the number of distinct prime ideals of the ring $\mathbb Q[x]/\langle x^4-1\rangle$. The number of distinct prime ideals of the ring $${ \mathbb Q[x]}/{I}$$ where $I$ is the ideal generated by $x^4-1$. its ans. is given $3$. I can't understand how? I know the ans. when in place of $4$ there is a prime number

These are the keys for looking for the ideals directly:


*

*$\Bbb Q[x]$ is a principal ideal domain, and so the ideals above $(x^4-1)$ are those generated by divisors of $x^4-1$. Since $x^4-1=(x^2+1)(x-1)(x+1)$ is a complete factorization, you can read all divisors from it (and hence see all the ideals above $I$).

*The quotient $\Bbb Q[x]/I$ for $I\neq 0$ has finite $\Bbb Q$ dimension (equal to the degree of a polynomial generating $I$). That means it's an Artinian ring, so all the prime ideals are maximal. 

*By correspondence, the maximal ideals of the quotient are the maximal ideals of $\Bbb Q[x]$ containing $I$.

*In principal ideal domains, the maximal ideals are the ones generated by nonzero irreducible polynomials.

*Combining these points, can you now see which prime ideals contain $(x^4-1)$?

Alternatively, you can apply the Chinese remainder theorem, if you know it. It says that (with $I_1=(x^2+1)$, $I_2=(x+1)$ $I_3=(x-1)$) since the $I_j$ are all pairwise comaximal, $\Bbb Q[x]/I=\Bbb Q[x]/(I_1\cap I_2\cap I_3)\cong (\Bbb Q[x]/I_1)\oplus (\Bbb Q[x]/I_2)\oplus (\Bbb Q[x]/I_3)$. The first quotient is isomorphic to $\Bbb Q(\sqrt{-1})$ and the other two are isomorphic to $\Bbb Q$. So, this is a product of three fields.
This is again Artinian, so searching for the prime ideals is searching for the maximal ideals. Luckily, the maximal ideals of a product of rings are easy to find: They are of the form $\prod I_i$ where each $I_i=R_i$ except exactly one $j$ where $I_j$ is a maximal ideal of $R_j$.
