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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Nov 30, 2022 at 18:45
  • $\begingroup$ @azerbajdzan Why? And how did you get your equation? Can you provide an answer, please? $\endgroup$
    – yode
    Commented Nov 30, 2022 at 18:50
  • $\begingroup$ 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. $\endgroup$ Commented Nov 30, 2022 at 19:15
  • $\begingroup$ 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 $\endgroup$ Commented Nov 30, 2022 at 19:18
  • $\begingroup$ 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. $\endgroup$ Commented Nov 30, 2022 at 19:22

1 Answer 1

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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 *)
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  • $\begingroup$ Does this theoretically guarantee that this polynomial is a polynomial of the lowest degree? $\endgroup$
    – yode
    Commented Nov 30, 2022 at 18:57
  • $\begingroup$ Yes, that is why it is called minimal polynomial. $\endgroup$ Commented Nov 30, 2022 at 18:59
  • $\begingroup$ As your answer, the extension degree is $6$, but extensionDegree[{Sqrt[2], Sqrt[3], Sqrt[5]}] is $8$ here. Is there any conflict? $\endgroup$
    – yode
    Commented Nov 30, 2022 at 19:20
  • $\begingroup$ 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. $\endgroup$ Commented Nov 30, 2022 at 19:35
  • $\begingroup$ Thanks for your patience. I just note a theorem here $\endgroup$
    – yode
    Commented Nov 30, 2022 at 20:02

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