Is the following set countable::
The set of the roots of all polynomials in one variable with integer coefficients.
Please show the mapping between $N$ and the above set if the set is so
I think it is not countable since for each positive integer we can get uncountably many polynomials of that degree
You have to add a clause: the set of roots of all nonzero polynomials in one variable, with integer coefficients.
Those numbers are the algebraic numbers. Denote by $X$ the set of algebraic integers. Then we have an injective map $\mathbb{N}\to X$, because $n\in\mathbb{N}$ is a root of the polynomial $x-n$.
For each nonzero polynomial $f$ with integer coefficients, denote by $R(f)$ the set of (complex) roots of $f$. Each $R(f)$ is finite, so $X$ is the union of a countable family of finite sets, hence it is at most countable.
Why is the family countable? Because the set of polynomials with integer coefficients is countable.
Proof. It is sufficient to show that the set of polynomials with coefficients in $\mathbb{N}$ is countable (why?).
For a polynomial $f$ with natural coefficients, define its weight $w(f)$ as the sum of its coefficients (which is just $f(1)$). Then, for each $m\in\mathbb{N}$, there is a finite number of polynomials with weight $m$. So the set of polynomials is a countable union of finite sets, hence it is countable.