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Let $A$ be a local ring with maximal ideal $m$ that is $m$-adically complete, and assume $1/2 \in A^\times$. I've read in several places that for any $x \in m$, a square root of $1 + x$ in $A$ is given by the binomial series

$$ \sum_{n = 0}^\infty {{1/2} \choose n} x^n, $$ where ${{1/2} \choose n} = \frac{(1/2)(1/2 - 1) \cdots (1/2 - n + 1)}{n!}$. I don't understand why these binomial coefficients make sense in $A$. We have $n!$ in the denominator, so why is $n! \in A^\times$?

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    A ratio being in $A$ does *not* require that the denominator has to be a unit. For example, 15/5 is an integer even though 5 is not a unit in the integers. Or, to give an example with a local ring, 15/5 is in ${\mathbf Z}_5$ (the 5-adic integers) even though 5 is not a unit in ${\mathbf Z}_5$. The point is that the rational number $\binom{1/2}{n}$ in reduced form has a denominator that is a power of 2, so it makes sense in $A$. See http://math.stackexchange.com/questions/136206/show-that-sqrt1t-lies-in-mathbbz1-2-t.2012-04-29
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    Ah yes, I see. Thanks!2012-04-29
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    @KCd Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-25
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    Julian: You could post my comment as an answer if you wish.2013-06-25

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