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- Feb 14, 2012

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- Thread starter anemone
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If $x = \sqrt{2} + \sqrt[3]{3}$ then $3 = (x-\sqrt2)^3 = x^3 - 3\sqrt2x^2 + 6x - 2\sqrt2$. Therefore $x^3 + 6x - 3 = \sqrt2(3x^2+2)$, and $(x^3 + 6x - 3)^2 = 2(3x^2+2)^2$. So $x^6 - 6x^4 -6x^3 + 12x^2 - 36x + 1 = 0.$

I hope that someone who knows a bit about Galois theory would be able to tell us why 6 is the minimum possible degree for such a polynomial.

- Jan 25, 2013

- 1,225

$\& (\sqrt[3]{3})^3=3 $ (both are integers )

$\therefore 6 =(2\times 3)$ is the minimum possible degree

for such a polynomial with integer coefficients

- Jan 26, 2012

- 37

Right, but to spell it all out:

$\& (\sqrt[3]{3})^3=3 $ (both are integers )

$\therefore 6 =(2\times 3)$ is the minimum possible degree

for such a polynomial with integer coefficients

Let $\alpha = \sqrt{2} + \sqrt[3]{3}$·

First, it is equivalent to having a polynomial over $\mathbb Q$, indeed if $\frac{a_0}{b_0}+\frac{a_1}{b_1}\cdot \alpha + \ldots + \frac{a_n}{b_n} \cdot \alpha^n = 0$ (with $a_i\in {\mathbb Z}$ and $b_i\in {\mathbb Z}^+$) then $a_0 \cdot \left(b_1 \cdot \ldots \cdot b_n\right) + \ldots + a_n \cdot \left(b_1\cdot \ldots \cdot b_{n-1}\right) \cdot \alpha^n = 0$ and $\alpha$ would also be a root of an integer polynomial.

Next we prove that

\[{\mathbb Q} (\sqrt 2,\sqrt[3]{3}) = {\mathbb Q}(\alpha) \]

We will prove that ${\mathbb Q} (\sqrt 2,\sqrt[3]{3}) \subseteq {\mathbb Q}(\alpha)$ since the other inclusion is trivial.From Opalg's calculations we have \[3 = (\alpha-\sqrt2)^3 = \alpha^3 - 3\sqrt2\alpha^2 + 6\alpha - 2\sqrt2\] from where we can solve for $\sqrt 2$ in terms of $\alpha$ (and its powers) and rationals. Hence $\sqrt 2 \in {\mathbb Q}(\alpha)$

Finally, once you know $\sqrt{2} \in {\mathbb Q}(\alpha)$, it follows trivially that $\alpha - \sqrt{2} = \sqrt[3]{3}\in {\mathbb Q}(\alpha)$ and so $\sqrt[3]{3} \in {\mathbb Q}(\alpha)$.

Now it is not difficult to see that $[{\mathbb Q}(\sqrt 2,\sqrt[3]{3}) : {\mathbb Q}(\sqrt[3]{3})] = 2$.

Indeed $\sqrt 2 \not\in {\mathbb Q}(\sqrt[3]{3})$, because $ 3 = [{\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}] = [{\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}(\sqrt{2})] \times [{\mathbb Q}(\sqrt{2}) : {\mathbb Q}] = 2 [{\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}(\sqrt{2})]$ is absurd since $2\not | 3$, and so $x^2 - 2$ has no roots in ${\mathbb Q}(\sqrt[3]{3})$. Thus $x^2 - 2$ is the irreducible polynomial for $\sqrt 2$ over ${\mathbb Q}(\sqrt[3]{3})$ and we have $[{\mathbb Q}(\sqrt 2,\sqrt[3]{3}) : {\mathbb Q}(\sqrt[3]{3})] = 2$.

So that the degree is exactly

\[[{\mathbb Q}(\alpha) : {\mathbb Q}] = [{\mathbb Q}(\sqrt 2, \sqrt[3]{3}) : {\mathbb Q}(\sqrt[3]{3})] \times [ {\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}] = 2 \times 3 = 6 \]

this means that the smallest possible degree is exactly 6.

Last edited:

- Jan 26, 2012

- 37

Indeed, remember ${\mathbb Q}(\alpha) = {\mathbb Q}(\sqrt 2, \sqrt[3]{3})$. And let $F = {\mathbb Q}(\sqrt 2, \sqrt[3]{3}, e^{2\pi i / 3})$. Note that $F/ {\mathbb Q}$ is a Galois extension, since it is the splitting field of $q(x) :=(x^2 - 2)\cdot (x^3 - 3)\in {\mathbb Q}[x]$

It follows now, since the extension $F/{\mathbb Q}$ is Galois, that the minimal polynomial $p(x) \in {\mathbb Q}[x]$ for $\alpha$ is the product \[p(x) = \prod_{a \in S} (x - a) \]

where $S = \{ \sigma (\alpha) : \sigma \in {\text{Gal}} \left(F / {\mathbb Q}\right)\}$, the set of the Galois conjugates of $\alpha$.

But what are the elements of $ {\text{Gal}} \left(F / {\mathbb Q}\right)$ ? Indeed, by the Fundamental Theorem of Galois Theory we have

\[ | {\text{Gal}} \left(F / {\mathbb Q}\right) | = [F : {\mathbb Q}] = 12\]

(the final part can be proved by using $6 = [{\mathbb Q}(\sqrt 2, \sqrt[3]{3}) : {\mathbb Q}]$, and noticing that $e^{2\pi i / 3}$ is a root of $x^2+x+1$ which doesn't have real roots, ando so has no roots in ${\mathbb Q}(\sqrt 2, \sqrt[3]{3}) \subseteq {\mathbb R}$).

Note that if $\sigma \in {\text{Gal}} \left(F / {\mathbb Q}\right)$, then $\sigma$ takes roots of $x^2 - 2$ to roots of $x^2 - 2$, and roots of $x^3 - 3$ to roots of $x^3 - 3$. Hence $\sigma (\sqrt 2) \in \{ \sqrt 2 , -\sqrt 2\}$, and $\sigma (\sqrt[3]{3}) \in \{\sqrt[3]{3}, \sqrt[3]{3}e^{2\pi i / 3},\sqrt[3]{3}e^{4\pi i / 3}\}$. Similarly, we note that $\sigma(e^{2\pi i / 3}) \in \{e^{2\pi i / 3},e^{4\pi i / 3}\}$ too, since it takes roots of $x^2+x+1$ to roots of $x^2+x+1$.

Since there are exactly twelve elements in the Galois group, this means that every choice for $\sigma(\sqrt 2)$, $\sigma(\sqrt[3]{3})$ and $\sigma(e^{2\pi i / 3})$ (by picking from the previous sets), the images of the generators, gives rise to an element of the Galois group. And conversly, this determines all of the elements of the Galois group.

Now it is easily checked then, that if we let $\omega = e^{2\pi i / 3}$ we have \[S = \{\sqrt 2 + \omega^0 \sqrt[3]{3}, \sqrt 2 + \omega \sqrt[3]{3}, \sqrt 2 + \omega^2 \sqrt[3]{3}, -\sqrt 2 + \omega^0 \sqrt[3]{3}, - \sqrt 2 + \omega^1 \sqrt[3]{3}, - \sqrt 2 + \omega^2 \sqrt[3]{3} \}\]

and now $p(x)$ is the product above :

\[p(x) = \prod_{j=0}^1 \prod_{k=0}^2 \left(x - (-1)^j \sqrt 2 - \omega^k \sqrt[3]{3} \right) \]

here, after doing operations for a while, we get

\[p(x) = x^6-6x^4-6x^3+12x^2-36x+1\]

as Opalg found earlier.