Question

Let A be an algebra over a field k such that every element of A is algebraic over k. Let B be a subalgebra of A, and b ∈ B. Show that b is a unit in B iff it is a unit in A.

Answer 1

Consider any nonzero element $b \in A$, and let $a_{n}b^{n} + \ldots + a_{m}b^{m} = 0$ ($a_{i} \in k$, $a_{n} \neq 0 \neq a_{m}$, $n \geq m$) be a polynomial of smallest degree satisfied by $b$. If $m = 0$, then, for $d = a_{n}b^{n-1} + \ldots + a_{1}$ we have $db = bd = -a_{0} \in k^{*}$. In this case, $b$ is a unit in $A$. If $b \in B$ and $b$ is a unit in $A$, the above also shows that $b^{-1} = -a_{0}^{-1}d \in k[b] \subseteq B$.