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I encountered the following passage in Matsumura's Commutative Ring Theory :

A a Noetherian ring, $B=A[[x]]$ a formal power series ring. $M\subset B$ a maximal ideal, $\mathfrak{m}=M\cap A$. Then $(B_{M})^{\mbox{^}}=(A_{\mathfrak{m}})\mbox{^}[[x]]$, where ^ indicate $M$-adic and $\mathfrak{m}$-adic completions, respectively.

It's not immediately clear to me why this is the case. How should I go about proving this? Thanks!

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    $\begingroup$ The universal property of the completion? $\endgroup$
    – Daniel Litt
    Commented Jun 23, 2010 at 15:41
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    $\begingroup$ The content is the assertion that $x \in M$, from which the rest is then clear (by either Litt's suggestion or inspection of constructions). So you should prove that $x$ lies in every maximal ideal (i.e., is in the "Jacobson radical"). There is a well-known criterion for such membership involving units... $\endgroup$
    – Boyarsky
    Commented Jun 23, 2010 at 15:48
  • $\begingroup$ "Daniel" is fine by the way, rather than "Litt." And if the Jacobson radical feels too high-powered, one might proceed by noting that if x is not in M, then it descends to a unit in the quotient (the quotient being a field). But it's easy to see this can't happen. (This is essentially the same, of course, but has essentially no prerequisites.) $\endgroup$
    – Daniel Litt
    Commented Jun 23, 2010 at 17:01
  • $\begingroup$ Daniel: Thanks for fleshing out the argument to which I was alluding in my final sentence. $\endgroup$
    – Boyarsky
    Commented Jun 23, 2010 at 21:31
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    $\begingroup$ @kwan: I meant localization at something other than units. The point is that when computing the completion of the localization at a maximal ideal, there is no need to invert anything new: one just forms the inverse limit of quotients $B/M^n$ (and similarly for $A$). Here we use the evident fact that $B/M^n \rightarrow B_M/(M B_M)^n$ is an isomorphism for maximal $M$ (and similarly for $A$) due to localization commuting with the formation of quotients. So you don't need to invert anything new when doing the calculation, regardless of whether $A$ or $B$ is local. $\endgroup$
    – Boyarsky
    Commented Jun 25, 2010 at 10:01

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