# Minimal polynomial of $\frac{\sqrt{2}+\sqrt[3]{5}}{\sqrt{3}}$

I am struggling to find the minimal polynomial for $\displaystyle \frac{\sqrt{2}+\sqrt[3]{5}}{\sqrt{3}}$. Does anyone have any suggestions?

Thanks,

Katie.

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@Chandru: somehow the last parenthesis is not within the reference. I've noticed this in comments before, but don't know how to make links in a comment. Wikipedia gets you there with an extra click, though. –  Ross Millikan May 31 '11 at 2:58
@Ross: Yes, I don't know how to rectify it. :x) –  user9413 May 31 '11 at 3:02
@Ross, use [text](url). So if the URL has a closing parenthesis, you need to put two. Like so: minimal polynomial –  lhf May 31 '11 at 3:04
@Katie: Over which field? –  user17762 May 31 '11 at 3:39

One possible procedure that works in general is to first find some polynomial that has this number as its root. E.g. you can start as $$\alpha=\frac{\sqrt{2}+5^{1/3}}{\sqrt{3}}\Rightarrow (\sqrt{3}\alpha-\sqrt{2})^3=5,$$ to get rid of the cube root, then factor out $\sqrt{2}$ and get rid of that, etc. Eventually, you will end up with some polynomial, which might not be irreducible. Decompose it into irreducible factors and check which one of these is satisfied by $\alpha$. If you are clever in deriving the initial polynomial, it will have very few irreducible factors. Also, think in advance what degree you might expect your minimal polynomial to have.

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@Katie: After you do the calculation described by Alex B., you might want to check into what Wolfram Alpha thinks the answer is. It gets (as is to be expected) a polynomial of degree $12$. But unfortunately it leaves the proof of irreducibility to you. –  André Nicolas May 31 '11 at 4:03

One can compute the minimal polynomial using resultants or Grobner bases. But that is a bit overkill here since it can be done fairly straightforwardly by hand. Namely, let $\rm\ y = \sqrt{2}+\sqrt[3] 5\:.\$ Then $\rm\: (y-\sqrt 2)^3 = 5\:,\:$ i.e. $\rm\:y^3 + 6\ y - 5 - (3\ y^2 + 2)\ \sqrt{2} = 0\:.\:$ Multiplying that by its conjugate yields $\rm\: y^6 - 6\ y^4 -10\ y^3 + 12\ y^2 -60\ y + 17 = 0\:.\:$ Putting $\rm\ y = \sqrt{3}\ x\:,\:$ then multiplying that by its conjugate yields $\rm\:729\ x^{12} -2916\ x^{10} + 4860\ x^8 - 5670\ x^6 -11340\ x^4 - 9576\ x^2 + 289 = 0\:.$

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There is a systematic procedure for finding a polynomial which annihilates an algebraic element of a field extension and which involves no ad hoc manipulation.Since it is rarely mentioned in textbooks (except sometimes in exercises), I'll describe it. It even works for the elements $a\in A$ of a finite-dimensional commutative algebra $A$ over the field $k$.

Consider on the subalgebra $k[a]\subset A$ the endomorphism $\mu_a:k[a]\to k[a]:x\mapsto ax$. It has a characteristic polynomial $\chi (X)=\chi _{\mu_a}(X)=det( XId-\mu_a) \in k[X]$ which, by Cayley-Hamilton's theorem, annihilates $\mu_a$ and hence also $a$. [Here is an example clarifying the last assertion. If $\mu_a^3+7\mu_a^4-2Id=0\in End(k[a])$, then, applying the endomorphisms on both sides of that equality to the unit element $1_A$, we get $a^3+7a^4-2.1_A=0 \in A$]

Now that we have found an annihilating polynomial $\chi (X)$, the minimal polynomial of $a$ is the same as that of $\mu_a$ and can be found, as has already been mentioned, by decomposing $\chi (X)$ into irreducible factors and making finitely many tests to see which divisors of $\chi (X)$ still kill $a$. [Beware that if the algebra $A$ is not a field, the minimal polynomial of $a$ needn't be irreducible!]

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What are the conjugates of this number? Their elementary functions are the coefficients of the minimal polynomial.

Hint: The conjugates of $\sqrt 2$ are $\pm \sqrt 2$. The conjugates of $\sqrt 3$ are $\pm \sqrt 3$. The conjugates of $\sqrt[3] 5$ are $\omega \sqrt[3]5$, where $\omega^3=1$. Combine all those and you get all conjugates of the number in question. The minimal polynomial is $\prod (X-\alpha)$, where $\alpha$ runs through the conjugates.

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$$289 - 9576x^2 - 11340x^4 - 5670x^6 + 4860x^8 - 2916x^{10} + 729x^{12}$$