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Say we have the field extension $\Bbb Q(w,\sqrt[3]{5})$ over $\Bbb Q$, where w is the primitive cubed root of unity. I know that the minimum polynomial of $\sqrt[3]{5}$ is $x^3-5$. I want to figure out the degree of the extension;

I know we can use the tower law $|\Bbb Q(w, \sqrt[3]{5});\Bbb Q|=|\Bbb Q(w, \sqrt[3]{5});\Bbb Q(\sqrt[3]{5})||\Bbb Q(\sqrt[3]{5});\Bbb Q|=x.3$ as $x^3-5$ has degree 3

But how does one find the minimum polynomial of the extension of degree x


marked as duplicate by Dietrich Burde abstract-algebra Apr 13 at 19:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See also this question with $n=3$. So $x=2$. $\endgroup$ – Dietrich Burde Apr 13 at 19:48
  • $\begingroup$ @DietrichBurde I edited my question to hone in the part of the question I was particularly interested in , I'm not sure the duplicate link properly addresses that point $\endgroup$ – Voltron Apr 13 at 19:52
  • $\begingroup$ The duplicate has an answer exaxctly to this, i.e., why the first extension has degree $x=2$. That's your question, right? $\endgroup$ – Dietrich Burde Apr 13 at 19:54
  • $\begingroup$ $w$ is a root of $x^3-1=(x-1)(x^2+x+1)$. Since it is a primitive root (hence not equal to $1$) it isn't a root of $x-1$, so it is a root of $x^2+x+1$. This is a polynomial of degree $2$ which has no real roots, hence it is irreducible over $\mathbb{Q(\sqrt[3]{5})}$. So this is the minimal polynomial. $\endgroup$ – Mark Apr 13 at 19:57
  • $\begingroup$ not quite, as an answer below stated $x^2+x+1$ is the minimum polynomial of the extension but I don't understand how this is found, maybe I'm not understanding something in the link but it didn't seem to show clearly how its assumed that this Is the minimum polynomial $\endgroup$ – Voltron Apr 13 at 19:57

Clearly $w \not \in \Bbb Q(\sqrt[3]{5})$ so the degree is at least $2$. Now notice that $x^2+x+1$ is the minimum polynomial.

  • $\begingroup$ Why is $x^2+x+1$ the minimum polynomial , how was it found ? $\endgroup$ – Voltron Apr 13 at 19:55
  • $\begingroup$ By noticing that $0=w^3-1=(w-1)(w^2+w+1)$. $\endgroup$ – Marco Vergamini Apr 13 at 20:04
  • $\begingroup$ An interesting generalisation of this problem could be this one: find the degree of the splitting field of $x^p-s$ over $\Bbb Q$ where $p$ is a prime number and $s>1$ is a squarefree integer. $\endgroup$ – Marco Vergamini Apr 13 at 20:24

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