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?
[Math] Algebraic Integer and its minimal polynomial
abstract-algebraintegersminimal-polynomialsnormed-spaces
Related Solutions
Let $\alpha,\beta\in\mathbb{C}$ be algebraic integers; e.g. $\alpha^3+\alpha+1=0$ and $\beta^2+1=0$.
Since the polynomials that $\alpha$ and $\beta$ satisfy are monic, we can, by division, write any element of $\mathbb{Z}[\alpha,\beta]$ as an integer combination of elements of the basis, $B$: $$ \left\{1,\alpha,\alpha^2,\beta,\alpha\beta,\alpha^2\beta\right\}\tag1 $$ That is, we can write any $p\in\mathbb{Z}[\alpha,\beta]$ as $$ p=\overbrace{\begin{bmatrix}p_0&p_1&p_2&p_3&p_4&p_5\end{bmatrix}}^{\mathbb{Z}^n}\overbrace{\ \begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}\ }^B\tag2 $$ which means we can write $$ \begin{align} p\ \begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix} &=\overbrace{\begin{bmatrix} p_{0,0}&p_{0,1}&p_{0,2}&p_{0,3}&p_{0,4}&p_{0,5}\\ p_{1,0}&p_{1,1}&p_{1,2}&p_{1,3}&p_{1,4}&p_{1,5}\\ p_{2,0}&p_{2,1}&p_{2,2}&p_{2,3}&p_{2,4}&p_{2,5}\\ p_{3,0}&p_{3,1}&p_{3,2}&p_{3,3}&p_{3,4}&p_{3,5}\\ p_{4,0}&p_{4,1}&p_{4,2}&p_{4,3}&p_{4,4}&p_{4,5}\\ p_{5,0}&p_{5,1}&p_{5,2}&p_{5,3}&p_{5,4}&p_{5,5} \end{bmatrix}}^{\mathbb{Z}^{n\times n}} \begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}\tag3\\[6pt] pB&=MB\tag4 \end{align} $$ Since $M$ satisfies its characteristic polynomial $$ {\large\chi}_M(M)=0\tag5 $$ we have that $p$ also satisfies ${\large\chi}_M$: $$ {\large\chi}_M(p)=0\tag6 $$ Since ${\large\chi}_M$ is a monic integer polynomial, $p$ must be an algebraic integer.
Example 1
For example, $$ (\alpha+\beta)\begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}=\begin{bmatrix} 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ -1 & -1 & 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & -1 & -1 & 0 \end{bmatrix}\begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix} $$ and ${\large\chi}_M(\lambda)=\det(M-I\lambda)=\lambda^6+5\lambda^4+2\lambda^3+4\lambda^2-4\lambda+1$.
Thus, $\alpha+\beta$ is an algebraic integer.
Example 2
Furthermore, $$ \alpha\beta\ \begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 & -1 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix} $$ and ${\large\chi}_M(\lambda)=\det(M-I\lambda)=\lambda^6-2\lambda^4+\lambda^2+1$.
Thus, $\alpha\beta$ is an algebraic integer.
Conclusions
The key points that can be extrapolated from the examples above are that
(1) since $\alpha$ and $\beta$ both satisfy monic integer polynomials, $\mathbb{Z}[\alpha,\beta]$ can be written as a $\mathbb{Z}$-module generated by the Kronecker product (or flattened outer product) of $\left\{1,\alpha,\alpha^2,\dots,\alpha^{k-1}\right\}$ and $\left\{1,\beta,\beta^2,\dots,\beta^{m-1}\right\}$.
(2) the action of $p\in\mathbb{Z}[\alpha,\beta]$ on the basis of $n=km$ generators, from the Kronecker product above, can be represented by $M\in\mathbb{Z}^{n\times n}$.
(3) since $M$ satisfies its own characteristic polynomial, ${\large\chi}_M(M)=0$, so does $p$, ${\large\chi}_M(p)=0$
(4) the characteristic polynomial of an integer matrix is a monic integer polynomial.
Therefore, if $\alpha$ and $\beta$ are algebraic integers, then each $p\in\mathbb{Z}[\alpha,\beta]$ is an algebraic integer.
Best Answer
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…
Conversely, if $a_0=1$ or $a_0=-1$, then either $\hat{f}(x)$ or $-\hat{f}(x)$…