When I study formal completion and formal schemes, on p.194 of Hartshorne's "Algebraic Geometry", he said "One sees easily that the stalks of the sheaf $\mathcal{O}_{\hat{X}}$ are local rings."
Notice that here $\mathcal{O}_{\hat{X}}$ is not the structure sheaf of X, there is a "hat" on the symbol $X$.
But I can't see the reason for that the stalks of the sheaf $\mathcal{O}_{\hat{X}}$ are local rings.
Could someone explains this for me, thanks.
Here is a self-contained explanation (hopefully without any blunder):
Locally, $\hat{X}$ is an affine formal scheme, so each point has a neighbourhood basis admitting of open sets $U$ admitting the following description: there is a ring $A$, with ideal $I$, such that the underlying topological space is $U_0 :=$ Spec $A/I$, and the structure sheaf is the projective limit of the sheaves $\mathcal O_{U_n},$ where $U_n :=$ Spec $A/I^{n+1}$.
(Note that the underlying topological space of all the $U_n$ coincide, so as topological spaces $$U = U_0 = \cdots = U_n = \cdots .$$ The sheaves $\mathcal O_{U_n}$ are all sheaves on this same underlying topological space, which form a projective system with obvious transition maps, corresponding to the surjections of rings $A/I^{n+1} \to A/I^n.$
Now if $x$ is a point of $\hat{X}$, and if we choose a neighbourhood $U$ of $x$ as above, then the natural map on sheaves $\mathcal O_{\hat{X}} \to \mathcal O_{U_0}$ induces a natural map on stalks $\mathcal O_{\hat{X},x} \to \mathcal O_{U_0,x}$. The target is a local ring. Let $\mathfrak m_x$ denote the preimage in $\mathcal O_{\hat{X},x}$ of the maximal ideal in $\mathcal O_{U_0,x}$; I claim that it is the unique maximal ideal of $\mathcal O_{\hat{X},x}$.
To see this, suppose that $f$ is an element of $\mathcal O_{\hat{X},x}$ which does not lie in $\mathfrak m_x$. Then by definition of the stalk, $f$ extends to a section of $\mathcal O_{\hat{X}}$ over some neighbourhood of $x$, which (by shrinking $U$ as necessary) we may as well assume is our affine neighbourhood $U$. Thus we may think of $f$ as a section of the projective limit of $\mathcal O_{U_n}(U)$, i.e. the projective limit of the rings $A/I^{n+1}$.
The assumption that $f$ is not in $\mathfrak m_x$ says that its image in $\mathcal O_{U_0}(U)$ is not in the maximal ideal at $x$, and so shrinking $U$ further, if necessary, we may assume that $f$ is a unit in $\mathcal O_{U_0}(U) = A/I$.
Thus $f$ is an element of the projective limit of $A/I^{n+1}$ which is a unit in $A/I$. One easily verifies that $f$ is then a unit in every $A/I^{n+1}$, and hence in the projective limit. Thus $f$ is a unit in the ring $\mathcal O_{\hat{X}}(U)$, and so in particular in the stalk $\mathcal O_{\hat{X},x}$.
I've shown that every element of the stalk $\mathcal O_{\hat{X},x}$ not lying in $\mathfrak m_x$ is a unit, which implies that $\mathcal O_{\hat{X},x}$ is local with maximal ideal $\mathfrak m_x$.
