Formal power series vs localization at non-constant polynomials

Question

Let $A$ be a commutative ring.

On one hand we have the completion $ A[\![ x ]\!]$, given by the ring of formal power series. Elements are of the form $\sum_k a_kx^k$. The Jacobson radical of $ A[\![ x ]\!]$ consists of power series whose value at zero is in the Jacobson radical of $A$.

On the other hand, we have the localization $A[x]_{1+\langle x\rangle }$ at the polynomials whose value at zero is invertible. Using the notation $(1-xf)^{-1}=\sum_k f^k x^k$ this localization is an $A$-subalgebra of $ A[\![ x ]\!]$. I think its Jacobson radical is generated by polynomials whose value at zero is in the Jacobson radical of $A$.

Geometrically, what is the difference between these constructions? I am thinking of the latter as thickening the affine line by adding an infinitesimal neighborhood of the origin. Is this correct?

The completion seems to thicken the entire line, so perhaps it is best compared with the localization $A[x]_{1+I}$ where $I$ is the ideal of non-constant polynomials?

Answer 1

Here is a concrete example which may be useful to think about. It is not exactly the situation you describe but is similar. Let $k$ be a field of characteristic zero and let $R = k[x, y]/(y^2 - x^3 - x^2)$ be the ring of functions on a nodal cubic, and consider the maximal ideal $m = (x, y)$ of functions vanishing at the origin. Near the origin the nodal cubic crosses itself; loosely speaking, locally (in an analytic sense, e.g. when $k = \mathbb{C}$) it looks like two curves crossing each other. This crossing behavior is not directly visible in the quotient $R/m$, which is too local; it is also not directly visible in the localization $R_m$, which is "not local enough." By "directly visible" I mean the ring of functions doesn't resemble $k[x, y]/xy$, the ring of functions on two copies of the affine line crossing at the origin, and in particular it is still an integral domain so has no zero divisors. Geometrically the localization still remembers enough about the global behavior of the curve to remember that it is irreducible; it hasn't "zoomed in enough" to forget this.

The crossing is directly visible in the completion $\widehat{R}_m = \lim R/m^n$ (I'm not sure what the standard notation is for this). This completion is no longer an integral domain, since

$$x^3 + x^2 = x^2(1 + x) = \left( x \sum_{i \ge 0} { \frac{1}{2} \choose i} x^i \right)^2$$

is now a square, so we can now factor $y^2 - x^2(1 + x)$ as a difference of two squares, corresponding to the two branches of the curve $y = \pm x \sqrt{1 + x}$ which cross at the origin. So this "formal neighborhood" is "smaller" than the "Zariski neighborhood" and in this case it is now small enough that it resembles the "analytic neighborhood."

Answer 2

This seems to be raised (and probably answered) before, but let me give a short summary.

Let $A$ be a commutative ring, and $I\subseteq A$ an ideal. The Zariski localization of $A$ at $I$ is the localization $S^{-1}A$ for the multiplicative subset $S=1+I\subseteq A$. Geometrical, it is the intersection (or limit, more precisely) of all Zariski neighborhoods of $V(I)$.

If we replace Zariski by étale, we get the henselization of $(A,I)$ — the category of étale $A$-algebras $B$ such that the canonical map $A/I\to B/IB$ is an isomorphism could be informally understood as the category of étale neighborhoods of $V(I)$, see Stacks Project Tag 0A02. Geometrically, this is slightly "smaller" than the Zariski localization.

The completion $A_I^\wedge$ is even smaller geometrically — $\operatorname{Spf}(A_I^\wedge)$ is the formal neighborhood of $V(I)$ in $\operatorname{Spec}(A)$.

Answer 3

This is philosophically similar to Qiaochu's answer, but one observation that helps to clarify this for me:

• If we take a local ring $R_{p}$ at a point $p$, this still remembers the fraction field of $R$. So, for example, if $R$ is the functions on an affine curve, then you can determine from $R_{p}$ the genus of the normalization of the closure of the curve. For example, the local rings of points of smooth projective curves with different genuses will never be isomorphic. The local ring knows about all the other points of the curve (which you can think of as valuations on the fraction field), but won't notice if you remove some of them.
• On the other hand, if we take a smooth point on a variety over an algebraically closed field, the completion is always a power series ring. So, from this perspective, all smooth points on curves all look the same. It doesn't know anything about the rest of the curve.