Determine the minimal polynomial of $v=\sqrt{3}+\sqrt[3]{2}$ over $\mathbb Q[x]$.

Can't find the right calculations. I am trying to find another way. I know the minimum polynomial of $w=\sqrt{3}+\sqrt{2}$ but still that does not help. I know it must be of degree 6 since the field extension of $\sqrt{3}$ + $\sqrt[3]{2}$ has degree 6.

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    $\begingroup$ You might find this answer helpful if you already know a bit of Galois theory: math.stackexchange.com/questions/825884/…. Otherwise, there's not much you can do other than get your hands dirty with the algebra: start computing powers of $\sqrt{3}+\sqrt[3]{2}$ and observe linear dependencies once you get to degree 6. Showing irreducibility isn't so easy, but the comments on this question give some insight there: math.stackexchange.com/questions/8409/… $\endgroup$ – Alex Wertheim Feb 22 '16 at 20:18
  • $\begingroup$ It is an exercise on fields extension im supposed to solve it without galois theory.. $\endgroup$ – Manolis Lyviakis Feb 22 '16 at 20:25
  • $\begingroup$ @ManolisLyviakis How do you know the field extension is of degree 6? $\endgroup$ – user26857 Feb 22 '16 at 21:59
  • $\begingroup$ you can check that is the same field extension as $Q(\sqrt{3},\sqrt[3]{2})$ which has a degree 6 which you can prove with the "tower" theorem and some number theory. $\endgroup$ – Manolis Lyviakis Feb 23 '16 at 11:06
  • $\begingroup$ That $[\mathbb Q(\sqrt 3,\sqrt[3]{2}):\mathbb Q]=6$ I can understand, but why $\mathbb Q(\sqrt 3,\sqrt[3]{2})=\mathbb Q(\sqrt 3+\sqrt[3]{2})$? (This seems a much more difficult job than finding a polynomial from $\mathbb Z[X]$ having $\sqrt 3+\sqrt[3]{2}$ as a root. Surprisingly, you claim that managed the hard job, but had troubles with the easy part.) $\endgroup$ – user26857 Feb 24 '16 at 0:21

Take the cube of both sides $v-\sqrt{3}=\sqrt[3]{2}$


We factorize: $3\sqrt{3}(v^2+1)=v^3+9v-2$.

A last squaring of both sides get rid of the last square root:



$v^6 - 9\,v^4 - 4\,v^3 + 27\,v^2- 36\,v -23 = 0$

The left hand side is the minimal polynomial.

  • $\begingroup$ ohhh that was easy i should have isolated the cube first such an idiot $\endgroup$ – Manolis Lyviakis Feb 22 '16 at 20:39

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