Show that two field extensions are equal

Question

I am new to this and I was hoping to get some help so that I can better understand field extensions.

Given $\omega=\frac{-1+\sqrt{3}i}{2}$, I want to show that $\mathbb{Q}(\omega,\sqrt[3]{2})=\mathbb{Q}(\omega+\sqrt[3]{2})$.

Thank you for your help.

Answer 1

To summarize the comments:

Since $\mathbb{Q}(\omega + \sqrt[3]{2}) \subset \mathbb{Q}(\omega, \sqrt[3]{2})$, and the latter has degree $6$ over $\mathbb{Q}$, it suffices to show that the minimal polynomial of $t=\omega + \sqrt[3]{2}$ has degree $6$.

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We can find a polynomial equation for $t$ as follows: $$2 = (t-\omega)^3 = t^3 - 3\omega t^2 + 3\omega^2 t - \omega^3$$ $$= t^3 - 3\omega t^2 + 3(-\omega - 1)t - 1$$ $$ = (t^3 - 3t - 1) + (-3t^2 - 3t)\omega$$

and so $\omega = (t^3 - 3t - 3)/(3t^2 + 3t)$. Plugging in to $\omega^2 + \omega + 1=0$ and clearing denominators gives us $t^6 + 3t^5 + 6t^4 + 3t^3 + 9t + 9 = 0$.

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The last step is to show that $p(X) = X^6 + 3X^5 + 6X^4 + 3X^3 + 9X + 9$ is irreducible. Suppose otherwise.

First, note that $\mathbb{Q}(\omega,t) = \mathbb{Q}(\omega, \sqrt[3]{2})$. It follows that $t$ cannot have degree $2$, so it must have degree $3$. It follows that $p(X)$ has a cubic factor.

It is easy to check that $p$ has no linear factor, and so $p(X)=q(X)r(X)$ is the product of two cubics.

Clearly, $p(X)\equiv X^6\pmod{3}$. Since $\mathbb{Z}/3\mathbb{Z}$ is a field, we have $q(X),r(X)\equiv X^3\pmod{3}$.

By considering the coefficient of $X^3$ in $p(X)$, modulo $9$, we find that the constant terms of $q$ and $r$ are both $-3$.

Write $q(X) = X^3 + aX^2 + bX - 3$. Looking at the $X^5$ and $X$ coefficients of $p(X)$, we find that $r(X) = X^3 + (3-a)X^2 + (-3-b)X - 3$.

From this, we can calculate the $X^2$ coefficient of $p(X)$ to be $b(-3-b)-9=0$, so $b^2 + 3b + 9 = 0$, which has no solutions in integers.

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Here is another method that also works by hand, and is much shorter, though thinking of it probably requires a computer:

Note that $X^2 + 2$ divides $p(X)\pmod{5}$. Since $p(X)$ has no root modulo $5$, $p(X)$ cannot be the product of two cubic factors.