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A complex number $z $ is called an algebraic number if there is $P \in\mathbb {Q }[X]\setminus\{0\} $ such that $P (z) = 0$.

We say that $x\in\mathbb {C} $ is an algebraic integer algebraic integer if there exists a monic polynomial $P\in\mathbb {Z} [X] $ unitary such that $P (x) = 0$.

we fix an algebraic number $z $. The set $$ I (z) = \{P \in\mathbb {Q}[X]\ :P (z) = 0\} $$ is an ideal of $\mathbb {Q}[X]$. There is therefore a unique monic polynomial $\Pi_z \in\mathbb {Q}[X]$, called minimal polynomial of $z$, such that $$I (z) = \{\Pi_z Q :Q \in\mathbb {Q}[X]\}.$$

We admit the following results:

(1) The set of algebraic integers is a subring of $\mathbb {C} $.

(2) If $x \in\mathbb {Q}$ is an algebraic integer, then $x\in\mathbb {Z}$.

Problem

Show that if $z\in\mathbb { C } $ is an algebraic integer then $\Pi_z \in\mathbb {Z} [X] $.

An idea please

Milman
  • 97

2 Answers2

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There’s actually a simple proof not involving some nontrivial considerations on $\mathbb{Z}[X]$ given what you admit.

The roots of $\Pi_z$ are all algebraic integers (because $\Pi_z$ divides a monic polynomial with integral coefficients). So the coefficients of $\Pi_z$ (by the Vieta relations) are algebraic integers as well, and rational numbers. So they’re integers.

Aphelli
  • 34,439
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Fix a nonzero algebraic integer $k$.

Show that there is some monic $p\in \mathbf Z[x]$ which is irreducible (in $\mathbf Z[x]$) and such that $p(k)=0$ (to see this, notice that a proper factor of a monic polynomial over $\mathbf Z[x]$ is also monic, up to sign, of lower degree, and use induction with respect to the degree).

Then by Gauss, it follows that $p$ is irreducible in $\mathbf Q[x]$, so $p=\Pi_k$.

tomasz
  • 35,474