Can every polynomial with algebraic coefficients be transformed into a polymonial with integer coefficients and at least some of the same roots?
No; the polynomial $x-\sqrt{2}$ has algebraic coefficients and unique root $\sqrt{2}$. But any polynomial with integer coefficients that has $\sqrt{2}$ as a root also has $-\sqrt{2}$ as a root.
To the edited question; yes. You already note two key facts:
- The field of algebraic numbers is algebraically closed.
- Algebraic numbers are roots of polynomials with integer coefficients.
The first means that every root of a polynomial with algebraic coefficients is itself an algebraic number. The second means that it is a root of a polynomial with integer coefficients.
So given a polynomial $f$ with algebraic coefficients, for every root $\alpha$ of $f$ there exists a polynomial $f_{\alpha}$ with integer coefficients and with root $\alpha$. The product of all these polynomials for all roots $\alpha$ of $f$ then is a polynomial with integer coefficients, with all roots of $f$ as its roots.
To go into a bit more detail than Servaes has, you can start with your original polynomial $f(X)=\sum_na_nX^n$, with the coefficients algebraic. Then for any automorphism $\sigma$ of $\Bbb C$ (there are uncountably many such), you can take $f^\sigma$, by which I mean for you to apply $\sigma$ to all the coefficients of $f$. It’s crucial that there are actually only finitely many different $f^\sigma$ that occur by this process, and that their product has its coefficients in $\Bbb Q$. Then multiply this $\prod f^\sigma$ by an integer sufficient to kill all the denominators, and you’ll have a $\Bbb Z$-polynomial that has all the roots you were originally interested in.