2

Here's what I've tried:

I've checked that there's exactly one real root for this polynomial (that's easy to prove i.e. there's at least one real root since it has odd degree and it is easy to show that the derivative is always $\geq 0$). I've noticed that if $g(X) = X^3 - 2dX^2 + 3d^2 X - d^3$, then $f(X) = g(X^3)$. Also, $g(X)$ is irreducible over $\mathbb{Q}$ if and only if $h(X) = d^3 g(X/d)$ is irreducible over $\mathbb{Q}$, but $h(X) = X^3 - 2X^2 + 3X - 1$ has no rational root and, hence, is not reducible. I was trying to conclude the problem by saying that if $g(X)$ is irreducible over $\mathbb{Q}$, then $g(X^3)$ is also irreducible over $\mathbb{Q}$, but I'm not sure if this is true. Can someone help me conclude this problem?

Arthur Queiroz Moura
  • 573
  • 1
    Let $g(x)= x - 1$. Clearly, $g(x)$ is irreducible over $\mathbb{Q}$ but $g(x^3)= x^3 - 1= (x - 1)(x^2 + x + 1)$ is reducible. – mathematics2x2life Jan 19 '24 at 00:28
  • Can you find a counter example for when $\text{deg}(g) \geq 2$? (i.e. the conjecture about the irreducibility becomes "if $g(X)$ has no rational roots and is irreducible, then $g(X^3)$ is irreducible over $\mathbb{Q}$") – Arthur Queiroz Moura Jan 19 '24 at 00:30
  • $g(x)= x^2 + 4$ is irreducible over $\mathbb{Q}$ but $g(x^2)= x^4 + 4= (x^2 - 2x + 2)(x^2 + 2x + 2)$ is reducible. – mathematics2x2life Jan 19 '24 at 00:33
  • For $g(x^3)$, a counterexample is $g(x) = x^2 + 64$ – Arthur Queiroz Moura Jan 19 '24 at 01:09
  • It seems the given polynomial for $d=1$ is itself a counterexample to the conjecture on $g(X)$ irreducible implying $g(X^3)$ irreducible: $X^3 - 2X^2 + 3X - 1$ is irreducible, but $X^9 - 2 X^6 + 3 X^3 - 1 = (X^3 + X^2 - 1) (X^6 - X^5 + X^4 - 2X^3 + X^2 + 1)$ (credit to Sagemath for finding that factorization). – Daniel Schepler Jan 19 '24 at 01:12
  • 1
    I've imposed $d>1$ on the title of the question, but this factorization is interesting – Arthur Queiroz Moura Jan 19 '24 at 01:13

1 Answers1

1

An outline of a proof, as a series of exercises:

Let $\alpha$ be a root of $f(X)$.

  1. Note that $f(X) = (X^3-d)^3 + dX^6$. From this, show that $\sqrt[3]{d} \in \Bbb{Q}(\alpha)$.
  2. Show that $\Bbb{Q}(\alpha)$ also contains a root of $X^3 - 2X^2 + 3X - 1$; call this root $\beta$.
  3. Show that if $f(X)$ is reducible, then $\sqrt[3]{d} \in \Bbb{Q}(\beta)$. (Look at the degrees of the relevant extensions.)
  4. Compare the discriminant of the minimum polynomial of an arbitrary element of $\Bbb{Q}(\beta) \setminus \Bbb{Q}$ with the discriminant of the minimum polynomial of $\beta$. In particular, show that the product of these two discriminants must be a perfect square.
  5. Use this to show $\sqrt[3]{d} \not \in \Bbb{Q}(\beta)\setminus \Bbb{Q}$, and hence $f(X)$ is irreducible.

Note that this works for any $d$ that is not a perfect cube, which is slightly different from $d$ being cube-free.

The relationship between the irreducibility of $g(X)$ and the irreducibility of $g(X^3)$ is a special case of the Capelli Lemma and the Vahlen-Capelli irreducibility criterion; together they say (for this special case) that if $\gamma$ is a root of the irreducible polynomial $g(X)$, then $g(X^3)$ is also irreducible if and only if $\gamma$ is not a perfect cube in $\Bbb{Q}(\gamma)$.

Benjamin Wright
  • 1,611