Does a finite dimensional algebra over real numbers with no zero divisors necessarily have a multiplicative unit element?
Non-examples:
- If algebra is not finite, it is possible, for example algebra of polynomials with zero constant term (with the usual multiplication), i.e. polynomials of the type $a_1x+a_2x^2+\ldots+a_nx^n$. Multiplying two polynomials of degree $m$ and $n$ results in a polynomial of degree $m+n$, so there are no zero divisors, and there is no unit because constant term is excluded.
- An example of non-unital finite dimensional algebra is cross product. However, not only does it have zero divisors, but every element is a zero divisor in that case, since $x\times x = 0$, $\forall x\in\mathbb R^3$.
Is there any finite non-unital algebra $A$ over $\mathbb R$ such that $ab=0$ implies $a=0$ or $b=0$ and such that there is no $e\in A$ such that $ea=ae=a$, $\forall a\in A$? If yes, what is an example of it? If no, how to prove it?
My attempt:
Multiplication by $a \in A$ is a linear transformation $\mathbb R^n \to \mathbb R^n$. If $a$ is not a zero divisor, then that linear transformation is injective, otherwise, there would exist $b_1,b_2\in A$, such that $b_1\ne b_2$ and $ab_1=ab_2$, but that would imply $a(b_1-b_2)=0$, which contradicts the assumption that $a$ is not a zero divisor, since $b_1-b_2\ne0$. Every injective linear transformation $\mathbb R^n \to \mathbb R^n$ is also surjective and therefore bijective. This means that the linear transformation of multiplying by $a$ has an inverse.
However, just because the inverse transformation exists, it does not by itself seem to imply that it corresponds to multiplication by an element of $A$ (which would then be the inverse element $a^{-1}$ and the unit element would be constructed simply as $e=a^{-1}a$).
