Why can't $\alpha - 1$ be a unit here?

Question

I've been reading a the book "Distribution Modulo one and Diophantine Approximation" by Yann Bugeaud. Bugeaud is proving a statement due to Toufic Zaïmi about Salem numbers $\alpha$. A Salem number is an algebraic integer such that one of its Galois conjugates is $\frac{1}{\alpha}$ and all of the other Galois conjugates lie on the unit circle.

Bugeaud shows that for Salem numbers $\mathbb{Q}$ that satisfy a certain additional property, we have that $|P(1)| \neq 1$, where $P$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Bugeaud concludes that $\alpha - 1$ cannot be a unit. I don't understand where this conclusion comes from. Is this a general fact about algebraic integers?

More specifically, if $\alpha$ is an algebraic integer with minimal polynomial $P$, and $\alpha - 1$ is a unit, is it necessarily true that $|P(1)| = 1$? If so, why?

Answer 1

Note Let $P(X)$ be the minimal Polynomial of $\alpha$. Then $Q(X)=P(X+1)$ is irreducible and satisfies $Q(\alpha-1)=0$. Therefore, $Q$ is teh minimal polynomial of $\alpha-1$.

Now, $\alpha-1$ is an unit if and only if $|Q(0)|=|P(1)|=1$.

Answer 2

The norm of $\alpha-c$, where $c\in\Bbb Z$ is $(-1)^dP(c)$ where $d=\deg P$. If $\alpha-1$ is a unit, then its norm is $\pm1$ so $P(1)=\pm1$.