# Gauss' Lemma and algebraic integers

#### 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.

#### NonCommAlg

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

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

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

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#### TheArtofSymmetry

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

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 $$\displaystyle \beta \in \mathbb{Z}$$.

#### NonCommAlg

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

#### Bruno J.

MHF Hall of Honor
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 $$\displaystyle \beta \in \mathbb{Z}$$.
It's true that $$\displaystyle \beta$$ should be the root of some monic polynomial with integer coefficients, but why should $$\displaystyle bx-a$$ have to be that polynomial? Saying "there exists a monic polynomial with integer coefficients having $$\displaystyle \beta$$ as a root" is far from saying "all polynomials with integer coefficients having $$\displaystyle \beta$$ as a root are monic". The second statement is not true!

chiph588@

#### chiph588@

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

#### Bruno J.

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

chiph588@

#### NonCommAlg

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

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

#### TheArtofSymmetry

that is not true. what we can say is this: $$\displaystyle g(x)=f(x)h(x)$$ and that's because $$\displaystyle f$$ is irreducible and $$\displaystyle f(a)=g(a)=0.$$
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.

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

#### Bruno J.

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