Is $x$ a unit if it is a unit in every localization at a maximal ideal?

Question

Let $R$ be a commutative unital ring. For every maximal ideal $\mathfrak{m}$ of $R$ there is a localization map \begin{eqnarray} l_\mathfrak{m}: R \to R_\mathfrak{m}\\ x\mapsto \frac{x}{1}. \end{eqnarray} Is it true that an element $x\in R$ is a unit if $l_\mathfrak{m}(x)$ is a unit in $R_\mathfrak{m}$ for every maximal ideal $\mathfrak{m}$?

Answer 1

If $x$ isn't a unit, then $x\in\mathfrak{m}$ for some maximal ideal. Could then $x/1\in R_{\mathfrak m}$ be a unit?

Answer 2

If $x$ is a unit, then $x_m$ is a unit too with inverse $x_m^{-1}$.

Assume $x$ is not a unit. Let $m$ be a maximal ideal that contains $x$. Such a maximal ideal exists because the ideal generated by $x$ is proper, and we can use Zorn's lemma on the set of proper ideals that contain $x$.

Then $x_m$ belongs to the maximal ideal of $R_m$. Therefore it cannot be invertible.

Answer 3

Suppose $x$ is not a unit, then $x$ is contained in a maximal ideal $M$, consider the localization $R_M$, $x$ is not invertible in $R_M$, otherwise you have $x{a\over b}=1$ in $R_M$, and $xa-b\in M$ this implies $b\in M$ contradiction

Answer 4

Since $x$ locally unit, we have $<x>R_m=R_m$ for every maximal ideal of $R$. Hence, $<x>=R $ and so $x$ is unite in $R$.