Suppose that $x$ is a non-unit in a non-commutative ring $R$ with unity $1$. Is it true that $x$ is contained in some maximal left ideal of $R$. I tried to prove the statement by Zorn's Lemma by defining $\mathcal{L}= \lbrace I \subsetneqq R \,\text{ left ideal} : x\in I \rbrace$. I just need to show that $\mathcal{L} \neq \emptyset$. Since $x$ is non-unit, at least one of $Rx$ and $xR$ is proper. It $Rx$ is proper, then $\mathcal{L} \neq \emptyset$ and we can proceed. But what if $xR$ is proper (and $Rx$ is not guaranteed to be proper) ?!.
I need any help. Thanks in a advance.
The statement as claimed is false. It can happen that $x$ has a left inverse, but no right inverse. For example, if $R=M_{\mathbb N}(\mathbb R)$ is the ring of infinite matricies with coefficients in $\mathbb{R}$, consider the element $$x=\begin{pmatrix}0&0&0&\cdots\\1&0&0&\cdots\\0&1&0&\cdots\\\vdots&\vdots&\vdots&\ddots\end{pmatrix}.$$ Then $$y=\begin{pmatrix}0&1&0&\cdots\\0&0&1&\cdots\\0&0&0&\cdots\\\vdots&\vdots&\vdots&\ddots\end{pmatrix}$$ is a left-inverse of $x$. But if $a\in R$ is arbitrary, we have $$xa=\begin{pmatrix}0&*&*&\cdots\\0&*&*&\cdots\\0&*&*&\cdots\\\vdots&\vdots&\vdots&\ddots\end{pmatrix},$$ so in particular $xa\neq 1$. Hence $x$ is not a unit (since it has no right inverse), but $Rx=R$, so $x$ isn't contained in any left maximal ideal.