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It was an exercise in Isaac's character theory:

Let $\alpha$ be an algebraic integer (so $\alpha$ satisfies monic irreducible polynomial $g(x)$ over $\mathbb{Z}$).

Let $\alpha$ satisfies a monic irreducible polynomial $f(x)$ over $\mathbb{Q}$.

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

My question is about conclusion. This $f(x)$ is actually $g(x)$ (which is in $\mathbb{Z}[x]$), which I want to conclude. This is done as follows.

Since $\alpha$ is algebraic integer, it satisfies a monic irreducible polynomial $g(x)$ over $\mathbb{Z}$; by Gauss lemma, the polynomial $g(x)$ is irreducible over $\mathbb{Q}$; then finally, monic irreducible polynomial over a field satisfied by an element (of some extension field) should be unique, hence $f(x)=g(x)$ [and so $f(x)\in\mathbb{Z}[x]$].

Q. Is this conclusion right?

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    $\begingroup$ I found no flaw in your reasoning. $\endgroup$ – Toast May 5 '17 at 4:28
  • $\begingroup$ @Beginner -- Yes, it's fine. $\endgroup$ – quasi May 5 '17 at 4:36

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