I saw in a textbook the following assertion:
Let $R$ be a commutative ring with unity, and $R[[X]]$ be the ring of power series in one indeterminate $X$. If the homomorphism $\phi∶ R[[X]] \to R$ sending $X$ to $0$ takes a maximal ideal $M$ of $R[[X]]$ to a maximal ideal $\mathfrak{m}$ in $R$, then the two localizations $R[[X]]_M$ and $R_{\mathfrak{m}}[[X]]$ are equal.
I think this problem is some fishy, and would appreciate anyone helping me solve it. Thanks in advance.
If $X\notin M$, then $M+XR[[X]]=R[[X]]$, so one can write $1=u+Xv$ with $u\in M$ and $v\in R[[X]]$. If apply $\phi$ one gets $1=u(0)\in\mathfrak m$, a contradiction. So $X\in M$ and one can easily prove that $M=\mathfrak m[[X]]+XR[[X]]$. This shows that $R[[X]]_M\subseteq R_{\mathfrak m}[[X]]$.
However, the converse is wrong.
Let $R=\Bbb Z$, and $M=2\Bbb Z[[X]]+X\Bbb Z[[X]]$. We get $\mathfrak m=2\Bbb Z$. Set $p_0=3$, $p_1=5$, $p_2=7$ (prime numbers), and so on. Let $f(X)=\frac{1}{p_0}+\frac{1}{p_1}X+\cdots+\frac{1}{p_n}X^n+\cdots\in\Bbb Z_{(2)}[[X]]$. If $f\in\Bbb Z[[X]]_M$ there exists $g\in\Bbb Z[[X]]-M$ such that $f(X)g(X)\in\Bbb Z[[X]]$. Write $g(X)=a_0+a_1X+\cdots a_nX^n+\cdots$ and let's see what we get: $a_0\in\Bbb Z-2\Bbb Z$, $\frac{a_0}{p_0}\in\Bbb Z$, $\frac{a_0}{p_1}+\frac{a_1}{p_0}\in\Bbb Z$, and so on. In particular, $p_1\mid a_0$. Step by step we find $p_n\mid a_0$ for all $n\ge 0$, so $a_0=0$, a contradiction.