Question

Let A be an algebraic algebra over a field k. Show that A is Dedekind-finite.

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 $c = a_{n}b^{n-1} + \ldots + a_{m}b^{m-1} \neq 0$, and we have $cb = bc = 0$. In this case, $b$ is both a left 0-divisor and a right 0-divisor. 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 $A$.