# Properties of Completions

First note that taking an inverse limit is a functor. I won’t need the functorial properties in the immediate future, but it would be good to state some of them. First off, the functor is not exact, but it is left exact. So given an exact sequence of inverse systems $0\to \{A_n\}\to \{B_n\}\to \{C_n\}\to 0$ (it is exact at each level and all the diagrams commute) we get an exact sequence $\displaystyle 0\to \lim_{\longleftarrow}A_n\to \lim_{\longleftarrow}B_n\to \lim_{\longleftarrow} C_n$.

It turns out that if the first system $\{A_n\}$ has the property that the homomorphisms $\theta_n$ are surjective, then the inverse limit is exact. So in our case with completions, this always happens.

The properties I’d really like to prove are the ones listed in Hartshorne without proof. Suppose for the rest of this post that $(R, \frak{m})$ is a Noetherian local ring and $\hat{R}$ is its completion with respect to the $\frak{m}$-adic topology. The numbers will refer to Hartshorne numbering:

5.4A(a) $\hat{R}$ is a local ring with maximal ideal $\hat{\frak{m}}=\frak{m}\hat{R}$ and there is a natural injective homomorphism $R\to \hat{R}$.

We already discussed the second part, since the kernel of the hom is just $\cap \frak{m}^n=0$. Using right exactness of tensoring and exactness of completions, we get that for any finitely generated $R$-module $M$, $\hat{R}\otimes_R M\to \hat{M}$ is an iso (if we remove Noetherian on R, we only get surjective). This gives us that $\hat{R}\otimes_R \frak{m}\to \hat{\frak{m}}$ is an isomorphism and since the image is $\frak{m}\hat{R}$ we get the first part of the statement.

Now we need that it is a unique maximal ideal. But applying the above result to any ideal (which is finitely generated since R is Noetherian) we get that $\widehat{\frak{a}^n}=\frak{a}^n\hat{R}=(\hat{R}\frak{a})^n=(\hat{\frak{a}})^n$. Thus $R/\frak{a}^n\cong \hat{R}/\hat{\frak{a}}^n$. Taking inverse limits gives that $R/\frak{m}\cong \hat{R}/\hat{\frak{m}}$ and hence $\hat{\frak{m}}$ is a maximal ideal since the quotient is a field. But for any $x\in\hat{m}$, we can define an inverse for $(1-x)$ formally by $(1-x)^{-1}=1+x+x^2+\cdots$. Since we are in the completion, this converges in $\hat{R}$ and hence $x\in J(\hat{R})$. But a maximal ideal $\hat{\frak{m}}\subset J(\hat{R})$ means that it is the Jacobson radical and hence the unique maximal ideal.

5.4A (b) If $M$ is a finitely generated $R$-module, its completion $\hat{M}$ is isomorphic to $M\otimes_R \hat{R}$. Well, I needed this to prove (a) and briefly described how it would go, but since I didn’t prove the exactness properties, it seems needlessly detailed to do a full proof using them. For more details, see posts at delta epsilons.

5.4A (c) $\dim R=\dim \hat{R}$.

Let’s use some of the machinery we developed. The dimensions are equal to the degree of the Hilbert polynomial, but $R/\frak{m}\cong \hat{R}/\hat{\frak{m}}$ says precisely that $\chi_m(n)=\chi_{\hat{m}}(n)$. So the polynomials are actually the same.

5.4A (d) $R$ is regular if and only if $\hat{R}$ is regular.

We’ll hold off on this until I cover regularity (which will either be next time or the time after).

1. A typo: You want $R/\frak{m}^n \cong \hat{R}/\hat{\frak{m}^n}$ in 5.4c.