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.



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> > **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 $V$?**
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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).



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


