# Proof, that factor of irreducible polynomial is irreducible over field defined by root of irreducible

I work on my homework from abstract algebra and i stuck on this problem.

Let $f(x)=x^3-2$ and $\rho_1,\rho_2,\rho_3$ its roots. Construct field $\mathbb{Q}(\rho_1)$ and proof, that $g(x)=$$f(x)\over x-\rho_1 is irreducible over \mathbb{Q}(\rho_1). So i know, that f(x) is irreducible over \mathbb{Q} and i found its roots: \rho_1=-$$\sqrt[3]{2}\over 2$$+$$\sqrt[3]{2}\sqrt{3}\over 2$$i \rho_2=-$$\sqrt[3]{2}\over 2$$-$$\sqrt[3]{2}\sqrt{3}\over 2$$i \rho_3=$$\sqrt[3]{2}$

then i construct field $\mathbb{Q}(\rho_1)$, but i'm not sure which element contains. I know, that elements are polynomials of $deg(f)-1=2$ with coefficients from $\mathbb{Q}$.

$\mathbb{Q}(\rho_1)= \bigg\{\sum_{i=0}^2a_i\rho_1^i$ $\bigg|$ $a_i\in\mathbb{Q} \bigg\}=$

$= \bigg\{ a+b\big(-\sqrt[3]{2}+\sqrt[3]{2}\sqrt{3}i\big)+c\big(-\sqrt[3]{4}-\sqrt[3]{4}\sqrt{3}i\big)$ $\bigg|$ $a,b,c\in\mathbb{Q}\bigg\}$

is it correct? And the proof... I know, that $f(x)=(x-\rho_1)(x-\rho_2)(x-\rho_3)$ and polynomial $g(x)=(x-\rho_2)(x-\rho_3)$, so i must proof, that $\mathbb{Q}(\rho_1)$ not contain elements $\rho_2$ and $\rho_3$? Is it this right way?

• In addition to @user103697's remark that changing the indexing so that the simplest of the roots is $\rho_1$, you can additionally simplify the arithmetic by picking a primitive third root of unity (call it $\omega$), and write numbers in terms of $\omega$ rather than in terms of $\sqrt{3}$ and $i$. i.e. the three roots are $\sqrt[3]{2}$, $\omega \sqrt[3]{2}$, and $\omega^2 \sqrt[3]{2}$. For arithmetic, $\omega^2 = \overline{\omega} = -(1 + \omega)$ if you need to rewrite formulas. – user14972 Mar 25 '18 at 22:02

As $g(x)\in\mathbb{Q}(\rho_1)[x]$ is a polynomial of degree $2$ over a field, it is irreducible if and only if it has no roots in $\mathbb{Q}(\rho_1)$. So you are correct, proving that $\rho_2,\rho_3\notin\mathbb{Q}(\rho_1)$ would indeed prove $g(x)\in\mathbb{Q}(\rho_1)[x]$ is irreducible.
By the way, there is another solution that does not need explicit calculations: You know that $f(x)$ is irreducible over $\mathbb{Q}$ (for example by using Gauß' lemma and the criterion of Eisenstein for the prime $2$ in $\mathbb{Z}$). So the degree of the field extension $\mathbb{Q}\subset \mathbb{Q}(\rho_3)$ is $$[\mathbb{Q}(\rho_3):\mathbb{Q}]=\deg(f)=3.$$ Then, as $\rho_1,\rho_2\in\mathbb{C}\setminus\mathbb{R}$, but $\mathbb{Q}(\rho_3)\subset\mathbb{R}$, we must have $\rho_1,\rho_2\notin\mathbb{Q}(\rho_3)$, so $$[\mathbb{Q}(\rho_1,\rho_2,\rho_3):\mathbb{Q}(\rho_3)]>1.\tag{1}$$ On the other hand, $\mathbb{Q}(\rho_1,\rho_2,\rho_3)=\mathbb{Q}(\rho_1,\rho_3)$, and if $h(x)\in\mathbb{Q}(\rho_3)[x]$ is the minimal polynomial of $\rho_1$ over $\mathbb{Q}(\rho_3)$, $$[\mathbb{Q}(\rho_1,\rho_3):\mathbb{Q}(\rho_3)]=\deg(h).$$ According to (1), we must have $\deg(h)>1$. But $\tilde h(x):=f(x)/(x-\rho_3)\in\mathbb{Q}(\rho_3)[x]$ is a monic polynomial with $\tilde{h}(\rho_1)=0$ and $\deg(\tilde h)=2$, so $h=\tilde h$ and $$[\mathbb{Q}(\rho_1,\rho_2,\rho_3):\mathbb{Q}]=[\mathbb{Q}(\rho_1,\rho_2,\rho_3):\mathbb{Q}(\rho_3)]\cdot[\mathbb{Q}(\rho_3):\mathbb{Q}]=6.$$ Writing $$6=[\mathbb{Q}(\rho_1,\rho_2,\rho_3):\mathbb{Q}]=[\mathbb{Q}(\rho_1,\rho_2,\rho_3):\mathbb{Q}(\rho_1)]\cdot[\mathbb{Q}(\rho_1):\mathbb{Q}].$$ we thus know that $[\mathbb{Q}(\rho_1,\rho_3):\mathbb{Q}(\rho_1)]=2$, so $g$, satisfying $g(\rho_3)=0$ and being monic and of degree $2$ must be the minimal polynomial of $\rho_3$ over $\mathbb{Q}(\rho_1)$, hence irreducible.
Remark: Depending of wheter or not the question let you choose $\rho_1$ freely, you might speed up the argument by choosing $\rho_1=\sqrt[3]{2}$. (For example, concerning your original idea, you could in this case immediately get $\rho_2,\rho_3\notin\mathbb{Q}(\rho_1)$ from $\rho_2,\rho_3\in\mathbb{C}\setminus\mathbb{R}$, but $\mathbb{Q}(\rho_1)\subset\mathbb{R}$.)