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In the book Algebraic Geometry - A First Course by Harris, it is given a proof for the following theorem:

Theorem 14.9. Let $\pi:X\rightarrow Y$ be a finite map of varieties. Then $\pi$ is an isomorphism if and only if it is bijective and the map $d\pi:T_p(X)\rightarrow T_{\pi(p)}(Y)$ is an injection for all $p\in X$.

In the proof, they assume $X$ and $Y$ are affine and pass to an algebra question. By localizing at some maximal ideal, they get an integral extension of local rings $(A,\mathfrak{m})\subset(B,\mathfrak{n})$ where the map $\mathfrak{m}/\mathfrak{m}^2\rightarrow\mathfrak{n}/\mathfrak{n}^2$. Using Nakayama's lemma they get $\mathfrak{m}B=\mathfrak{n}$. Then they apply the same lemma together with $B=\mathfrak{n}+A$ to get $A=B$. My question is: why is $B$ equal to $\mathfrak{n}+A$? This is in general not true for finite extensions of local ring, but I don't see how they use the other information to get this equality.

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  • $\begingroup$ Harris works over complex numbers and then the natural map $A\to B/\mathfrak{n}=\mathbb{C}$ is onto. $\endgroup$
    – Mohan
    May 13 '20 at 16:16
  • $\begingroup$ Thank you for your answer! So he just uses that $A$ and $B$ are algebras over an algebraically closed field. $\endgroup$
    – o---o
    May 13 '20 at 19:07
  • $\begingroup$ Finite type algebras over an algebraically closed field and $\mathfrak{n}$ is a maximal ideal. $\endgroup$
    – Mohan
    May 13 '20 at 20:02
  • $\begingroup$ Can someone give a counterexample where $\pi$ is finite, bijective but the differential is not an injection? $\endgroup$ May 14 '20 at 12:33
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    $\begingroup$ @IoannisZolas your query would probably would fare better as a new question instead of a comment on this post. Anyways, a projection of a curve in $\Bbb P^3$ from a point on a tangent line should do the trick, I think? $\endgroup$
    – KReiser
    May 15 '20 at 3:37
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This is a community wiki answer in order to record the discussion from the comments so that this question may be marked as answered (once this post is upvoted or accepted).

Harris works over complex numbers and then the natural map $A\to B/\mathfrak{n}=\Bbb C$ is onto. – Mohan

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