# Thread: Gauss' Lemma and algebraic integers

1. ## Gauss' Lemma and algebraic integers

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.

2. Originally Posted by robeuler
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.
since $f(x)$ is irreducible, every root $a_j$ of $f(x)$ (conjugates of $a$) is also an algebraic integer. now recall that we can write $f(x)=x^n - s_1x^{n-1} + s_2x^{n-2} - \cdots + (-1)^n s_n,$ where

$s_1=\sum_{j=1}^n a_j, \ s_2 = \sum_{1 \leq i < j \leq n} a_ia_j , \ \cdots , s_n=a_1a_2 \cdots a_n.$ thus each $s_j$ is an algebraic integer. we also have $s_j \in \mathbb{Q}$ and so $s_j \in \mathbb{Z},$ because a rational algebraic integer is an integer.

one thing, you should've posted this question in number theory subforum.

3. Originally Posted by robeuler
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 $\alpha \in K$ is called an algebraic integer if $\alpha$ is the root of some monic polynomial with coefficients in $\mathbb{Z}$, where K is an extension field of $\mathbb{Q}$. Let $g(x) \in \mathbb{Z}[x]$ be such a monic polynomial with $g(\alpha)=0$.

If $f(\alpha)=0$, then $f(x)=g(x)h(x)$, where $f(x), h(x) \in \mathbb{Q}[x]$ and $g(x) \in \mathbb{Z}[x]$. If f(x) is irreducible in $\mathbb{Q}[x]$, then $g(x) \in \mathbb{Z}[x]$ should also be irreducible in $\mathbb{Q}[x]$ and h(x) should be a unit in $\mathbb{Q}[x]$, i.e., a unit in $\mathbb{Q}$. Further, if f(x) is monic, then f(x)=g(x). Thus $f(x) \in \mathbb{Z}[x]$.

A corollary from the above definition is that the algebraic integers in $\mathbb{Q}$ are integers $\mathbb{Z}$. For instance, if $\beta$ is an algebraic integer in $\mathbb{Q}$, then the minimal polynomial of $\beta=a/b \in \mathbb{Q}$, where a and b are integers and $b \neq 0$, is $bx - a$. Since $\beta$ is an algebraic integer by hypothesis, $\beta$ should be the root of a monic polynomial with coefficients in $\mathbb{Z}$. Thus b is 1 and $\beta \in \mathbb{Z}$.

4. Originally Posted by TheArtofSymmetry

If $f(\alpha)=0$, then $f(x)=g(x)h(x)$, where $f(x), h(x) \in \mathbb{Q}[x]$ and $g(x) \in \mathbb{Z}[x]$.
that is not true. what we can say is this: $g(x)=f(x)h(x)$ and that's because $f$ is irreducible and $f(a)=g(a)=0.$

5. Originally Posted by TheArtofSymmetry
A corollary from the above definition is that the algebraic integers in $\mathbb{Q}$ are integers $\mathbb{Z}$. For instance, if $\beta$ is an algebraic integer in $\mathbb{Q}$, then the minimal polynomial of $\beta=a/b \in \mathbb{Q}$, where a and b are integers and $b \neq 0$, is $bx - a$. Since $\beta$ is an algebraic integer by hypothesis, $\beta$ should be the root of a monic polynomial with coefficients in $\mathbb{Z}$. Thus b is 1 and $\beta \in \mathbb{Z}$.
It's true that $\beta$ should be the root of some monic polynomial with integer coefficients, but why should $bx-a$ have to be that polynomial? Saying "there exists a monic polynomial with integer coefficients having $\beta$ as a root" is far from saying "all polynomials with integer coefficients having $\beta$ as a root are monic". The second statement is not true!

6. Originally Posted by Bruno J.
It's true that $\beta$ should be the root of some monic polynomial with integer coefficients, but why should $bx-a$ have to be that polynomial? Saying "there exists a monic polynomial with integer coefficients having $\beta$ as a root" is far from saying "all polynomials with integer coefficients having $\beta$ as a root are monic". The second statement is not true!
I can't for the life of me remember why $\frac ab$ is not an algebraic integer for $b\geq2$. Can anyone say why?

7. Originally Posted by chiph588@
I can't for the life of me remember why $\frac ab$ is not an algebraic integer for $b\geq2$. Can anyone say why?
Because of Gauss's lemma : we know that if $f(x) = g(x)h(x) \in \mathbb{Z}[x]$ is monic, and $g,h \in \mathbb{Q}[x]$, then in fact $g,h \in \mathbb{Z}[x]$. So if $f$ has the rational root $\alpha$, we can write $f(x)=(x-\alpha)q(x)$; we have $x-\alpha, q(x) \in \mathbb{Q}[x]$ and therefore $x-\alpha \in \mathbb{Z}[x] \Rightarrow \alpha \in \mathbb{Z}$. (It might not be immediately obvious that $q(x)$ has rational coefficients, but think about it!)

8. Originally Posted by chiph588@
I can't for the life of me remember why $\frac ab$ is not an algebraic integer for $b\geq2$. Can anyone say why?
you can see it directly: suppose $r = \frac{a}{b}, \ a \in \mathbb{Z}, \ b \in \mathbb{N}, \ \gcd(a,b)=1,$ is an algebraic integer. then, by definition, there exists $p(x)=x^n + c_1x^{n-1} + \cdots + c_n \in \mathbb{Z}[x]$ such that $p(r)=0.$

thus $a^n + bc_1a^{n-1} + \cdots + b^na_n=b^n p(r)=0$ and so $b \mid a^n,$ which is impossible unless $b=1$, because $\gcd(a,b)=1.$

9. Originally Posted by NonCommAlg
that is not true. what we can say is this: $g(x)=f(x)h(x)$ and that's because $f$ is irreducible and $f(a)=g(a)=0.$
No, this is true.

Irreducibility of f forces the choice of g to be irreducible in Q[x], and g(x) in Z[x] is a monic polynomial satisfying g(a)=0, where a is an algebraic integer.

g should always exists for sure in this case, otherwise a is not an algebraic integer.

Originally Posted by Bruno J.
It's true that $\beta$ should be the root of some monic polynomial with integer coefficients, but why should $bx-a$ have to be that polynomial? Saying "there exists a monic polynomial with integer coefficients having $\beta$ as a root" is far from saying "all polynomials with integer coefficients having $\beta$ as a root are monic". The second statement is not true!
Because bx-a is the minimal polynomial of $\beta$ and the minimal polynomial of an algebraic integer has integer coefficients and monic for sure.

10. Originally Posted by TheArtofSymmetry
Because bx-a is the minimal polynomial of $\beta$ and the minimal polynomial of an algebraic integer has integer coefficients and monic for sure.
It's true that the minimal polynomial over $\mathbb{Q}$ of an algebraic integer has integer coefficients - but this is a consequence of Gauss's lemma. It's not trivial!