Is (x^2+1) a maximal ideal in Z7[x]?

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SUMMARY

The polynomial \( p(x) = x^2 + 1 \) in \( \mathbb{Z}_7[x] \) is established as a maximal ideal. It is irreducible since it has no roots in \( \mathbb{Z}_7 \), as demonstrated by evaluating \( p(x) \) at all elements of the field. Given that \( p(x) \) is a degree 2 polynomial and irreducible, the ideal generated by \( p(x) \), denoted as \( (p) \), is a prime ideal. In a principal ideal domain, all prime ideals are maximal, confirming that \( (p) \) is indeed a maximal ideal.

PREREQUISITES
  • Understanding of polynomial rings, specifically \( \mathbb{Z}_7[x] \)
  • Knowledge of irreducibility criteria for polynomials
  • Familiarity with the concepts of prime and maximal ideals
  • Basic understanding of finite fields and their properties
NEXT STEPS
  • Study the properties of polynomial rings over finite fields
  • Learn about the structure of principal ideal domains
  • Explore examples of irreducible polynomials in \( \mathbb{Z}_p[x] \)
  • Investigate the relationship between prime and maximal ideals in commutative algebra
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Mathematicians, algebra students, and anyone studying abstract algebra, particularly those focusing on ring theory and polynomial ideals.

Fernando Revilla
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I quote an unsolved question from Yahoo! Answers

Give a maximal ideal of Z7[x] with justification

I have given a link to the topic there so the OP can see my complete response.

P.S. I have posted there only a hint.
 
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Consider the polynomial $p(x)\in\mathbb{Z}_7[x]$, $p(x)=x^2+1$. This polynomial has no roots: $$p(0)=1,\;p(1)=2,\;p(2)=5,\;p(3)=3,\;p(4)=3,\;p(5)=5,\;p(6)=2$$ and has degree 2 which implies that $p(x)$ is irreducible, as a consequence $(p)$ is a prime ideal. But in a principal ideal domain all prime ideals are maximal, hence, $(p)$ is a maximal ideal.
 

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