Algebraic Integer and its minimal polynomial Given k is a nonzero algebraic integer, I am trying to show 1/k is algebraic as well iff  f(0) = 1 or -1 where f(t) is the minimal polynomial over the rationals. Think I can show (<=) by the fact that the constant term of its minimal polynomial gives 1, from where 1/k will be a root of some monic polynomial with the terms in the minimal polynomial reversed.. For (=>) : I am not sure, do we make use of the norm of 1/k?
 A: A number $k$ is an algebraic integer if and only if its (monic) minimal polynomial over the rationals has integer coefficients.
Indeed, if $k$ is the root of a monic polynomial with integer coefficients, it will be the root of one of its irreducible factors, which are monic and with integer coefficients as well (Gauss’ lemma).
The reverse of a degree $n$ polynomial with nonzero constant term $f(x)=a_0+a_1x+\dots+a_nx^n$ is
$$
\hat{f}(x)=a_n+a_{n-1}x+\dots+a_0x^n=x^nf(1/x)
$$
It follows that $f$ is irreducible if and only if $\hat{f}$ is irreducible. (Prove it.)
Now, suppose $k$ is an algebraic integer with minimal polynomial $f(x)=a_0+a_1x+\dots+a_{n-1}x^{n-1}+x^n$. Then $1/k$ is a root of $\hat{f}(x)$, which is irreducible, so $\hat{f}(x)=cg(x)$, where $g(x)=b_0+b_1+\dots+b_{n-1}x^{n-1}+x^n$ is the (monic) minimal polynomial for $1/k$. (Why the degree has to be the same?)
Thus
$$
a_0x^n+a_1x^{n-1}+\dots+a_{n-1}x+1=
cb_0+cb_1x+\dots+cb_{n-1}x^{n-1}+cx^n
$$
In particular, $cb_0=1$ and $c=a_0$, so $a_0b_0=1$.
Suppose $1/k$ is an algebraic integer. Then…

 $b_0$ is an integer, so $a_0=\pm1$.

Conversely, if $a_0=1$ or $a_0=-1$, then either $\hat{f}(x)$ or $-\hat{f}(x)$…

 is a monic polynomial with integer coefficients having $1/k$ as root.

A: Let $k$ be an algebraic integer and consider the field $\mathbb{Q}(k)$. It is a well-known fact that the (field-theoretic) minimal polynomial of $k$ is the same as the (linear algebra) minimal polynomial of the $\mathbb{Q}$-linear map $\varphi: \mathbb{Q}(k) \to \mathbb{Q}(k)$ $x \mapsto kx$. 
Note that if we use the standard basis $B = (1,k,k^2, \dots k^{n-1})$, where $n = [\mathbb{Q}(k):\mathbb{Q}]$, then the matrix of $\varphi$ has integral coefficients. To be precise, if the minimal polynomial of $k$ is $T^n+a_{n-1}T^{n-1}+\dots+a_0$, we have $\operatorname{Mat}_B^B(\varphi)= \begin{pmatrix}
0 & 0 & \dots & 0 & -a_0 \\
1 & 0 & \dots & 0 & -a_1 \\
0 & 1 & \ddots & \vdots & -a_2 \\
\vdots & \ddots & \ddots & 0 & \vdots \\
0 & \dots & 0 & 1 & -a_{n-1} \\
\end{pmatrix}$.
Suppose that $\frac{1}{k}$ is also an algebraic integer, say with minimal polynomial $T^n+b_{n-1}T^{n-1}+\dots+b_0$, then we have $k^{-n}+b_{n-1}k^{-n+1}+\dots+b_0=0$ Multiplying through by $k^{n-1}$, we see that $\frac{1}{k}=-b_{n-1}-b_{n-2}k-\dots-b_0k^{n-1}$, so if we consider the function $\psi: \mathbb{Q}(k) \to \mathbb{Q}(k)$ $x \mapsto \frac{x}{k}$, we have
$\operatorname{Mat}_B^B(\psi)=\begin{pmatrix}
-b_{n-1} & 1 & \dots & 0 & 0 \\
-b_{n-2} & 0 & 1 & 0 & 0 \\
-b_{n-2} & \vdots & \ddots & \ddots & \vdots \\
\vdots   & 0 & \dots &0 & 1 \\
-b_0     & 0 &  \dots &0 &0 \\
\end{pmatrix}$ which again has integral coefficients. But the functions $\varphi$ and $\psi$ are inverses of each other. This means that $\operatorname{Mat}_B^B(\varphi) \in \operatorname{GL}_n(\mathbb{Z})$, but this implies that the determinant of $\varphi$, which is just $a_0$, must be a unit in $\mathbb{Z}$. Note that conversely, if $a_0$ is a unit in $\mathbb{Z}$, $\operatorname{Mat}_B^B(\varphi) \in \operatorname{GL}_n(\mathbb{Z})$, and by Cayley-Hamilton the characteristic polynomial of the inverse matrix is a monic integral polynomial with root $\frac{1}{k}$.
