# Prove that if $\alpha$ is an algebraic number and a root of $f(x)$ where $f$ has leading coefficient $a$, then $a\alpha$ is an algebraic integer. [duplicate]

I am currently learning Number Theory and one of the appendices of the book I am currently using introduces algebraic numbers and algebraic integers. One of the exercices is to prove the statement in the title: Prove that if $$\alpha$$ is an algebraic number and a root of $$f(x) \in \mathbb{Z}[x]$$ where $$f$$ has leading coefficient $$a$$, then $$a\alpha$$ is an algebraic integer.

However, that statement does not make sense to me when looking at the definition of these objects. If $$\alpha$$ is a root of $$f(x)$$ then $$\alpha$$ is an algebraic number, but since $$f(x)$$ has leading coefficient $$a$$ then $$f(x)$$ is not monic sur $$\alpha$$ is not an algebraic integer.

Now, I thought about taking the function $$g(x) = \frac{1}{a}f(x)$$ so that $$g(x)$$ is monic, but then $$g(x)$$ might not be in $$\mathbb{Z}[x]$$ but we would have $$\alpha$$ an algebraic integer. However this does not solve the issue of $$a\alpha$$ being an algebraic integer. I have tried coming up with an example but even with simple ones it does not make sense to me. Take $$f(x) = 2x^2 - 4$$

then $$\alpha = \sqrt{2}$$ is a solution and $$f(x)$$ has leading coefficient $$a = 2$$ so we can apply the statement I am trying to prove and get that $$2\sqrt{2}$$ is an algebraic number but, even if we forget about the monic part of the definition: $$f(2\sqrt{2}) = 2(2\sqrt{2})^2 - 4 = 16 - 4 = 12 \neq 0$$

So $$2\sqrt{2}$$ is not a root of $$f(x)$$ therefore it is not an algebraic number, even less an algebraic integer.

What am I doing wrong? Is my understanding of these objects wrong in some way?

• But $2 \sqrt{2}$ is solution of $g(x)=x^2-8$. Then it is algebraic Sep 7, 2020 at 3:45
• That makes sense, so it is not necessarily a root of the given $f(x)$ function, we just need to show that we can construct a new monic function $g(x)$ from $f(x)$ such that $a\alpha$ is a root of $g(x)$? Sep 7, 2020 at 3:49
• Yeah. You just need to find some polynomial with integer coefficients such that $a \alpha$ is a root Sep 7, 2020 at 3:53
• I will try that thanks! The textbook does not provide actual examples of these concepts, so I am trying to figure this all by myself. Sep 7, 2020 at 3:56
• See this answer in the dupe for a more conceptual view. Sep 7, 2020 at 5:11

Let $$f(x)=\sum_{i=0}^n a_i x^i$$ and suppose $$a_n>0$$. Since $$\alpha$$ is a root of f(x) we have that it is also a root of $$a_n^{n-1}f(x)=\sum_{i=0}^n a_i a_n^{n-i-1} (a_n x)^i$$. Define $$g(x)=\sum_{i=0}^n a_i a_n^{n-i-1} x^i$$. By construction we have that $$a_n \alpha$$ is root of $$g$$. $$g$$ is monic because $$a_n a_n^{n-n-1}=1$$. So $$a_n \alpha$$ is an algebraic integer. If $$a_n>0$$ consider $$-f$$ and use the same construction to prove for $$-a_n$$. Once you have that $$-a_n \alpha$$ is an algebraic integer it is clear that $$a_n \alpha$$ also is.