# Minimal polynomial for normal closure

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 $$\omega$$?

• $|M:K|=6$, not $9$. – Lord Shark the Unknown Jan 1 '19 at 17:47
• @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) – Yadati Kiran Jan 1 '19 at 17:52
• I said no such thing. – Lord Shark the Unknown Jan 1 '19 at 17:57
• @LordSharktheUnknown: Could you give me a hint as to how you said $[M:K]=6$? – 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 $$[\mathbb Q(\sqrt[3]{2}, \omega) : \mathbb Q]$$.

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