Let a be an algebraic integer and suppose that f(a)=0 for a polynomial f, with f in Q[x] (a polynomial with rational coefficients). Suppose f(x) is irreducible and monic. Show that f(x) is in Z[x].
I feel kind of stupid... I know that Gauss' lemma reduces problems in Q[x] to the integers but I don't see exactly how to draw this conclusion.
This is another approach directly from the definition of an algebraic integer.
By definition, an element \(\displaystyle \alpha \in K\) is called an algebraic integer if \(\displaystyle \alpha\) is the root of some monic polynomial with coefficients in \(\displaystyle \mathbb{Z}\), where K is an extension field of \(\displaystyle \mathbb{Q}\). Let \(\displaystyle g(x) \in \mathbb{Z}[x]\) be such a monic polynomial with \(\displaystyle g(\alpha)=0\).
If \(\displaystyle f(\alpha)=0\), then \(\displaystyle f(x)=g(x)h(x)\), where \(\displaystyle f(x), h(x) \in \mathbb{Q}[x]\) and \(\displaystyle g(x) \in \mathbb{Z}[x]\). If f(x) is irreducible in \(\displaystyle \mathbb{Q}[x]\), then \(\displaystyle g(x) \in \mathbb{Z}[x]\) should also be irreducible in \(\displaystyle \mathbb{Q}[x]\) and h(x) should be a unit in \(\displaystyle \mathbb{Q}[x]\), i.e., a unit in \(\displaystyle \mathbb{Q}\). Further, if f(x) is monic, then f(x)=g(x). Thus \(\displaystyle f(x) \in \mathbb{Z}[x]\).
----------(Some additional remarks)-----------
A corollary from the above definition is that the algebraic integers in \(\displaystyle \mathbb{Q}\) are integers \(\displaystyle \mathbb{Z}\). For instance, if \(\displaystyle \beta\) is an algebraic integer in \(\displaystyle \mathbb{Q}\), then the minimal polynomial of \(\displaystyle \beta=a/b \in \mathbb{Q}\), where a and b are integers and \(\displaystyle b \neq 0\), is \(\displaystyle bx - a \). Since \(\displaystyle \beta\) is an algebraic integer by hypothesis, \(\displaystyle \beta\) should be the root of a monic polynomial with coefficients in \(\displaystyle \mathbb{Z}\). Thus b is 1 and \(\displaystyle \beta \in \mathbb{Z}\).