Let $F$ be a subfield of $K$. It is easy to see that if $\alpha $ is algebraic over $F$, then its minimal polynomial $p(x)$ is (monic) unique and irreducible.
Taking the quotient, $F[x]/p(x)F[x]$ is a field (as $p(x)$ is irreducible), and any element $g\in F$ can be represented uniquely as $p(x)F(x)$+ a polynomial of $degree<n$, where $n=deg \ p(x).$
I can immediately see that if instead of taking the minimal polynomial $p(x)$, we take any polynomial $f$ and quotient by its ideal $(f)$, we get a ring (not necessarly a field).
In this case, the representation does not need to be unique, right?