Does every finite dimensional real nil algebra admit a multiplicative basis? We say that a finite dimensional real commutative and associative algebra $\mathcal{A}$ is nil if every element $a \in \mathcal{A}$ is nilpotent. 
By multiplicative basis, I mean a basis $\{ v_1, \dots , v_n \}$ for $\mathcal{A}$ as a real vector space such that for each $v_i$ and $v_j$, the algebra multiplication $v_i \star v_j = c v_k$ for some $c \in \mathbb{R}$, and some other element $v_k$ of the basis.
Given such a nil algebra $\mathcal{A}$, does it always admit a multiplicative basis in the sense described above? If not, what is an example of a nil algebra which does not admit a multiplicative basis?
 A: No. The following constructs a counterexample.
Let $R$ be the graded ring $\mathbb{R}[x_1, \ldots, x_5]$ and $I = (x_1, \ldots, x_5)$. In any homgeneous degree $d$, define the "pure" polynomials to be the the products of $d$ linear polynomials, and let the rank of a homogeneous polynomial $f$ be the minimum number of terms needed to express $f$ as a sum of pure polynomials.
I assert the following:


*

*The grade one piece $R_1$ is isomorphic to the vector space of $1 \times 5$ matrices

*The grade two piece $R_2$ is isomorphic to the vector space of symmetric $5 \times 5$ matrices

*The product $R_1 \times R_1 \to R_2$  corresponds to the symmetrized outer product $(v,w) \mapsto \frac{1}{2}(v^Tw + w^Tv)$


(phrased differently: $R_1$ is the space of linear forms, and $R_2$ is the space of symmetric bilinear forms)
The rank of a matrix has a similar characterization: $\text{rank}(A)$ is the smallest number of terms you need to express $A$ as a sum of outer products $\sum_i v_i^T w_i$.
Of particular note is that if a homogeneous quadratic polynomial $f$ corresponds to the matrix $A$, then $\text{rank}(A) \leq 2 \text{rank}(f)$.
Consequently, there exists a homogeneous quadratic polynomial $f$ such that $\text{rank}(f) \geq 3$. One such example is $f = \sum_i x_i^2$.
Now, consider the graded algebra $A = I / (I^3 + fR)$. Its grade 1 piece is 5-dimensional and its grade 2 piece is 14-dimensional.
Suppose we have a collection of polynomials of $I$ that form a multiplicative basis for $A$. The basis must consist of at least five polynomials that span $I/I^2$. There are 15 products of pairs of these polynomials, and they are all distinct elements of $I^2 / I^3$.
Suppose two of these products were the same in $A$. That would imply we have two rank one quadratic polynomials $g$ and $h$ with the property that $rg = sh + tf$ for some scalars $r,s,t$. However, we would have $rt^{-1}g + (-st^{-1})h = f$ which is impossible, because the left hand side has rank at most 2, but the right hand side has rank 3.
Consequently, the 15 pairwise products of the multiplicative basis for $A$ are all distinct (and nonzero) elements of $A$, and they are distinct from the original $5$ polynomials as well. Consequently, the basis must have at least 20 elements, contradicting the fact that $A$ is 19-dimensional.
