If $\alpha$ is an algebraic element of $\mathbb{C}$, then there is a unique non-zero polynomial $f \in \mathbb{Q}[x]$ with leading coefficient $1$ such that $f(\alpha) = 0$, and $f$ is irreducible.
The first part of this proof would be proving that $f$ is not a unit, but what does the concept of a unit mean in the set of polynomials? I can't see how a polynomial would have a $2$ sided inverse under multiplication?
A unit in the ring of polynomials over some structure (say, a field) is the same as in any other ring: an element that has a multiplicative inverse within that ring.
If you want to focus on polynomials over fields then the units in that ring are precisely the constant polynomials $\,f(x)=k\,,\,0\neq k\,$ a constant.
Hint $\rm\:\ f\:\!$ unit (i.e. invertible) $\rm\: \Rightarrow\: 1 = f(x)g(x)\:\Rightarrow\: 1 = f(\alpha)g(\alpha) = 0\cdot g(\alpha) = 0$
The special case $\rm\:f(x) = x\:$ is one of my most popular posts (due to its universal view?)
Generally $\rm\:\sum a_i\:\!x^i\:$ is a unit in $\rm R[x] \iff a_0\:$ is a unit$\rm\:R\:$ and $\rm\:a_1,...,a_n$ are nilpotent in $\rm R.$
