Show Q[x]/(x^2-4x+4) has infinite nilpotent elements

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SUMMARY

The discussion focuses on the ring R = Q[x]/(x^2-4x+4), which is isomorphic to Q[x]/((x-2)^2). It is established that R contains infinitely many nilpotent elements, specifically those of the form (x-2)^n for n ≥ 1. The nilpotent condition is satisfied since (x-2)^2 = 0 in this quotient ring. The quotient map φ: Q[x] → Q[x]/(x^2-4x+4) is utilized to demonstrate the nilpotency of elements derived from (x-2).

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Homework Statement



Let R = Q[x]/(x^2-4x+4). Show that R has infinitely many nilpotent elements.

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The Attempt at a Solution



I see that x^2-4x+4 = (x-2)(x-2) and that a nilpotent means a^2 = 0. Other than that, I'm not sure where to start.
 
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Let \varphi:\mathbb{Q}[x]\rightarrow \mathbb{Q}[x]/(x^2-4x+4) be the quotient map. Can you show that \varphi(x-2) is nilpotent? Can you in some way use this to obtain more nilpotents?
 
Thanks for the direction, I think I got something that will work.
 

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