We know that for any algebraic number $\alpha$ $\exists$ $m\in\mathbb{Z}\setminus\{0\}$ such that $m\alpha$ is an algebraic integer. If $\alpha$ is an algebraic integer then $m=1$ suffices. But if $\alpha$ is not an algebraic integer but an algebraic number then we have the following theorem.

Theorem: Let $$f(X)=a_nX^n+a_{n-1}X^{n-1}+\ldots+a_1X+a_0\in\mathbb{Z}[X]\;(a_n>0)$$ be the unique irreducible polynomial with $\gcd(a_n,a_{n-1},\ldots,a_1,a_0)=1$ and $\alpha$ as a root. Then $a_n\alpha$ is an algebraic integer.

Proof: Consider the monic polynomial $$P(X)=X^n+a_{n-1}X^{n-1}+a_na_{n-2}X^{n-2}+\ldots+a_n^{n-2}a_1X+a_n^{n-1}a_0\in\mathbb{Z}[X]$$

Then $$P(a_n\alpha)=(a_n\alpha)^n+a_{n-1}(a_n\alpha)^{n-1}+a_na_{n-2}(a_n\alpha)^{n-2}+\ldots+a_n^{n-2}a_1(a_n\alpha)+a_n^{n-1}a_0\\=a_n^{n-1}(a_n\alpha^n+a_{n-1}\alpha^{n-1}+a_{n-2}\alpha^{n-2}+\ldots+a_1\alpha+a_0)=a_n^{n-1}f(\alpha)=0$$ Hence $a_n\alpha$, being a root of the monic polynomial $P(X)$ in $\mathbb{Z}[X]$, is an algebraic integer.

My question: Denote the set of algebraic integers by $\mathbb{A}$. Then the theorem says for a particular algebraic number $\alpha$ the set $$S_{\alpha}=\{|m|:m\in\mathbb{Z},m\alpha\in\mathbb{A}\}\setminus\{0\}\neq\emptyset$$

Consider the algebraic number $\frac{\sqrt{2}}{3}$. Clearly $3\in S_{\frac{\sqrt{2}}{3}}$. The minimal polynomial in $\mathbb{Z}[X]$ for $\frac{\sqrt{2}}{3}$ is $9X^2-2$. Hence by the theorem $9\in S_{\frac{\sqrt{2}}{3}}$. Moreover since $\frac{\sqrt{2}}{3},\frac{2\sqrt{2}}{3}$ are not algebraic integers we have $\min(S_{\frac{\sqrt{2}}{3}})=3$.

This example shows that $a_n$ is not necessarily $\mathrm{min}(S_{\alpha})$. But by Well-ordering principle $\min(S_{\alpha})$ exists. Can we compute $\min(S_{\alpha})$ in terms of $\alpha$?

  • 1
    $\begingroup$ Isn't $S_\alpha$ always a subgroup of $\Bbb Z$? $\endgroup$ Jun 15, 2020 at 19:31
  • $\begingroup$ Yes and your answer tells you: one can find a leading coefficient $a_n$ in terms of $\alpha$ by finding its minimal polynomial and clearing denominators. Then simply check each of $1,2,...,a_n$ to see which (if any) smaller one exists. (ok, not elegant or efficient but certainly doable in "practical" cases) $\endgroup$
    – user208649
    Jun 15, 2020 at 19:31
  • $\begingroup$ @AnginaSeng Yes if we define $S_{\alpha}$ to be $S_{\alpha}=\{m:m\in\mathbb{Z},m\alpha\in\mathbb{A}\}$ then it's a subgroup of $\mathbb{Z}$. $\endgroup$ Jun 15, 2020 at 19:39
  • $\begingroup$ What kind of answer are you seeking "in terms of $\alpha$"? It surely depends on how $\alpha$ is given. $\endgroup$
    – lhf
    Jun 15, 2020 at 19:39
  • $\begingroup$ I meant when you're told that the symbol "$\alpha$" represents an algebraic number then what is $\min(S_{\alpha})$? I am in search of an answer in closed form. maybe we can use the coefficients of the minimal polynomial of $\alpha$ or some other parameters regarding it. But I want to know if there's any way to express this minimum. $\endgroup$ Jun 15, 2020 at 20:18

1 Answer 1


Note that $S_\alpha$ is an ideal in the integers that contains $a_n$ so it's generator is a divisor of $a_n$. It is easy to figure out the minimal polynomial of $\lambda\alpha$ for any integer $\lambda$, it is simply:

$$a_nx^n/\lambda^n + a_{n-1}x^{n-1}/\lambda^{n-1} + \dots + a_0 = 0$$ and after dividing through by the leading term, the coefficients are $a_k\lambda^{n-k}/a_n$ and we would like all of these to be integers.

In other words, the ideal $S_\alpha$ is generated by that $\lambda$ so that $a_k\lambda^{n-k}/a_n$ is integral for all $k$. This is about as explicit as you can hope to get.


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