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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