Number of distinct prime ideals

Question

How many distinct prime ideals are there for $\mathbb Q[x]/(x^5 - 1)$?

I've got $x^5 - 1$ as a reducible polynomial in $\mathbb Q[x]$. Also $x^5 - 1 =(x-1)(x^4 + x^3 +x^2 + x + 1)$. Here $(x^4 + x^3 +x^2 + x + 1)$ is irreducible in $\mathbb Q[x]$. Is it possible to say anything from this?

Answer 1

Notice that $(x-1,\,x^4+x^3+x^2+x+1)=(1)$. Moreover, since $\mathbb{Q}[x]$ is a PID, $$(x-1)\cap(x^4+x^3+x^2+x+1)=(x-1)\cdot(x^4+x^3+x^2+x+1)=(x^5-1)$$

Hence, by Chinese Remainder Theorem,

$$\mathbb{Q}[x]/(x^5-1)\cong{\mathbb{Q}[x]\over(x-1)}\times{\mathbb{Q}[x]\over(x^4+x^3+x^2+x+1)}\cong\mathbb{Q}\times\mathbb{Q}[\zeta_5]$$

But now you can use the characterisation of ideals of a product of rings and the fact that the two factors are fields...

Answer 2

If I am not mistaken, your ring is isomorphic to $\mathbf{Q} \times \mathbf{Q}(\zeta)$ where $\zeta$ is a primitive fith root of unity. Any ideal in this product ring is of the form I $\times$ S, where I is an ideal in the first component, and S in the second component. But the two components are fields! What can you then say?

Answer 3

The prime ideals of $\mathbb Q[x]/(x^5-1)$ are of the form $P/(x^5-1)$ with $P\subset\mathbb Q[x]$ a prime ideal containing $x^5-1$. Since $\mathbb Q[x]$ is a principal ideal domain, $P=(p)$ with $p\in\mathbb Q[x]$ an irreducible (monic) polynomial. From $(x^5-1)\subseteq(p)$ we get $p\mid x^5-1$. It follows now that $p=x-1$ or $p=x^4+x^3+x^2+x+1$, so there are exactly two prime ideals satisfying the requirement.