Must nonunit in group algebra of free group generate proper two-sided ideal? Let $F$ be a free group and $k$ be a field.  If $x$ is an element of
the group algebra $k[F]$ that is not a unit (equivalently, that is not
a nonzero scalar multiple of an element of $F$), must the 2-sided ideal
$k[F]\,x\,k[F]$ generated by $x$ be proper?  In other words, is it true
that for all $y_1,\dots, y_n, z_1,\dots, z_n \in k[F]$, we have
$$y_1 x z_1 + y_2 x z_2 +\dots + y_n x z_n \neq 1 ?$$
This question was asked by George Bergman.  For the application he has
in mind, it would actually be enough to show that if $a$ is one of
the free generators of $F$, and $x$ is an element of $k[F]$ whose support 
consists of elements of $F$ having total degree $0$ in $a$, then
$1\notin k[a]\,x\,k[F]$.  (For example, $x$ might be $b + abaca^{-2}$, where
$b$ and $c$ are two other members of a free generating set for $F$.) 
Equivalently, he would like to know that the right ideal of
$k[F]$ generated by the conjugates of $x$ by the non-negative powers of $a$
is a proper ideal. 
 A: For one-sided ideals it's true (see for example Lemma 5.9 in this paper). The following is an unsuccessful attempt to handle two-sided ideals in the same method:
A free group is bi-orderable, so let $<$ be a bi-order on $F$. 
(It means $w_1 < w_2\implies w_1 u < w_2 u, \,\,\,u w_1 < u w_2$ for every $w_1, w_2, u\in F$.
See this post for details).
Assume now we have a non-unit: $x = \sum_{j=1}^d \lambda_j x_j$ for $\lambda_j\in k-\{0\}, \,\,x_1 < \ldots < x_d\in F$ with $d\ge 2$, and some
$c_i\in k,\,\,\, y_i, z_i\in F$ such that
$$1 
= \sum_{i=1}^n c_i y_i x z_i 
= \sum_{i=1}^n \sum_{j=1}^d c_i \lambda_j \cdot y_i x_j z_i. $$
Let $i_{\max}, i_{\min}$ be such that
$$y_{i_{\min}} x_1 z_{i_{\min}} = \min\left(\left\{ y_i x_1 z_i \right\}_{i=1}^n\right), \,\,\,
y_{i_{\max}} x_d z_{i_{\max}} = \max\left(\left\{ y_i x_d z_i \right\}_{i=1}^n\right).$$
Then for every $j\neq 1$ and every $i$ we have
$$y_{i_{\min}} x_1 z_{i_{\min}} \le y_{i} x_1 z_{i} < y_{i} x_j z_{i}$$
 so the total coefficient of $y_{i_{\min}} x_1 z_{i_{\min}}$ is non-zero, and similarly for $y_{i_{\max}} x_d z_{i_{\max}}$.
Thus the sum cannot be only $1$ - a contradiction. 
I still believe this kind of argument should work, but as is it doesn't suffice (see YCor's comment).
