How can I determine the prime ideals in $\mathbb{Q}[x]$ containing the ideal $(f)$, with $f$ being the polynomial $f = x^6-2x^3-3$.

Some properties I have found so far:

  • $f$ can be be written as a product of irreducible polynomials: $f= (x+1)(x^2-x+1)(x^3-3)$
  • Ideal $(f)$ is not a prime ideal

I would really appreciate some help.

Also just found this question where they also state the same prime ideals: Find idempotent elements in the ring $\mathbb{Q}[x]/(x^6-2x^3-3)$

  • $\begingroup$ Can you think of another polynomial $g$ such that $f\in (g)$? In that case, we would have $(f)\subseteq (g)$. $\endgroup$ – Arthur Dec 10 '17 at 20:52

An ideal $P$ is prime, if

$$AB\subseteq P\Longrightarrow A\subseteq P \:\lor\: B\subseteq P$$

In your case you want

$$\left<f\right>=\left<x+1\right>\left<x^{2}-x+1\right>\left<x^{3}-3\right>\subseteq P$$

so you must have

$$\left<x+1\right>\subseteq P \:\lor\: \left<x^{2}-x+1\right>\subseteq P \:\lor\: \left<x^{3}-3\right>\subseteq P$$

In fact

$$P=\left<x+1\right> \:\lor\: P=\left<x^{2}-x+1\right> \:\lor\: P=\left<x^{3}-3\right>$$

because those ideals are maximal.


Hint: If $\mathfrak{p}$ is a prime ideal of $\mathbb{Q}[x]$ containing $f(x) = (x + 1)(x^2 - x + 1)(x^3 - 3)$, then $\mathfrak{p}$ contains one of those three polynomials, say $g(x)$, because $\mathfrak{p}$ is prime.

I claim that $\mathfrak{p} = (g(x))$. We already know $\mathfrak{p} \supset (g(x))$, and because $g$ is irreducible, $(g(x))$ is a maximal ideal.


You just have to factor $x^6-2x^3-3$ as a product of irreducible polynomials since $\mathbf Q[x]$ is a P.I.D. so irreducible polynomials generate principal prime ideals.

Now this is easy: $$x^6-2x^3-3ˆ=(x^3-1)^2-4=\cdots$$ Remember a polynomial of degree $2$ or $3$ is irreducible over $\mathbf Q\,$ if and only if it has no rational root.

  • 3
    $\begingroup$ He already has the factorization, and of course, your second factor is factored into $(x+1)(x^2-x+1)$. I think you need to add more to your answer to help the OP out. I gave the OP a way to prove the irreducibility of his polynomials over $\mathbb{Q}$ without the need to resort to Eisenstein's lemma for example. $\endgroup$ – stressed out Dec 10 '17 at 21:05
  • $\begingroup$ @stressed-out: You're right. I've changed the end of my answer – hope this isn't too much. Thanks! $\endgroup$ – Bernard Dec 10 '17 at 21:18
  • $\begingroup$ It's pretty much the same as my answer (:P), but there aren't many ways to answer the same question anyway. You still need to finish the factorization or just remove it, I guess. $\endgroup$ – stressed out Dec 10 '17 at 21:23

We know that the polynomial ring over a field is a PID because it is a Euclidean ring. Moreover, in a PID, $a \mid b \iff \langle b\rangle \subseteq \langle a \rangle$ .

Now, remember that a polynomial of degree at most $3$ in any field $F$ is irreducible if and only if it doesn't have a root in the field. This simple fact helps you check the irreducibility of your polynomials over $\mathbb{Q}$ more efficiently.

We know that the quotient ring of the polynomial ring $F[X]$ modulo the ideal generated by an irreducible polynomial is a field. We also know that $R/I$ is a field if and only if $I$ is a maximal ideal. Therefore, an irreducible polynomial in $F[X]$ is maximal. You have the factorization, now check the factors and see where it leads you.

  • 1
    $\begingroup$ There a typo: the root criterion is rather a criterion of reducibility, the way it is formulated. $\endgroup$ – Bernard Dec 10 '17 at 21:38
  • $\begingroup$ @Bernard: I have emphasized at most $3$ in bold. Or is there a typo somewhere else? $\endgroup$ – stressed out Dec 10 '17 at 21:43
  • 1
    $\begingroup$ It's irreducible if it has no root (or it's reducible if it has one). Freudian slip ? ;o) $\endgroup$ – Bernard Dec 10 '17 at 21:46
  • $\begingroup$ @Bernard: Haha. I noticed it just now. I fixed it. Thanks for pointing it out. You also taught me a new psychological term. So, thanks two times. $\endgroup$ – stressed out Dec 10 '17 at 21:49
  • $\begingroup$ Thank you. Very nice to see the definitions from class being combined :D $\endgroup$ – Gross Aera Dec 10 '17 at 21:57

The first thing to notice is that $\mathbb Q[x]$ is a principal ideal domain. In particular, this means that a polynomial $g$ is irreducible if and only if $(g)$ is prime, and all the prime ideals are of this form.

So the question becomes:

For which irreducible polynomials $g\in\mathbb Q[x]$ is $(f)\subset(g)$.

But saying that $(f)\subset (g)$ is exactly the same as saying $f\in (g)$, or equivalently that $g\mid f$.

Can you finish from here?


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