Is every element of a finite dimensional commutative non-unital $\mathbb{R}$-algebra nilpotent?

Considering a few examples of finite dimensional non-unital algebras over the reals, I tried coming up with an example of such an algebra with non-nilpotent zero divisor elements. In all the examples I came up with to try to construct such an algebra, everything turned out to actually be unital.

For example, if we consider the free commutative $\mathbb{R}$-algebra over the generators $\alpha, \beta$ mod the ideal generated by the relations $\alpha^2=0,\beta^2=0,\alpha^2=2\alpha, \beta^2 = 2\beta$ in an attempt to create an algebra where $\alpha, \beta$ have infinite order, and hence are not nilpotent, but $\alpha \beta = 0$ so they are zero divisors, it turns out this algebra is in fact unital.

Since it is finite dimensional, we can of course consider such an algebra $\mathcal{A}$ to be a real matrix algebra, so I thought the fact that every singular matrix is the product of nilpotent matrices would help me prove such an assertion, but the issue is that such a product of nilpotent matrices do not have to be from $\mathcal{A}$ itself, and thus may not be commutative, meaning that the product itself which is in $\mathcal{A}$ may not be nilpotent.

Are there any counterexamples to my claim that every element of a finite dimensional commutative non-unital $\mathbb{R}$-algebra is nilpotent? If not, how might one prove that this property holds for all finite dimensional commutative non-unital $\mathbb{R}$-algebras?

Let $R=\Bbb R[x]/(x^2)$, and $I$ be the ideal $(x)\lhd R$. Then $(x)\times R$ has dimension $3$ as a real vector space.
• @Najibldrissi $\mathbb{R} \times \mathbb{R}$ is unital. – Nathan BeDell Jun 15 '16 at 17:06
• @NajibIdrissi The last time I checked, $\Bbb R\times \Bbb R$ was a unital algebra... But perhaps you are thinking of some other operation. – rschwieb Jun 15 '16 at 17:07
• Woops. Sorry.${}$ – Najib Idrissi Jun 15 '16 at 17:37