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I have a expr like following:

expr = Collect[Expand[(x - Sqrt[2]) (x - Sqrt[3]) (x - Sqrt[5])], x]

$x^3+\left(-\sqrt{2}-\sqrt{3}-\sqrt{5}\right) x^2+\left(\sqrt{6}+\sqrt{10}+\sqrt{15}\right) x-\sqrt{30}$

How to rationalize irrational equations and keep it with same roots($\sqrt{2},\sqrt{3},\sqrt{5}$)? I'd like to get all the coefficients to integers eventually.

yode
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  • It is not possible unless more roots are added. For example $-30 + 31 x^2 - 10 x^4 + x^6$ has the same roots as your expression plus three more. – azerbajdzan Nov 30 '22 at 18:45
  • @azerbajdzan Why? And how did you get your equation? Can you provide an answer, please? – yode Nov 30 '22 at 18:50
  • I only read a bit of Galois theory but I believe that if you define the set of numbers of the form $m+r\sqrt{n}$ where $m, r, n$ are integers and you define a conjugation action $S$ that operates as a conjugation for $\sqrt{n}$ that is $S(m+r\sqrt{n})=m-r\sqrt{n}$ then if the polynomial coefficients are invariant to $S$ then the roots must come in conjugate pairs related by a change of sign in $-\sqrt{n}$. It is similar to the fact that if the coefficients of a polynomial are real then roots appear as complex conjugate pairs. – userrandrand Nov 30 '22 at 19:15
  • Multiply by the algebraic conjugate: `In[287]:= exprC = (x + Sqrt[2]) (x + Sqrt[3]) (x + Sqrt[5]); Expand[expr*exprC]

    Out[288]= -30 + 31 x^2 - 10 x^4 + x^6`

     – Daniel Lichtblau Nov 30 '22 at 19:18
  • That said, I repeat that I only read some Galois theory out of curiosity and it was years ago so I might be wrong but if there is a theorem it is probably there in the field extension section or something. – userrandrand Nov 30 '22 at 19:22

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As I said in the comment it is impossible but you can have polynomial with the same roots plus roots that make the coefficients integers.

PolynomialLCM @@ 
  MinimalPolynomial[{Sqrt[2], Sqrt[3], Sqrt[5]}, x] // Expand
(* -30 + 31 x^2 - 10 x^4 + x^6 *)
azerbajdzan
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  • Does this theoretically guarantee that this polynomial is a polynomial of the lowest degree? – yode Nov 30 '22 at 18:57
  • Yes, that is why it is called minimal polynomial. – azerbajdzan Nov 30 '22 at 18:59
  • As your answer, the extension degree is $6$, but extensionDegree[{Sqrt[2], Sqrt[3], Sqrt[5]}] is $8$ here. Is there any conflict? – yode Nov 30 '22 at 19:20
  • The degree of the polynomial is 6, it has nothing to do with degree of extension. Extended field works with roots that have been used as extension, so coefficients of polynomials in that field are not all integers as you requested for your polynomial. – azerbajdzan Nov 30 '22 at 19:35
  • Thanks for your patience. I just note a theorem here – yode Nov 30 '22 at 20:02
  • As I see it... the theorem only applies to $Q(r)$, i.e. in cases of a single root. Say $Q(\sqrt{\sqrt{2}+2}+2)$ has extension degree $4$ and also the degree of the minimal polynomial of that root is $4$. But once you have more roots like $Q(\sqrt{2},\sqrt{3})$ then no longer the degree of polynomial matches the degree of extension. – azerbajdzan Nov 30 '22 at 20:15
  • Oh, sorry, that's probably the last doubt. I notice that your method will get a reducible polynomial. Can I get an irreducible polynomial with minimal degree? – yode Nov 30 '22 at 20:24
  • Whether a polynomial is called irreducible depends on the chosen field. $x^2-2$ is irreducible over $Q$ but in NOT irreducible over $Q(\sqrt{2})$. So your question has no sense to me. – azerbajdzan Nov 30 '22 at 20:28
  • I mean $Q$ here – yode Nov 30 '22 at 20:33