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If $a$ is an algebraic number, show that there exists a positive integer $n$ such that $na$ is an algebraic integer.

I'm supposing that $a$ is algebraic over a field $F$, that is, there exists a polynomial $p(x)$ over $F$ such that $p(a) = 0$ . If we think about $F$ being $\mathbb{Q}$, then $a = p/q$. And if we think that $p(x) = x-a$, we have that $p(x) = x-p/q$ is our polynomial. But I don't know if it helps. I don't even know if we can suppose that $x-a$ is our polynomial

Magdiragdag
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Guerlando OCs
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    Hmm, usually "algebraic integer" means a complex number that is a root of a monic polynomial with integer coefficients. Besides, is your statement right? you can take $n=1$. – Xam Sep 19 '17 at 05:24
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    @Xam yes, I think I misunderstood the definition of algebraic integer – Guerlando OCs Sep 19 '17 at 05:25

2 Answers2

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Let $f(a)=0$ be monic, degree $n$. If the polynomial is not already a polynomial over the integers, it has a term whose coefficient is rational, with denominator $q$ when fully reduced. Suppose that's the coefficient of $x_k$.

Let $g(x)=q^nf(x/q)$. Note that the coefficient of $x^k$ in $g$ is a whole number, $g$ is monic, and $g(qa)=0$.

If $g$ still has some non-integer rational coefficients, repeat until you do have a polynomial whose coefficients are all integers.

Michael Hartley
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  • https://en.wikipedia.org/wiki/Algebraic_number, https://en.wikipedia.org/wiki/Algebraic_integer#Definitions – user26857 Sep 19 '17 at 17:43
  • if the polynomial is not already a polynomial over the integers, it has a term whose coefficient is rational, with denominator $q$. Suppose that's the coefficient of $x^k$. Then $g$ is a polynomial with $g(qa)=0$, and the coefficient of $x^k$ is an integer. Repeat until you have a coefficient whose coefficients are all integers. – Michael Hartley Sep 21 '17 at 12:34
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Hint $\ $ Recall from high-school the AC-method that transforms factoring any polynomial into one that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $=1)$ e.g. as follows:

$$\quad\ \ \begin{eqnarray} f &\,=\,& \ \ 9\ x^2-\ 80\ x\ -\,\ 9\\ \Rightarrow\ 9f &\,=\,& (9x)^2\! -80(9x)-81\\ &\,=\,& \ \ \ \ \color{#c00}{X^2\!- 80\ X\ -\,\ 81},\,\ \ X\, =\, 9x\\ &\,=\,& \ \ \ \,(X-81)\ (X+\,1)\\ &\,=\,& \ \ \ (9x-81)\,(9x+1)\\ \Rightarrow\ f\,=\, 9^{-1}(9f) &\,=\,& \ \ \ \ \ (x\ -\ 9)\,(9x+1)\\ \end{eqnarray}$$

If we denote our factoring algorithm by $\,\cal F,\,$ then the above transformation is simply

$$\cal F f\, = a^{-1}\cal F\, a\,f\quad\,$$

Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. This is sometimes called the AC method. It works for higher degree polynomials too. As above, we can reduce the problem of factoring a non-monic polynomial to that of factoring a monic polynomial by scaling by a $ $ power of the lead coefficient $\rm\:a\:$ then changing variables: $\rm\ X = a\:x$

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\rm\: X^2 + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\smash[t]{\overbrace{ac}^{\rm\qquad\ \ \ \ \ {\bf AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\ \\ \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c \end{eqnarray}$$ After factoring the monic $\rm\,g(X)\, =\, a^{n-1}f(x),\,$ we are guaranteed that the transformation reverses to yield a factorization of $\rm\:f,\ $ since $\rm\ a^{n-1}$ must divide into the factors of $\rm\ g\ $ by Gauss' Lemma, i.e. primes $\,p\in\rm\mathbb Z\,$ remain prime in $\rm\,\mathbb Z[X],\,$ so $\rm\ p\ |\ g_1(x)\:g_2(x)\,$ $\Rightarrow$ $\,\rm\:p\:|\:g_1(x)\:$ or $\rm\:p\:|\:g_2(x).$

Clearly this resolves your problem (and more). See this answer for further discussion.

Bill Dubuque
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