Is it possible to complete a basis for a free module over a finite-dimensional associative unital real algebra?

Question

Let $\mathbb F$ be a finite-dimensional associative unital real algebra. Consider $V:=\mathbb F^n$ and let's say $p \in V$ is good if $xp=0$ only has $x=0$ as solution.

Question: If $p_1$ is good, are there $p_2,\ldots, p_n \in V$ such that $p_1,\ldots,p_n$ is a basis for $V$?

I know the result is true for $\mathbb F$ commutative (because $\mathrm{GL}(n,\mathbb F)$ acts transitively on good points) and for the quaternions.

I believe that the proof of this statement, if true, uses the following lemma: A vector $(x_1,\ldots,x_n) \in \mathbb F^n$ is good if, and only if , $x_1\mathbb F + x_2\mathbb F+\cdots + x_n\mathbb F=\mathbb F$. At least for $\mathbb F$ commutative that is the case. (This is false).

Answer 1

Not in general, no.

Let $\mathbb{F}$ be the algebra of upper triangular $2\times 2$ matrices, let $n=2$, and let $$p_1=(x_1,y_1)=\left(\begin{pmatrix}0&0\\0&1\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix}\right),$$ so that $$\mathbb{F}p_1=\left\{\left( \begin{pmatrix}0&b\\0&d\end{pmatrix},\begin{pmatrix}0&a\\0&0\end{pmatrix}\right): a,b,d\in\mathbb{R}\right\}$$

If $p_2=(x_2,y_2)\in\mathbb{F}^2$, with $p_1,p_2$ a basis for $\mathbb{F}^2$, then $\mathbb{F}p_1\cap\mathbb{F}p_2=\{0\}$. But $$\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\0&d\end{pmatrix} =\begin{pmatrix}0&d\\0&0\end{pmatrix},$$ so for $x=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, $xp_2\in \mathbb{F}p_1$ for every $p_2\in\mathbb{F}^2$.

The way I designed this example is that the answer to the question is no if there is a nonsplit injective (left module) homomorphism $\mathbb{F}\to\mathbb{F}^n$, or equivalently, if there is a nonprojective left $\mathbb{F}$-module $M$ with a free resolution of the form $$0\to\mathbb{F}\to\mathbb{F}^n\to M\to0.$$ For the algebra of upper triangular $2\times 2$ matrices, there is a semisimple module with such a resolution.

By the way, the proposed lemma fails in this example.