Finding the minimal monic polynomial given a root

Question

Given $z = \sqrt[3]{7} + \sqrt{5} \in \mathbb{R}$ a root, compute its minimal monic polynomial $f(x)\in \mathbb{Q}[x]$. Is there a clever way to do this without having to compute $z^2$, $z^3$, $z^4$, $z^5$ and $z^6$?

Answer 1

Let $z=\sqrt[3]{7}+\sqrt{5}$.

\begin{align*} \text{Then}\;\;& z=\sqrt[3]{7}+\sqrt{5} \\[4pt] \implies\;& z-\sqrt{5}=\sqrt[3]{7} \\[4pt] \implies\;& (z-\sqrt{5})^3=7 \\[4pt] \implies\;& z^3-3z^2\sqrt{5}+15z-5\sqrt{5}=7 \\[4pt] \implies\;& z^3+15z-7=(3z^2+5)\sqrt{5} \\[4pt] \implies\;& (z^3+15z-7)^2=\bigl((3z^2+5)\sqrt{5}\bigr)^2 \\[4pt] \implies\;& z^6+30z^4-14z^3+225z^2-210z+49 = 45z^4+150z^2+125 \\[4pt] \implies\;& z^6-15z^4-14z^3+75z^2-210z-76=0 \\[4pt] \end{align*} hence $z$ is a root of the polynomial $$ x^6-15x^4-14x^3+75x^2-210x-76 $$ To show that the above polynomial is actually the minimal monic polynomial for $z$ over $\mathbb{Q}$, we can argue as follows . . .

By the rational root test, the polynomials $x^3-7$ and $x^2-5$ are irreducible in $\mathbb{Q}[x]$, hence we get $$ [\mathbb{Q}(\sqrt[3]{7}):\mathbb{Q}]=3 \;\;\text{and}\;\; [\mathbb{Q}(\sqrt{5}):\mathbb{Q}]=2 $$ Now let $K=\mathbb{Q}(\sqrt[3]{7},\sqrt{5})$.

Then from $$ [K:\mathbb{Q}]=[K:\mathbb{Q}(\sqrt[3]{7})][\mathbb{Q}(\sqrt[3]{7}):\mathbb{Q}] $$ we get that $[K:\mathbb{Q}]$ is a multiple of $3$, and from $$ [K:\mathbb{Q}]=[K:\mathbb{Q}(\sqrt{5})][\mathbb{Q}(\sqrt{5}):\mathbb{Q}] $$ we get that $[K:\mathbb{Q}]$ is a multiple of $2$.

Hence $[K:\mathbb{Q}]$ is a multiple of $6$, so $[K:\mathbb{Q}]\ge 6$.

But then from $$ K = \mathbb{Q}(\sqrt[3]{7},\sqrt{5}) = \mathbb{Q}(\sqrt[3]{7})(\sqrt{5}) $$ we get \begin{align*} & [K:\mathbb{Q}] \\[4pt] =\;& [\mathbb{Q}(\sqrt[3]{7})(\sqrt{5}):\mathbb{Q}] \\[4pt] =\;& [\mathbb{Q}(\sqrt[3]{7})(\sqrt{5}):\mathbb{Q}(\sqrt[3]{7})][\mathbb{Q}(\sqrt[3]{7}):\mathbb{Q}] \\[4pt] =\;& 3[\mathbb{Q}(\sqrt[3]{7})(\sqrt{5}):\mathbb{Q}(\sqrt[3]{7})] \\[4pt] \le\;& 3{\,\cdot\,}2 \\[4pt] =\;& 6 \\[4pt] \end{align*} and then, since we already have $[K:\mathbb{Q}]\ge 6$, we get $[K:\mathbb{Q}]=6$.

Now let $F=\mathbb{Q}(z)$.

From $z=\sqrt[3]{7}+\sqrt{5}$, we get $z\in K$, hence $F\subseteq K$.

But from the equation $$ z^3+15z-7=(3z^2+5)\sqrt{5} $$ we get $$ \sqrt{5} = \frac {z^3+15z-7} {3z^2+5} $$ so $\sqrt{5}\in F$.

Then since $\sqrt[3]{7}=z-\sqrt{5}$, we get $\sqrt[3]{7}\in F$.

Since $\sqrt{5}\in F$ and $\sqrt[3]{7}\in F$, it follows that $K\subseteq F$.

Then, since we already have the inclusion $F\subseteq K$, we get $F=K$, so $ [F:\mathbb{Q}] = [K:\mathbb{Q}] = 6 $.

From $[F:\mathbb{Q}]=6$, it follows that $$ x^6-15x^4-14x^3+75x^2-210x-76 $$ is the minimal monic polynomial for $z$ over $\mathbb{Q}$.

Answer 2

There is an easy way to calculate using galios group of $\mathbb{Q}(\sqrt{5},\sqrt[3]{7})$, Suppose that the Galios group is $G = \{\sigma \mid automorphisms\}$, then look at the orbit of $z =\sqrt{5} + \sqrt[3]{7}$ under $G$, that will be all the conjugates of $z$, using all the conjugates we get the minimal polynomial of $z$