If $a$ is algebraic over $\mathbb Z$, every polynomial in $a$ can be expressed in a low degree? I ran into something when working on a problem in Artin's Algebra.

If $a$ is algebraic over $\mathbb Z$ with order $n$ (i.e., the smallest degree integer polynomial with $a$ as a root has degree $n$), is it true that the value of every integer polynomial in $a$ can be expressed as an integer polynomial of degree at most $n$?

I feel like it is true from earlier problems determining the elements of Q or Z with some algebraic number, but I am not sure if this has a simple proof or not.
This makes the solution to the problem I am working on solvable.
Thanks!

No, consider $A=\mathbb{Z}[x]/(2x)$, the element $x$, and $n=1$.
Notice that in ring theory, the notion of "integral element" is better behaved than the notion "algebraic element".
