I am a student whose major interest is not algebraic geometry, but for various reasons I began to re-read Mumford's Red Book, and I found myself stuck in a quite simple theorem. Since I had studied some algebraic geometry before, (based on my remaining intuitions) this theorem seems very simple, but I am having trouble understanding the last part of the proof. The theorem is as follows:

(Theorem 2, page 177) Let $\mathcal O$ be a complete local ring, with maximal ideal $\mathfrak m$, residue field $k$. Let $f:X\rightarrow Y = \mathrm {Spec} (\mathcal O )$, $x\mapsto y = [\mathfrak m ]$. Assume $k=\kappa (y) \simeq \kappa (x)$. Then $f$ etale near $x$ $\Rightarrow $ $f$ a local isomorphism near $x$.

In this book, Mumford first defines etale morphisms non-intrinsically: he first defines that the morphism $X=\mathrm {Spec} R[x_1 , \cdots , x_n ] / (f_1 , \cdots , f_n ) \rightarrow \mathrm {Spec} R$ induced by the inclusion is etale at a point $x\in X$ when the Jacobian of $f_1 , \cdots , f_n $ does not vanish at $x$.

Using this definition, he reduces this theorem to the following situation: $$X=\mathrm {Spec} \mathcal O [x_1 , \cdots , x_n ] / (f_1 , \cdots , f_n )$$ $$ x=\mathrm m + (x_1 -a_1 , \cdots , x_n - a_n )$$ $$\mathrm {det} (\partial f_i / \partial x_j )(x)\neq 0,$$ where $a_1 , \cdots , a_n \in \mathcal O$. Then by Hensel's lemma, he finds elements $\alpha _1 , \cdots , \alpha _n \in \mathcal O$ such that $$Z=\mathrm {Spec} \mathcal O [x_1 , \cdots , x_n ] / (x_1 -\alpha _1 , \cdots , x_n -\alpha _n )$$ is a subscheme of $X$ through $x$, which is clearly isomorphic to $Y$. Now he tries to prove that $X=Z$ near $x$, by first defining $\mathcal Q$ to be the $\mathcal O _X $-ideal defining $Z$ and then as follows, which I cannot understand:

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If this is shown, then the rest is clear from Nakayama's lemma.

I think that I am missing something fundamental and this might be a stupid question, but from the start, I can't see why $\mathcal O_{y,Y}=\mathcal Q$. Can someone please explain this thoroughly?

This might be an unappropriate question for mathoverflow; if so, I will migrate this down to MSE.

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    $\begingroup$ This republished and retypeset book is notorious for its many misprints. In the original redbook, that local ring is stated to be not the ideal of Z, but the original complete local ring O. I have not read this proof, so do not know if this solves the problem, but possibly this may help. $\endgroup$ – roy smith Sep 11 '17 at 18:08
  • $\begingroup$ @roy smith Thank you, I figured this out; the only error here seems to be what you mentioned. I was puzzled beforehand since the proceeding argument did not seem to be clear at first glance... $\endgroup$ – Gheehyun Nahm Sep 12 '17 at 2:05

To make this question non-active, I will post a simple answer: If one replaces $ \mathcal Q$ by $\mathcal O$ in the diagram, we have $\mathcal O _{x,X} = f^* (\mathcal O ) \oplus \mathcal Q _x $ and $\mathcal m _{x} = f^* (m_y ) \oplus \mathcal Q _x $ (they are $k$-algebras). Now we have $$ f^* (m_y) \oplus \mathcal Q _x = m_x = f^* (m_y ) \mathcal O _{y,Y} =f^* (m_y ) (f^* (\mathcal O ) \oplus \mathcal Q _x ) . $$ The rest is clear.


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