Let α be a complex number that satisfies α3 + βα2 + γα + δ = 0

β, γ, and δ satisfy cubics with rational coefficients. For example, β satisfies β3 + aβ2 + bβ + c = 0. However, it is not stated that these do not satisfy polynomials of smaller degree, so their degrees are ≤ 3.

I am trying to find an upper bound for α over the rational numbers, as well as show that the minimal polynomial of α over the rationals does not have degree of precisely five.

  • $\begingroup$ Are you looking for any upper bound or a strict one? $\endgroup$
    – Mathmo123
    Feb 25, 2015 at 23:46
  • $\begingroup$ I'm looking for a strict upper bound, but any upper bound would be appreciated as a nudge in the right direction :) $\endgroup$ Feb 25, 2015 at 23:49

1 Answer 1


For an upper bound, can you find an upper bound for $[\mathbb Q(\beta, \gamma): \mathbb Q]$? Observe that $[\mathbb Q(\beta): \mathbb Q] \le [\mathbb Q(\beta, \gamma):\mathbb Q(\gamma)]$, since a polynomial over $\mathbb Q$ is also a polynomial over $\mathbb Q(\gamma)$.

Since you have an upper bound for $[\mathbb Q(\alpha, \beta, \gamma): \mathbb Q(\beta, \gamma)]$, you can obtain a bound for $[\mathbb Q(\alpha, \beta, \gamma): \mathbb Q]$ using the tower law. This will also be an upper bound for $[\mathbb Q(\alpha): \mathbb Q]$.

If $[\mathbb Q(\alpha): \mathbb Q] = 5$, then

$$5 \mid [\mathbb Q(\alpha, \beta, \gamma): \mathbb Q]=[\mathbb Q(\alpha, \beta, \gamma): \mathbb Q(\beta, \gamma)]\cdot [\mathbb Q(\beta, \gamma) :\mathbb Q(\beta)]\cdot [\mathbb Q(\beta):\mathbb Q]$$

So since $5$ is prime, it divides one of $[\mathbb Q(\alpha, \beta, \gamma): \mathbb Q(\beta, \gamma)],\ [\mathbb Q(\beta, \gamma) :\mathbb Q(\beta)], \ [\mathbb Q(\beta):\mathbb Q]$. Can this occur?

  • $\begingroup$ It cannot! All extensions are of degree ≤ 3. Thank you so much! $\endgroup$ Feb 26, 2015 at 0:07

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