When asked to find the minimal polynomial of $\sqrt[3]{3}+\sqrt[2]{5}$ over $ \mathbb Q$, I easily found out that $X^6-15X^4-6X^3+75X^2-90X-116$ has $\sqrt[3]{3}+\sqrt[2]{5}$ as a root. It's very likely that it is indeed the minimal polynomial.

All I need to prove now is that it is irreducible over $\mathbb Q$ and equivalently that it is irreducible over $\mathbb Z$. The rational root theorem at least proves (computations needed though) that the polynomial has no root in $\mathbb Q$.

What to do next ? Eisenstein criterion is of no use here.

  • $\begingroup$ Have you studied primitive element theorem? Or for a more elementary approach, have you studied field extensions? $\endgroup$ Commented Jan 4, 2014 at 12:20
  • $\begingroup$ I have only basic knowledge about field extensions. $\endgroup$ Commented Jan 4, 2014 at 12:21
  • $\begingroup$ do you think going modulo $2$ help somehow? $\endgroup$
    – user87543
    Commented Jan 4, 2014 at 12:29
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    $\begingroup$ @Gabriel R. In that case, try to prove that $\mathbb{Q}(\sqrt[3]{3}+\sqrt[2]{5})=\mathbb{Q}(\sqrt[3]{3},\sqrt[2]{5})$ and that $|\mathbb{Q}(\sqrt[3]{3},\sqrt[2]{5}):\mathbb{Q}|=6$. Both of this can be done by elementary means. $\endgroup$ Commented Jan 4, 2014 at 12:33
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    $\begingroup$ The answer by Ayman Hourieh is perfect, but probably there is also a direct calculation? What about mathematicians back in the 18th century, they would have been able to solve this, right? But how? $\endgroup$ Commented Jan 4, 2014 at 12:53

3 Answers 3


We have $[\Bbb Q(\sqrt[3] 3, \sqrt 5) : \Bbb Q(\sqrt 5)] \le 3$ since $\sqrt[3] 3$ is a root of $x^3 - 3$. It's easy to verify that $x^3 - 3$ is irreducible in $\Bbb Q(\sqrt 5)$ for otherwise it would have a root of the form $a + b\sqrt 5$. It follows that the degree considered is indeed $3$. From this we obtain $[\Bbb Q(\sqrt[3] 3, \sqrt 5) : \Bbb Q] = 6$.

Now let $\alpha = \sqrt[3] 3 + \sqrt 5$. Since $\alpha \in \Bbb Q(\sqrt[3] 3, \sqrt 5)$, we have $\Bbb Q(\alpha) \subset \Bbb Q(\sqrt[3] 3, \sqrt 5)$. We claim that they are equal: Since $\alpha - \sqrt 5 = \sqrt[3] 3$, we have $\alpha^3 - 3\alpha^2 \sqrt 5 + 15 \alpha - 5\sqrt 5 - 3 = 0$. This gives an expression for $\sqrt 5$ in $\Bbb Q(\alpha)$. Since $\sqrt[3] 3 = \alpha - \sqrt 5$, we can write both $\sqrt[3] 3$ and $\sqrt 5$ in $\Bbb Q(\alpha)$. It follows that $\sqrt[3] 3, \sqrt 5 \in \Bbb Q(\alpha)$. Hence $\Bbb Q(\sqrt[3] 3, \sqrt 5) = \Bbb Q(\alpha)$ as claimed.

Since $[\Bbb Q(\alpha) : \Bbb Q] = 6$, we conclude that the given polynomial is indeed minimal.

  • $\begingroup$ I can follow your proof until the point where you state "This gives an expression for 5√ in Q(α)". Can you provide a reference which explains what $\mathbb Q(α)$ is, and what roles it plays in field theory ? Somebody used it also on an earlier question of mine: math.stackexchange.com/a/626438/66096 $\endgroup$ Commented Jan 4, 2014 at 13:06
  • $\begingroup$ @GabrielR. $\mathbb{Q}(\alpha) = \mathbb{Q}[\alpha]$ in this situation, and more generally when $\alpha$ is algebraic over the base field. Your textbook should give more details. $\endgroup$ Commented Jan 4, 2014 at 13:10
  • $\begingroup$ @GabrielR. I expanded my answer. $\Bbb Q(a)$ is just a shorthand for $\Bbb Q(\sqrt[3] 3 + \sqrt 5)$. It is the smallest extension that contains $\sqrt[3] 3 + \sqrt 5$. $\endgroup$ Commented Jan 4, 2014 at 13:18
  • $\begingroup$ @AymanHourieh You keep using $a$ when I think you mean $\alpha$. $\endgroup$
    – heropup
    Commented Jan 4, 2014 at 13:19
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    $\begingroup$ Note that $X^6-15X^4-6X^3+75X^2-90X-116=((X-\sqrt 5)^3-3)((X+\sqrt 5)^3-3)$ also provides a proof... $\endgroup$ Commented Jan 4, 2014 at 13:49

Another approach: let $\alpha=\sqrt[3]{3}, \beta=\sqrt{5}$. Writing $(\alpha+\beta)^i$ for $i=0, \ldots, 5$ in the $\{1, \alpha, \alpha^2, \beta, \alpha \beta, \alpha^2 \beta\}$ basis, one finds

$$ \left(\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 5 & 0 & 1 & 0 & 2 & 0 \\ 3 & 15 & 0 & 5 & 0 & 3 \\ 25 & 3 & 30 & 12 & 20 & 0 \\ 150 & 125 & 3 & 25 & 13 & 50 \end{matrix}\right) \left(\begin{matrix} 1 \\ \alpha \\ \alpha^2 \\ \beta \\ \alpha\beta \\ \alpha^2\beta \end{matrix}\right) = \left(\begin{matrix} 1 \\ \alpha+\beta \\ (\alpha+\beta)^2 \\ (\alpha+\beta)^3 \\ (\alpha+\beta)^4 \\ (\alpha+\beta)^5 \end{matrix}\right) $$

The matrix is non-singular, so we can invert it and in particular write $1$, $\alpha$, and $\beta$ as a $\mathbb{Q}$-linear combination of the first six powers of $\alpha+\beta$. This would allow us to compute a (proposed) minimal polynomial for $\alpha+\beta$ and also shows $\mathbb{Q}(\alpha+\beta) \supset \mathbb{Q}(\alpha), \mathbb{Q}(\beta)$, from which it follows quickly that your polynomial is the minimal polynomial.

While Ayman's approach is certainly briefer and cleaner, I find the generality of this approach appealing and illuminating.


Known Result :

Let $I$ be a proper ideal in the integral domain $R$ and let $p(x)$ be a non constant polynomial in $R[x]$. If the image of $p(x)$ in $(R/I)[x]$ is irreducible then $p(x)$ is irreducible in $R[x]$.

This is actually slightly general than Eisenstein. If you want some reference I would suggest you to look at "Dummit Foote -Abstract Algebra -Polynomial Rings -Irreducibility".

Now, $X^6-15X^4-6X^3+75X^2-90X-116$ after going modulo $2$ gives you $x^6-x^4+x^2$ which is seen to be reducible in the first sight.

Now, $X^6-15X^4-6X^3+75X^2-90X-116$ after going modulo $3$ gives you (??). This can be seen to be reducible (??) (Trust me i will not make you do laborious work )

Now, $X^6-15X^4-6X^3+75X^2-90X-116$ after going modulo $5$ gives you (??). This can be seen to be reducible (??) (Trust me i will not make you do laborious work )

Now, $X^6-15X^4-6X^3+75X^2-90X-116$ after going modulo $7$ gives you (??). As claimed by Mr.Martin, This is Irreducible.

I prefer going modulo some prime because this would give a finite field $\mathbb{Z}_p$ and it would not be so difficult to find irreducible polynomials of given degree.

Once you know all irreducible polynomials of degree $2$ and $3$ in $\mathbb{Z}_7$ then you are done...

Just make sure no irreducible polynomial you got of degree $2$ and $3$ divides your $p(x)$ (on $\mathbb{Z}_7$).

Thus, you found some prime $p$ in this case it is $7$ such that $p(x)$ is irreducible in $\mathbb{Z}_7$ and by above stated result $p(x)$ is irreducible in $\mathbb{Z}[x]$ and so is in $\mathbb{Q}[x]$ (why ??)

P.S : This is not any better than other ways but I felt it would be a good way to conclude irreducibility.. choosing a prime $p$ in this case is as difficult as choosing prime $p$ in eisenstein :D so this is not the worst way :).

  • $\begingroup$ why negative vote? $\endgroup$
    – user87543
    Commented Jan 4, 2014 at 14:19
  • $\begingroup$ Because you don't prove anything? $\endgroup$ Commented Jan 5, 2014 at 10:09
  • $\begingroup$ @MartinBrandenburg : If i prove everything that would not be seen as hint... I have already written "why irreducible" but then I thought It would be interesting for OP to do them.. That is why I changed answer for why to $???$ Is that a bad choice? $\endgroup$
    – user87543
    Commented Jan 5, 2014 at 10:12
  • $\begingroup$ I think you should only give a hint in an answer when you know the complete solution. Have you really tested all irreducible polynomials over $\mathbb{F}_7$ of degrees $1,2,3$ and tested for divisibility? Do you think that this kind of solution is doable? $\endgroup$ Commented Jan 5, 2014 at 10:20
  • $\begingroup$ This kind of solution is not doable always but as i have written in P.S choosing such prime is as difficult as choosing prime in eisenstein... This may be a bad choice I do not know! $\endgroup$
    – user87543
    Commented Jan 5, 2014 at 10:23

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