We have $\mathbb Q[x]$ be the ring of polynomial over $\mathbb Q$, then total number of maximal ideal in the quotient ring $\frac{\mathbb Q[x]}{x^4-1}$ is $2$ because $i$ and $-i$ are not in $\mathbb Q$. is it correct $?$.Thanlk you
Total number of maximal ideal in the quotient ring $\frac{\mathbb Q[x]}{x^4-1}$
1
$\begingroup$
abstract-algebra
proof-verification
ring-theory
2 Answers
2
The number is 3:
corresponding to the irreducible factors $x+1, x-1$ and $x^2+1$ of $x^4-1$.
2
Although this question is completely answered here, there is also an opportunity to address other problems for you. This question appeared in the related questions on the right, and therefore probably appeared in the list of possible duplicates whole you were entering your question. You really ought to actually pay attention and look for duplicates before submitting.
Anyway.
No, your solution is incorrect and aside from that, very incompletely expressed.
It has three maximal ideals because the maximal ideals correspond to prime divisors of $x^4-1$, of which there are three since $(x-1)(x+1)(x^2+1)$ is a complete factorization.
Be sure to include a complete explanation especially if it is brief. "Two maximal ideals because two random numbers aren't in $\mathbb Q$ " is not very enlightening, although we can tell you might be factoring.