Dtyjlui

I came across this problem while studying Normal closures.

Given $K=mathbb{Q}$ and the polynomial $x^3-2in K[x]$. $L/K$ is not normal where $L=mathbb{Q}(2^{frac13})$ since $omeganotin 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 $x^3-2in K[x]$.

$L/K$ 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 $omega$?

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 $[mathbb Q(sqrt[3]{2}, omega) : mathbb Q]$.