I am looking for a straightforward way to find all intermediate fields of a field extension. Let's take the splitting field of $X^3-2$ over $\Bbb Q$ as an example.

If we adjoin the roots of $X^3-2$ to $\Bbb Q$ we get $\Bbb Q(\sqrt[3]{2}, \zeta_3)$ with $\zeta_3$ a primitive root of unity different from $1$. My idea for finding the intermediate fields would be to look at the $\Bbb Q$-basis derived from the minimal polynomials of the successive extensions $\Bbb Q(\sqrt[3]{2}, \zeta_3)/\Bbb Q(\sqrt[3]{2})/\Bbb Q$.

This basis would be $\{1, \sqrt[3]{2}, (\sqrt[3]{2})^2, \zeta_3, \zeta_3 \sqrt[3]{2}, \zeta_3 (\sqrt[3]{2})^2\}$. Now I look for basis elements which can be turned into other basis elements by field operations like squaring, adding, etc. This is obviously the case for $(\sqrt[3]{2})^2$ but also $(\zeta_3 (\sqrt[3]{2})^2)^2 = -2\zeta_3 \sqrt[3]{2}$. So I get the following intermediate fields: $\Bbb Q(\sqrt[3]{2})$, $\Bbb Q(\zeta_3)$, $\Bbb Q(\zeta_3 \sqrt[3]{2})$.

Does this method give me all intermediate fields? If yes, what's the advantage of an analysis using Galois theory? If no, what am I missing?

  • 2
    $\begingroup$ Interesting "challenge" to Galois theory! But/and, the fatal issue is that intermediate fields could conceivably (in general, if not necessarily in your example) be generated by linear combinations of whatever basis elements one chooses. Maybe not, in some examples, as in your "simple method", but that needs proof... which is "field theory" and "Galois theory". $\endgroup$ Jun 10, 2015 at 22:16
  • $\begingroup$ @Marc Btw, you missed one intermediate field extension. $\endgroup$
    – user26857
    Jun 10, 2015 at 22:26
  • $\begingroup$ @paul: can you please give an example where this occurs? $\endgroup$
    – Marc
    Jun 11, 2015 at 6:44
  • $\begingroup$ @user26857: can you please tell me which one, so I can see where the method goes wrong? $\endgroup$
    – Marc
    Jun 11, 2015 at 6:45
  • 1
    $\begingroup$ Depending on what you mean about "turned into basis elements by field operations", inside $\mathbb Q(\zeta_5)$ (fifth root of unity) there is a unique subfield $\mathbb Q(\sqrt{5})$. Is this addressing your question? $\endgroup$ Jun 11, 2015 at 13:13

1 Answer 1


How do you prove you've found all possible relations between basis elements? This is especially difficult if you're unlucky in your choice of basis, or in how the field is given. For example, the field $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$ can just as validly be presented as $\mathbb{Q}(\sqrt[3]{4}-5\sqrt[3]{2} + \zeta_3)$. What do you do now?

More importantly, how do you know you've captured all intermediate fields $\mathbb{Q}\subset K\subset \mathbb{Q}(\sqrt[3]{2},\zeta_3)$ that are of the form $K=\mathbb{Q}(a)$ for some $a$ that's not a basis element?

  • $\begingroup$ I don't know how to prove what you ask for. But if the field extension is given the way you write it, I can still try to successively add the elements and analyze the minimal polynomials which leads me to $\Bbb Q(\sqrt[3]{2}, \zeta_3)$. $\endgroup$
    – Marc
    Jun 11, 2015 at 6:48
  • $\begingroup$ Your second objection seems fatal because it implies that the method wouldn't work even if one could prove what you asked me to. Can you please give an example of such a situation? I think I haven't seen this yet. $\endgroup$
    – Marc
    Jun 11, 2015 at 6:51

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