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  1. Let $u$ a algebraic number. Prove that exists a natural number $n\in \mathbb{Z}$ such that $nu$ is a algebraic integer
  2. If $u$ is algebraic integer and $n\in \mathbb{Z}$ then $u+n$ and $nu$ are algebraic integers.

I don't see how can I start.

Remember that: $u$ is algebraic integer if it is a root by a monic polynomial $f(x)\in \mathbb{Z}[x]$.

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I don't understand "if $u$ is algebraic integer y $n\in\mathbb Z$ so $u+n$ and $nu$ are algebraic integers." – Alex Becker Aug 26 '12 at 21:29
I think your $f(x)\in\mathbb{Z}[x]$ has to be monic. – Gregor Bruns Aug 26 '12 at 21:33
I think that "y" is Spanish for "and", and "so" should be "then" – Robert Israel Aug 26 '12 at 21:49
Even more is true, though a bit harder to prove: if $u$ and $v$ are algebraic integers, then so are $u\pm v$ and $uv$. – Gerry Myerson Aug 26 '12 at 23:47

Let me give you a hint for the first part. Let's take a concrete example here. Suppose that $x$ satisfies the monic polynomial equation

$$ x^2 - \frac{3}{4}x + \frac{1}{5} = 0 $$

Then you want to prove that there exists an integer $n$ such that $nx$ satisfies a similar monic polynomial equation but with integer coefficients. Well the idea is that you can clear denominators. For example since $4$ and $5$ appear as denominators we can multiply by $20$ to get

$$ 20x^2 - 15x + 4 = 0 $$

But now the problem is that this is no longer monic. But then we can multiply again by 20 so that we get a factor $20^2$ in the leading coefficient which then can be absorbed by the square as follows

$$ 20^2 x^2 - 15\cdot 20x + 4\cdot20 = 0 \implies (\color{red}{20x})^2 -15 \cdot(\color{red}{20x}) + 4\times 20 = 0 $$

so you see that now $\alpha = 20x$ satisfies

$$ \alpha^2 - 15\alpha + 80 = 0 $$

Thus $\alpha = 20x$ is an algebraic integer. And this same idea works in general.

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Yes, this is a good hint. Then, I should take the LCM of demoninators by each coefficient and multiply with f(x), and rewrite the polynomial. But, I didn't write this after, because I was not sure. If I can use LCM in any Field and this outcome is a integer. Sorry. – jonaprieto Aug 26 '12 at 21:58
It's a hit and trial method and not a good method to prove this. For eg. suppose $\beta$ is an algebraic integer satisfying $\beta^3+\beta +1$ and $\gamma$ is an algebraic integer satisfying $\gamma^2 +\gamma -3$ then it would be really hard to show that both $\beta + \gamma$ and $\beta \gamma$ are also algebraic integers! – david watson Sep 10 '15 at 21:38

$(i)$ Let $P(X)=a_mX^m+\cdots+a_1X+a_0 \in \mathbb{Z}[x]$ be such that $P(u)=0$.

Then $a_mu$ is a root of the polynomial.....

$(ii)$. If $P(X)=X^m+a_{m-1}X^{m-1}+\cdots+a_0$ is a polynomial so that $P(u)=0$, then $u+n$ is a root of $P(X-n)$. Also $n^mP(u)=Q(nu)$ for some polynomial $Q$.

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An algebraic number satisfies an equation like $$ P(x)=a_kx^k+a_{k-1}x^{k-1}+a_{k-2}x^{k-2}+\dots+a_1x+a_0=0\tag{1} $$ where $a_j\in\mathbb{Z}$ and $a_k\not=0$.

An algebraic integer satisfies $(1)$ with $a_k=1$.

Hint for 1.

Suppose $u$ satisfies $(1)$ but $a_k\not=1$. Figure out what polynomial $a_ku$ satisfies (and cancel all common factors that you can easily find).

Hint for 2.

Suppose $u$ satisfies $(1)$ with $a_n=1$. Use the binomial theorem to figure out what polynomial $u+n$ satsfies (further hint: note that $P((u+n)-n)=0$). Also, figure out what polynomial $nu$ satisfies (another further hint: note that $n^kP((nu)/n)=0$ and cancel all common factors that you can easily find).

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In $(1)$, technically $a_j \in \Bbb{Q}$, right? Multiplying through by LCM can clear the denominators. – user2468 Aug 27 '12 at 0:01
@JenniferDylan: The definition I learned has all coefficients in $\mathbb{Z}$. Of course, by dividing by the lead coefficient, you can get a monic polynomial with coefficients in $\mathbb{Q}$. – robjohn Aug 27 '12 at 0:06

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