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 September 28th, 2015, 01:49 PM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 localisation of ring Hi can someone please help me to prove this equivalence: Let A be a commutatif ring and a in A .we note Aa the localisation of A by S ={an, n dans N} knowing that A[x]/(ax-1) is isomorphic to Aa show that Aa #0 if and only if a is not nilpotent thaks in advance
 October 12th, 2015, 09:08 AM #2 Newbie   Joined: Oct 2015 From: Toronto Posts: 14 Thanks: 1 Suppose that $a$ is nilpotent, there exists an integer $n$ such that $a^n = 0$. Let $y$ be an element of $A[X]$, we denote by $\bar{y}$ the image of $y$ by the canonical surjection $A\rightarrow A_a =A[x]/(ax - 1)$. We have $\bar{a}\bar{x} =\bar{1}$. This implies that $\bar{1} = {\bar{a}}^n{\bar{x}}^n$. Since $a^n = 0$, we deduce that ${\bar{a}}^n =0$ and henceforth ${\bar{a}}^n{\bar{x}}^n =\bar{1} =0$. This is equivalent to saying that $A_a = 0$ since $\bar{1}$ is the identity of $A_a$. Suppose that $A_a=0$ this is equivalent to saying that $\bar{1} = 0$, or equivalently that $1\in (1-ax)$. We deduce the existence of a polynomial $P(x) = b_0 + b_1x+...+b_nx^n$ such that $(1-ax)P(x) = 1$. Lets determine the coefficients of $(1-ax)P(x)$ and deduce the coefficients of $P$ by using $(1-ax)P(x) = 1$. The coefficient of degree 0 of $(1-ax)P(x)$ is $1b_0 =1$ The degree 1 coeff. is $b_1-a = 0$, thus $b_1 = a$. Suppose that $b_i = a^i$, the $i+1$ coeff. of $(1-ax)P(x)$ is $b_{i+1} -a^{i+1} = 0$. We deduce that $b_{i+1} = a^{i+1}$ and recursively $b_n =a^n$. Since $(1-ax)P(x) = 1$, the biggest coefficient of $(1-ax)P(x)$ is 0.Since this coefficient is $a^{n+1}$, we deduce that $a^{n+1}=0$ and henceforth that $a$ is nilpotent. Last edited by Aristide Tsemo; October 12th, 2015 at 09:10 AM.

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