The notation $\mathbb{Z}[\alpha]$ Let $\alpha$ be a real number. I'm studying group theory from the notes of my brother (I'm 16) and I often jump into the notation $\mathbb{Z}[\alpha]$, which, however, is defined nowhere through the text. I've thought of the smallest subgroup of the additive group of the real field containing $\alpha$ and all of its powers (with positive and negative exponents), but I'm not that sure. Also, is there any special motivation for this notation (whatever it may mean)?
 A: The notation $\mathbb{Z}[\alpha]$ means the set of all expressions of the form
$$
b_0+b_1\alpha+b_2\alpha^2+\dots+b_n\alpha^n
$$
with $b_0,b_1,\dots,b_n\in\mathbb{Z}$ (and whatever non-negative integer $n$), so the “polynomial expressions in $\alpha$” with coefficients in $\mathbb{Z}$”.
When studying abstract algebra it is convenient to think to polynomials $b_0+b_1x+b_2x^2+\dots+b_nx^n$ as simply formal expressions that don't denote a number. They can be summed and multiplied with the usual rules forming what's called a ring whenever the set of coefficients is a ring itself: for example the integers or the reals; they're denoted by $\mathbb{Z}[x]$ and $\mathbb{R}[x]$.
When $f=b_0+b_1x+b_2x^2+\dots+b_nx^n$ is a polynomial with real coefficients (in particular they may be integers) and $\alpha$ is a fixed real number, we can define
$$
f(\alpha)=b_0+b_1\alpha+b_2\alpha^2+\dots+b_n\alpha^n
$$
This has the pleasant properties that
$$
f(\alpha)+g(\alpha)=(f+g)(\alpha),\quad
f(\alpha)g(\alpha)=(fg)(\alpha)
$$
where $f+g$ and $fg$ denote the “formal sum and product” in the ring of (abstract) polynomials.
Thus $\mathbb{Z}[\alpha]$ is just the set of all numbers that you get by computing  $f(\alpha)$ for all polynomials with integer coefficients.
A real number is algebraic if there exists a nonzero polynomial $f\in\mathbb{Z}[x]$ such that $f(\alpha)=0$. It is transcendental otherwise.
So $\sqrt{2}$ is algebraic because $f(\sqrt{2})=0$ where $f=x^2-2$. Giving examples of transcendental numbers is not easy; better, it's not easy to prove that a given real number is transcendental. In 1880 it was proved by Lindemann that $\pi$ is transcendental; together with a result by Wantzel, this finally established that squaring the circle with ruler and compass is not possible.
You may as well use $\mathbb{Q}$ for the coefficients; in this case one can see where the big difference between algebraic and transcendental number is: if $\alpha$ is algebraic, then all numbers
$$
b_0+b_1\alpha+b_2\alpha^2+\dots+b_n\alpha^n\ne0
$$
in $\mathbb{Q}[\alpha]$ have their inverse in $\mathbb{Q}[\alpha]$ again! This is not true for transcendental numbers: if one of the coefficients $b_1,b_2,\dots,b_n$ is non zero and $\alpha$ is transcendental, then the inverse of $b_0+b_1\alpha+b_2\alpha^2+\dots+b_n\alpha^n$ is not in $\mathbb{Q}[\alpha]$.
For instance, $1+2\sqrt{2}\ne0$ and
$$
(1+2\sqrt{2})^{-1}=\frac{1}{1+2\sqrt{2}}=
\frac{-1+2\sqrt{2}}{(-1+2\sqrt{2})(1+2\sqrt{2})}=
\frac{-1+2\sqrt{2}}{7}=-\frac{1}{7}+\frac{2}{7}\sqrt{2}
$$
which is again in $\mathbb{Q}[\sqrt{2}]$.
Conversely, $(1+\pi)^{-1}$ cannot be represented as a polynomial expression with coefficients in $\mathbb{Q}$, because $\pi$ is transcendental.
A: It usually stands for all polynomials in α with integer coefficients.
