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I was just wondering if anyone knew how one would prove that every $k$-th root of a rational number $r$ is algebraic? My thought would be that some $r$ is equal to $\frac{p}{q}$ but since you are taking the $n$-th root $p$ and $q$ aren't necessarily integers anymore which is a requirement for algebraic- integer coefficients.

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    $\begingroup$ It is an algebraic number for sure but not necessarily an algebraic integer $\endgroup$
    – marwalix
    Oct 23, 2016 at 5:00
  • $\begingroup$ Uh, how would you prove ever even number is an integer? K-th root of a rational number solves the polynomial $x^k=r $ so it is algebraic by definition. $\endgroup$
    – fleablood
    Oct 23, 2016 at 5:37
  • $\begingroup$ Okay, if you have the definition of integer coefficient (which is NOT nescessary ) then you have $nx^k=m $ where $r=m/n $. Integer coefficients are not nescessary as if you have rational coefficients you just multiply both sides by a common denominator. $\endgroup$
    – fleablood
    Oct 23, 2016 at 5:41

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If $r$ is a rational number, then $r=\frac{p}{q}$ for integers $p,q$. Then a $k$th root of $r$ is a root of the polynomial $f(X)=qX^k-p$, which has integer coefficients.

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  • $\begingroup$ This proves it is an algebraic number not an algebraic integer for which we need to display a monic polynomial $\endgroup$
    – marwalix
    Oct 23, 2016 at 5:05
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    $\begingroup$ @marwalix: It looks to me like the question says to show it is an algebraic number. Naturally it won't be an algebraic integer in general. $\endgroup$ Oct 23, 2016 at 5:07
  • $\begingroup$ It seems to me as well but I am confused with the requirement for integer coefficients $\endgroup$
    – marwalix
    Oct 23, 2016 at 5:08
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    $\begingroup$ You can either define algebraic numbers as roots of monic polynomials with rational coefficients, or (by clearing denominators) as roots of (not necessarily monic) polynomials with integer coefficients. $\endgroup$ Oct 23, 2016 at 5:10

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