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

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The polynomial \( p(x) = x^2 + 1 \) in \( \mathbb{Z}_7[x] \) has no roots in \( \mathbb{Z}_7 \), indicating it is irreducible. Since \( p(x) \) is irreducible and has degree 2, the ideal generated by \( p(x) \), denoted as \( (p) \), is a prime ideal. In a principal ideal domain, all prime ideals are also maximal. Therefore, \( (p) \) is confirmed to be a maximal ideal in \( \mathbb{Z}_7[x] \.
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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