# Turning An Algebraic Number Into An Algebraic Integer

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 $$\alpha$$?

• Isn't $S_\alpha$ always a subgroup of $\Bbb Z$? Jun 15, 2020 at 19:31
• 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)
– user208649
Jun 15, 2020 at 19:31
• @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 $\mathbb{Z}$. Jun 15, 2020 at 19:39
• What kind of answer are you seeking "in terms of $\alpha$"? It surely depends on how $\alpha$ is given.
– lhf
Jun 15, 2020 at 19:39
• 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. Jun 15, 2020 at 20:18

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.