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