Distinct polynomials with exactly one shared root

Question

If $\alpha_1,\ldots,\alpha_n$ are roots of a polynomial with rational coefficients is there a polynomial with rational coefficients for which $\alpha_1$ is a root but $\alpha_2,\ldots,\alpha_n$ are not?

Answer 1

No. $X^2 + 1$ is a counter example (to the edited question). For rational coeffiecents the polynomial must have both $i$ and the complex conjugate $-i$ as roots (if it has the other).

Answer 2

Hint $\ $ The polynomials $\,f\in \Bbb Q[x]\,$ having $\,\alpha\,$ as root are closed under subtraction, and closed under multiplication by any other polynomial, so they are closed under gcd. Hence a minimal degree polynomial $\,g\,$ having $\,\alpha\,$ as root (e.g. an irreducible one) divides all other such polynomials. Equivalently, $\,f(\alpha) = 0\iff g\mid f\,$ in $\,\Bbb Q[x],\,$ where $\,g = \,$ minimal (irreducible) polynomial of $\,\alpha.$

Remark $\ $ If you know about ideals, then you may find it instructive to translate the above into ideal-theoretic language, using that $\,\Bbb Q[x]\,$ is Euclidean $\,\Rightarrow\,$ PID.