Suppose $\alpha$ is a root of $8x^{4}+4x+3$, then find the integers $n\in \mathbb{Z}$ such that $n \alpha$ is an algebraic integer.

Question

Suppose $\alpha$ is a root of $8x^{4}+4x+3$, then find the integers $n\in \mathbb{Z}$ such that $n \alpha$ is an algebraic integer.

(1) An algebraic integer is a complex number that is integral over the integers. That is, an algebraic integer is a complex root of some monic polynomial (a polynomial whose leading coefficient is 1) whose coefficients are integers. The set of all algebraic integers A is closed under addition, subtraction and multiplication and therefore is a commutative subring of the complex numbers.

(2) It is easy to note that for all $m\in \mathbb{Z}$, we have that $2m^{4}$ satisfies that $n \alpha$ is an algebraic integer.

But i don't know how to continue.

Answer 1

Note that the set $I = \{n : n\alpha \text{ is an algebraic integer}\}$ is closed under addition and multiplication by arbitrary integers.

If $n,m \in I$ then $n+m \in I$ because $(n+m)\alpha = n\alpha + m\alpha$ is the sum of two algebraic integers, and thus an algebraic integer. So $n+m \in I$.

If $n \in I$ and $m \in \mathbb{Z}$ then $(mn)\alpha = m(n\alpha)$ is the product of two algebraic integers and thus an algebraic integer.

So $I$ is an ideal of $\mathbb{Z}$.

Given what $n$ you know are and aren't in $I$, you can figure out exactly what $I$ has to be.

Answer 2

It can be checked that the polynomial $8x^4+4x+3$ is irreducible, as multiplying by two gives $(2x)^4+4\cdot(2x)+6$, which is irreducible over $\mathbb{Q}$. Thus $f(x) = x^4 +\frac{x}{2} +\frac{3}{8}$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. If $n\alpha$ is an algebraic integer, it is necessary that its norm $N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(n\alpha) =3n^4/8 $ is an integer. But this is only possible iff $n$ is even.

It can be checked by hand that $2\alpha$ is an algebraic integer. Since the integrality is preserved under addition, we have that $$2k \cdot \alpha= \underbrace{2\alpha + \ldots + 2\alpha}_\textrm{$k$-times} \quad (k >0) $$ is also an algebraic integer. Clearly $-2k$ is also algebraic integer. So all even integers $n$ will work.