Polynomial Rings: Finding 8 Elements with r^2=r

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SUMMARY

The discussion focuses on finding eight elements in the polynomial ring \(\mathbb{Q}[x]/(x^4-16)\) that satisfy the equation \(r^2=r\). The initial solutions identified are \(0+(x^4-16)\) and \(1+(x^4-16)\). To find the remaining six elements, participants suggest using brute force by considering arbitrary elements of the form \([ax^3+bx^2+cx+d]\) and squaring them to check for compliance with the equation.

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  • Understanding of polynomial rings, specifically \(\mathbb{Q}[x]\)
  • Familiarity with the concept of equivalence classes in modular arithmetic
  • Knowledge of squaring polynomials and solving polynomial equations
  • Basic algebraic manipulation skills
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  • Explore the structure of polynomial rings, particularly \(\mathbb{Q}[x]/(x^4-16)\)
  • Learn techniques for finding roots in polynomial equations
  • Study the properties of idempotent elements in algebraic structures
  • Investigate computational methods for polynomial manipulation and simplification
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Students and mathematicians interested in abstract algebra, particularly those studying polynomial rings and their properties.

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Homework Statement



Find eight elements [itex]r \in \mathbb{Q}[x]/(x^4-16)[/itex] such that [itex]r^2=r[/itex].

Homework Equations



N/A

The Attempt at a Solution



The elements [itex]0+(x^4-16)[/itex] and [itex]1+(x^4-16)[/itex] clearly satisfy the desired properties, but I still need six more elements. Can anyone help me figure out a technique for finding a few more elements?

Thanks!
 
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You can always brute force it. An arbitrary element in your ring has the form [itex][ax^3+bx^2+cx+d][/itex]. Square it and see when the relation is satisfied.
 

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