# Equality in proof of Theorem 14.9 from Harris's book

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.

• Harris works over complex numbers and then the natural map $A\to B/\mathfrak{n}=\mathbb{C}$ is onto. May 13 '20 at 16:16
• Thank you for your answer! So he just uses that $A$ and $B$ are algebras over an algebraically closed field. May 13 '20 at 19:07
• Finite type algebras over an algebraically closed field and $\mathfrak{n}$ is a maximal ideal. May 13 '20 at 20:02
• Can someone give a counterexample where $\pi$ is finite, bijective but the differential is not an injection? May 14 '20 at 12:33
• @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? May 15 '20 at 3:37

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