphic to. I would stumble at the very same issue, this algebra is not unital. Perhaps they are unital and I just got it wrongly.
Best Answer
The algebra $A$ is not semisimple. Your identity $A\cong A/\langle a\rangle \oplus \langle a\rangle$ is not true, as you have essentially proven.
As an aside: you should decompose unital rings as direct products, not as direct sums. The difference is that a direct sum should be endowed with canonical maps from the constituent pieces to the big thing, while a direct product comes with canonical maps from the big thing to the constituent pieces. In the world of rings, the obvious inclusions from the constituent pieces to the big thing do not send $1$ to $1$, so they are not ring homomorphisms. The projection maps, on the other hand, are ring homomorphisms.
Armed with this you can try to revisit your ring again. If it were a direct product of two quotient rings, then you would have to exhibit two corresponding ideals by which you are taking the quotients, which would have to intersect trivially and have sum equal to $A$, and these just do not exist.