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Let $k$ be a field and $A = k[[x_1, \dots, x_n ]]$ be the ring of formal series in $n$ variables. Consider $g_1, \dots, g_m \in A$ such that $g_1(0) = \cdots = g_m(0) = 0$. For every $f \in k[[t_1, \dots, t_m]]$ we can consider the formal series $f(g_1, \dots, g_m) \in A$ because the $g_i$'s have no constant term. Now consider the subring $B$ of $A$ made up of series of the form $f(g_1, \dots, g_m)$ as $f$ varies in $k[[t_1, \dots,t_m]]$. It is clear that $B$ is a noetherian local domain with $\dim B \leq m$.

(1) What are the main properties of the ring extension $B \subseteq A$? When is $A$ flat over $B$?

(2) After reading Problem 3 at page 115 of Milnor's Singular points of complex hypersurfaces, I have made the following conjecture that is an algebraization of the book's problem.

Conjecture. If $m = n$ and $\sqrt{A g_1 + \cdots + A g_n } = A x_1 + \cdots + A x_n$, then $A$ is a finite free $B$-module of rank $\mu$ and $\dim_k A/(g_1, \dots, g_n) = \mu$.

May someone prove or disprove this conjecture or tell me a good reference for complete noetherian rings?

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2 Answers 2

up vote 2 down vote accepted

Let $\mathfrak m$ be the maximal ideal of $A$ and $\mathfrak n$ the maximal ideal of $B$. Let $\phi:B \to A$ be the given map. Your hypothesis is that $rad(A\phi(n)) = \mathfrak m,$ i.e. that $A/A\phi(\mathfrak n)$ is an Artin ring, i.e. that the fibre of the map Spec $A \to $ Spec $B$ over the closed point of the target is zero-dimensional.

This implies that the image of $B \to A$ must have dimension equal to $n$ (which is the image of the domain $B$), thus that this map is an injection. The so-called "Miracle flatness theorem" (Theorem 23.1 of Matsumura's CRT) shows that $\phi$ is necessarily flat. Combined with the finite fibre assumption, this should then also imply that $A$ is actually finite flat, hence finite free, over $B$.

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Dear Matt, shouldn't the map $\phi$ in your first line go from $B$ to $A$? –  Georges Elencwajg Sep 29 '11 at 8:45
    
@Georges: Dear Georges, Thanks! Best wishes, –  Matt E Sep 29 '11 at 12:16

By your assumption, there is some $k$ such that $x_i^k\in(g_1,\ldots,g_n)$. Hence, we know $x_i^k - b=0$ for some $b\in B=k\left[[g_1,\ldots,g_n\right]]$. In other words, $x_i$ is integral over $B$, therefore $A$ is integral over $B$. This means that $A$ is finite as a $B$-module.

For reading, I recommend Eisenbud (Commutative Algebra with a View towards Algebraic Geometry), Chapter 7.

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I think you are wrong: the ideal $Ag_1 + \cdots + A g_n$ is different from $B$. –  Andrea Sep 11 '11 at 7:53

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