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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 $f(x)\in \mathbb{Z}[x]$.

  • 1
    I don't understand "if $u$ is algebraic integer y $n\in\mathbb Z$ so $u+n$ and $nu$ are algebraic integers."2012-08-26
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    I think your $f(x)\in\mathbb{Z}[x]$ has to be monic.2012-08-26
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    I think that "y" is Spanish for "and", and "so" should be "then"2012-08-26
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    Even more is true, though a bit harder to prove: if $u$ and $v$ are algebraic integers, then so are $u\pm v$ and $uv$.2012-08-26

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