Gauss' Lemma and algebraic integers

**robeuler** · May 24th 2010, 03:29 PM

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.

**NonCommAlg** · May 24th 2010, 04:05 PM

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.

**TheArtofSymmetry** · May 25th 2010, 05:15 AM

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

----------(Some additional remarks)-----------
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 $\beta \in \mathbb{Z}$ .

**NonCommAlg** · May 25th 2010, 05:52 PM

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 $f(a)=g(a)=0.$

**Bruno J.** · May 25th 2010, 06:03 PM

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!

**chiph588@** · May 25th 2010, 06:16 PM

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?

**Bruno J.** · May 25th 2010, 06:24 PM

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!)

**NonCommAlg** · May 25th 2010, 07:20 PM

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 $\gcd(a,b)=1.$

**TheArtofSymmetry** · May 25th 2010, 07:57 PM

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.

**Bruno J.** · May 25th 2010, 08:23 PM

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!

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