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I am reading the book Introduction to commutative algebra by Atiyah and Macdonald. On Page 104, I have some questions about the proof that $\{A_n\}$ is surjective implies $d^A$ is surjective. We have to prove that given $(a_n) \in A = \varprojlim A_n$, there is $(x_n) \in\varprojlim A_n$ such that $d^A((x_n))=(a_n)$. By definition, $d^A(x_n)=x_n-\theta_{n+1}(x_{n+1})$. So we have to solve the system of equations $x_n-\theta_{n+1}(x_{n+1})=a_n, n=0, 1, \ldots$, for $x_0, x_1, \ldots$. We have $x_0-\theta_1(x_1)=a_0$. But there are two unknowns $x_0, x_1$ in this equation. How could we solve this equation? Thank you very much.

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martini
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LJR
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  • Is $A_{n+1}\subset A_n$? Then is $\theta_{n+1}: A_{n+1}\rightarrow A_n$ the canonical injection? – Avitus Jun 14 '13 at 14:24

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As $(A_n)$ is surjective, the $\theta_n$ are onto. So let $x_0\in A_0$ be arbitrary, choose $x_1 \in A_1$ with $\theta_1(x_1) = x_0 - a_0$, which is possible as $\theta_1$ is onto. Now inductively we find $x_n \in A_n$ with $\theta_n(x_n) = x_{n-1} - a_{n-1}$ for each $n$, as wished.

martini
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