I came across this problem while studying Normal closures.

Given $K=\mathbb{Q}$ and the polynomial $x^3-2\in K[x]$. $L/K$ is not normal where $L=\mathbb{Q}(2^{\frac13})$ since $\omega\not\in L $ where $\omega$ is the cube root of unity. The normal closure of $L/K$ is $M/L$ where $M=L(\omega)=\mathbb{Q}(2^{\frac13},\omega)$ and $M/K$ is normal since $M$ is the splitting field for $x^3-2\in K[x]$.

$L/K$ is finite extension and $[L:K]=3$ $\left(\deg(m_{2^{\frac13}})=3,\; m_{2^{\frac13}}=x^3-2 \right)$ and $M/L$ is finite extension and $[M:L]=3$ $\left(\deg(m_\omega)=3, \; m_\omega=x^3-1 \right)$.

$[M:K]=[M:L][L:K]=9$ seems strange. Minimal polynomial associated with $2^{\frac13}$ and $\omega$ is $x^3-2$ which does not have degree $9$. Where am I wrong?

My question: What is the minimal polynomial of degree 9 associated with $2^{\frac13}$ and $\omega$?

  • 1
    $\begingroup$ $|M:K|=6$, not $9$. $\endgroup$ – Lord Shark the Unknown Jan 1 '19 at 17:47
  • $\begingroup$ @LordSharktheUnknown: So you are saying the minimal polynomial is $(x^3-1)(x^3-2)$. But isn't the product reducible in $\mathbb{Q}$? (Minimal=irreducible) $\endgroup$ – Yadati Kiran Jan 1 '19 at 17:52
  • $\begingroup$ I said no such thing. $\endgroup$ – Lord Shark the Unknown Jan 1 '19 at 17:57
  • $\begingroup$ @LordSharktheUnknown: Could you give me a hint as to how you said $[M:K]=6$? $\endgroup$ – Yadati Kiran Jan 1 '19 at 18:00

The minimal polynomial of $\omega$ over $\mathbb Q(\sqrt[3]{2})$ is actually $$ m(X) = 1 + X + X^2 \in \mathbb Q(\sqrt[3]{2})[X].$$ Indeed, $m(X)$ is irreducible over $\mathbb Q(\sqrt[3]{2})$, and has $\omega$ as one of its roots. $m(X)$ has degree two, which implies that $$[\mathbb Q(\sqrt[3]{2}, \omega) : \mathbb Q(\sqrt[3]{2})] = 2,$$ and so, by the tower law,$$[\mathbb Q(\sqrt[3]{2}, \omega) : \mathbb Q] = [\mathbb Q(\sqrt[3]{2}, \omega) : \mathbb Q(\sqrt[3]{2})]\times[ \mathbb Q(\sqrt[3]{2}) : \mathbb Q] = 2 \times 3 = 6.$$

As for your second remark, the splitting field of a degree $d$ polynomial $f(X) \in \mathbb Q[X]$ does not need to be a degree $d$ extension of $\mathbb Q$. It can be anything from a degree $1$ extension to a degree $d!$ extension.

Take a look at these examples of degree three polynomials in $\mathbb Q[X]$:

  • $ f(X) = X^3$ splits completely over $\mathbb Q$, so its splitting field is simply $\mathbb Q$, which is (trivially) a degree one extension of $\mathbb Q$.

  • $f(X) = X^3 - 1$ has splitting field $\mathbb Q(\omega)$, which is a degree two extension of $\mathbb Q$.

  • $f(X) = X^3 - 3X + 1$ is discussed here, where it is shown that its splitting field is a degree three extension of $\mathbb Q$.

  • $f(X) = X^3 - 2$ is the example you're interested in. We showed that its splitting field is $\mathbb Q(\sqrt[3]{2}, \omega) $, a degree six extension of $\mathbb Q$.

So there is nothing to worry about.

Finally, a point about terminology: there is no such thing as "the minimal polynomial of $\sqrt[3]{2}$ and $\omega$". You can talk about the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb Q$ (which is $X^3 - 2$), and you can talk about the minimal polynomial of $\omega$ over $\mathbb Q$ (which is $1 + X + X^2$). You can also talk about the minimal polyomial of $\omega$ over $\mathbb Q(\sqrt[3]{2})$ (which also happens to be $1 + X + X^2$). But it doesn't make sense to talk about "the minimal polynomial of $\sqrt[3]{2}$ and $\omega$". Nor is such a concept required in order to find the degree of the extension $[\mathbb Q(\sqrt[3]{2}, \omega) : \mathbb Q]$.

  • $\begingroup$ Thank you for the detailed answer. I shall take note of your point regarding the terminology. $\endgroup$ – Yadati Kiran Jan 2 '19 at 12:13

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