# Radical extension with root of cubic polynomial

If I take $f(x)$ is an irreducible cubic over $\mathbb{Q}$ with a root $\alpha$ in a splitting field and given that $\mathbb{Q}(\alpha)$ is a radical extension is it true that $\mathbb{Q}(\alpha) = \mathbb{Q}(b^{1/3})$ for some $b \in \mathbb{Q}$?

We have that $\alpha^q \in \mathbb{Q}$ for some $q$. I.e. $\alpha$ satisfies an equation $g(x) = x^q - b$ with $\alpha^q = b$. As $f$ is the minimal polynomial of $\alpha$ we must have $f$ divides $g$.

Now if it's the case that $q = 3$ then we'd have $f=g$ but that's clearly not the case in general. I'm not quite sure where I am going wrong here - or perhaps the original statement is false.

Another slight point of confusion is that it seems that $g$ is irreducible for any $b$ because it's splitting field is $\mathbb{Q}(\omega, b^{1/q})$ where $\omega$ is a primitive qth root of unity - that extension is Galois and acts transitively on the roots of $g$? But if that were the case then again we'd have $f=g$?

Thanks for any help clearing up this confusion!

• Precisely what is a radical extension? Isn’t the field of seventh roots of unity a radical extension? It has a subfield cubic over $\Bbb Q$ that definitely is not gotten by adjoining the third root of something. If you’d like me to expand this in an answer, just give the word. – Lubin May 28 '15 at 2:37

The original statement is true.

We have that if $f$ is an irreducible cubic, and then $\mathbb{Q}(\alpha): \mathbb{Q}$ is radical. Then there exist, from the definition, $\beta_1, \ldots, \beta_n$ with $$\mathbb{Q}(\alpha) = \mathbb{Q}(\beta_1, \ldots , \beta_n),$$ and for each $i$ we may assume there is some prime $p_i$ such that $$\beta_i^{p_i} \in \mathbb{Q}(\beta_1, \ldots, \beta_{i-1}).$$ Then by the Tower Law applied repeatedly to $[\mathbb{Q}(\beta_1,\ldots, \beta_n):\mathbb{Q}]$ we get $$[\mathbb{Q}(\beta_1,\ldots, \beta_n):\mathbb{Q}] = \prod_{i=1}^n[\mathbb{Q}(\beta_1,\ldots, \beta_i):\mathbb{Q}(\beta_1,\ldots, \beta_{i-1})] = \prod_{i=1}^n p_i.$$ But then as $[\mathbb{Q}(\beta_1,\ldots, \beta_n):\mathbb{Q}] = [\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, we see that, by prime factorisation, there is only one element $\beta$ with $\beta^3 \in \mathbb{Q}$ and $$\mathbb{Q}(\alpha) = \mathbb{Q}(\beta).$$ But the important point is that this does not imply that $\alpha = \beta$, but instead that there are rational numbers $q_0, q_1, q_2$ with $$\alpha = q_0 + q_1 \beta + q_2 \beta^2.$$

EDIT: Because $\mathbb{Q}(\alpha)$ is a radical extension of degree 3, by the tower law analysis above we know that there is some $\beta$ which generates $\mathbb{Q}(\alpha)$.

Then the minimal polynomial of $\beta$ divides $x^3 - \beta^3$. But since $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, so the minimal polynomial of $\beta$ must be of degree 3 and so equals $x^3 -\beta^3$.

• Thanks for your answer - so by the tower law any subfield of $\mathbb{Q}(\alpha)$ must be a divisor of 3 but then why is $\mathbb{Q}(\beta)$ a degree 3 extension? – Wooster May 27 '15 at 14:25
• Since $\beta$ is not in $\mathbb{Q}$, but $\beta^3$ is. So $x^3 - \beta^3$ is the minimal polynomial of $\beta$ over the rationals. – CameronJWhitehead May 27 '15 at 14:35
• I'll edit my answer, maybe the use of the tower law isn't clear – CameronJWhitehead May 27 '15 at 14:35
• Okay yes that is clear - how do we know that $f(x) = x^3 - \beta^3$ is irreducible? I think this is causing some of the confusion in my original post as well – Wooster May 27 '15 at 14:56
• We know that $x^3 - \beta^3$ is irreducible because we chose $\beta$ so that $\mathbb{Q}(\alpha) = \mathbb{Q}(\beta)$, then we must have $[\mathbb{Q}(\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha):\mathbb{Q}] = 3$. Then this degree is the degree of the minimal polynomial of $\beta$ which clearly divides $x^3 - \beta^3$. – CameronJWhitehead May 27 '15 at 15:09