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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

Mark Bennet
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excalibirr
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  • See also this question with $n=3$. So $x=2$. – Dietrich Burde Apr 13 '19 at 19:48
  • @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 – excalibirr Apr 13 '19 at 19:52
  • The duplicate has an answer exaxctly to this, i.e., why the first extension has degree $x=2$. That's your question, right? – Dietrich Burde Apr 13 '19 at 19:54
  • $w$ 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. – Mark Apr 13 '19 at 19:57
  • 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 – excalibirr Apr 13 '19 at 19:57
  • @Mark ah okay that makes sense, thank you. One question though, how does one figure out the first factorisation you made in your comment ? – excalibirr Apr 13 '19 at 19:59
  • Well, this is a very known factorization. For a general $n$ we have $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$. – Mark Apr 13 '19 at 20:00
  • @Mark ah okay, I was sick for a lot of rings and fields last year so I missed a lot of the more basic elements. So it's just a general fact we can use, got ya ! thanks :) – excalibirr Apr 13 '19 at 20:02
  • Check the edited text for the way to get cube (or other) roots in your text (\sqrt[3] is what you need) – Mark Bennet Apr 13 '19 at 20:09

1 Answers1

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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.

Marco Vergamini
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