How to show that a number is an algebraic Integer

Question

Show that $[\root3\of{100} + \root3\of{10}+ 1]/3$

My attempt is this:

Set it equal to x, multiply both sides by 3, and subtract both sides by 1. Then I cube both sides. However, I am stuck after that. Here's my math below:

$$x=[\root3\of{100} + \root3\of{10}+ 1]/3$$ $$\implies 3x=\root3\of{100} + \root3\of{10}+ 1$$ $$\implies3x-1=\root3\of{100} + \root3\of{10}$$ $$\implies(3x - 1)^3 = (\root3\of{100} + \root3\of{10})^3$$ $$\implies27x^3 - 27x^2 + 9x - 1 = 110 + 3(\root3\of{100})^2(\root3\of{10}) + 3(\root3\of{100})(\root3\of{10})^2$$

If I were to factor the right hand side, then I would get $110 + 3(\root3\of{100})(\root3\of{10})[\root3\of{100} + \root3\of{10}]$. If I were to cube both again, then my right hand side would have something like that. Help would be greatly appreciative.

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Sep 08 '21 at 03:17
Thank you @JoséCarlosSantos, I am pretty much new to math stack exchange and I apologize if my question looks sloppy. — bgj123, Sep 08 '21 at 03:38
Find its minimal polynomial over $\Bbb{Q}$ (a cubic), and check whether the coefficients of that polynomial are integers. That's all there is to it. A surefire way of finding the minimal polynomial is to write $1,x,x^2$ and $x^3$ in terms of the basis ${1,\root3\of{10},\root3\of{100}}$ and solve the coefficients $a,b,c$ from the linear system resulting from $$x^3+ax^2+bx+c=0.$$ — Jyrki Lahtonen, Sep 08 '21 at 03:43
A trick way is to observe that $1+\root3\of{10}+\root3\of{100}$ is, by the geometric sum formula, equal to $(1-10)/(1-\root3\of{10})$. This lets you find the minimal polynomial using that of $1-\root3\of{10}$ and the reciprocal. — Jyrki Lahtonen, Sep 08 '21 at 03:46
That would definitely change my math. Thank you @JyrkiLahtonen for showing me this! — bgj123, Sep 08 '21 at 03:52
@bgj123 "then I would get..." $;-;$ Hint: $;110 + 3\underbrace{(\root3\of{100})(\root3\of{10})}{\displaystyle =, 10}[\underbrace{\root3\of{100} + \root3\of{10}}{\displaystyle=,3x-1}],$. — dxiv, Sep 08 '21 at 04:40
The algebra looks much simpler if you avoid writing radicals and just use $a=\sqrt[3]{10}, a^3=10$ and the number in question is $b=(1+a+a^2)/3$. — Paramanand Singh, Sep 08 '21 at 06:21
The algebraic integer in question is part of an integral basis of $K=\mathbb{Q} (\sqrt [3]{10})$, the basis being ${1,a,b} $ where $a, b$ are as in my previous comment. — Paramanand Singh, Sep 08 '21 at 07:55
The general question is answered here. See also this related example. More discussion. — Jyrki Lahtonen, Sep 08 '21 at 09:13

score 6 · Answer 1 · answered Sep 08 '21 at 04:53

The number $$x=\frac{1+\root3\of{10}+\root3\of{100}}3=\frac3{\root3\of10-1}$$ by the formula for a geometric sum (or by $a^2+ab+b^2=(a^3-b^3)/(a-b)$).

The minimal polynomial of $\root3\of{10}$ is $f(T)=T^3-10$.
Therefore the minimal polynomial of $\root3\of{10}-1$ is $$g(T):=f(T+1)=T^3+3T^2+3T-9.$$
Therefore $1/(\root3\of{10}-1)$ is a zero of the reciprocal polynomial $$h(T):=T^3 g(1/T)=1+3T+3T^2-9T^3.$$
Therefore $x$ is a zero of the polynomial $$m(T):=-3h(T/3)=-3-3T-T^2+T^3.$$

This is monic and has integer coefficients. The claim follows.

The exact same calculation works for the general interesting case of $\root3\of{m}$, $m\equiv1\pmod9$ in place of $m=10$. Somehow I had avoided seeing this approach before, banal as it is. Or it only made it into my subconscious? Live and learn, I suppose :-) — Jyrki Lahtonen, Sep 08 '21 at 12:19

How to show that a number is an algebraic Integer

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