Is it true that for a ring $R$ and a noninvertible element $x\neq 0$ in it, can always find a ring $R'$, such that $R\leqslant R'$ and $x$ is invertible in $R'$ ?
If $R$ is an integral domain, the answer is of course that $R'$ is the field of fractions. But what happens, if $R$ is "less" than an integral domain ?
No. Suppose $x$ is a zero-divisor in $R$, i.e. there's a $y \neq 0$ such that $xy=0$. Then $x$ cannot be invertible in any overring $R'$: if it was, we'd have $x^{-1}xy=0$ in $R'$, and so $y=0$.
If $x$ is a regular element, meaning that multiplication by $x$ on $R$ is injective, i.e., $xr=0$ implies $r=0$, then $R$ injects into the localization $R^\prime=R[x^{-1}]$. On the other hand, if $x$ is not a regular element, then $R\rightarrow R[x^{-1}]$ is not injective, and if $R^\prime$ is any $R$-algebra such that the image of $x$ in $R^\prime$ is a unit, $R\rightarrow R^\prime$ factors through $R\rightarrow R[x^{-1}]$, and therefore is not injective.
So, it's possible if and only if $x$ is a regular element.
EDIT: I should add that I'm assuming $R$ is commutative.
This answer follows from general properties of localizations, but that's overkill. First, recall that a zero-divisor $\rm\:s\:$ is never a unit since $\rm\:rs = 0\:$ times $\rm\:s^{-1}$ yields $\rm\:r = 0.\:$ Thus if $\rm\:s\:$ is invertible in an extension ring it must be a non-zero-divisor. If so, then $\rm\:s\:$ is invertible in the ring $\rm\:R[x]/(s\,x -\!1)\:$ which naturally is an extension ring of $\rm\:R\:$ since the natural map $\rm\:r \to r + (s\,x-\!1)R[x]\:$ is an embedding: if it had nonzero kernel then $\rm\:s\,x-\!1\:$ would divide some nonzero constant $\rm\:r\in R,\:$ say $\rm\: (s\,x -\!1)\,f(x) = r\:$ in $\rm\:R[x],\:$ contra LHS has degree $\,> 0\,$ since $\rm\,s\,$ is not a zero-divisor.
