An algebraic element is invertible iff its minimal polynomial has a nonzero constant term

Let $𝐴$ be a [[K-monoid| $𝕂$ -ring]] and $𝑎 \in 𝐴$ be an algebraic element with minimal polynomial $𝑚 𝑎 (𝑥) \in 𝕂 [𝑥]$ . Then the following are equivalent¹

$𝑎$ is invertible in $𝐴$
$𝑎$ is not a left Zero-divisor
$𝑎$ is not a right Zero-divisor
$𝑎$ is not a (two-sided) Zero-divisor
$𝑚 𝑎 (𝑥)$ has a nonzero constant term, i.e. $𝑚 𝑎 (0) \neq 0$ .

Proof

If $𝑎 \in 𝐴$ is invertible but $𝑚 𝑎 (𝑥) = 𝑥 𝑝 (𝑥)$ for some $𝑝 (𝑥) \in 𝕂 [𝑥]$ , then $𝑎 𝑝 (𝑎) = 𝑚 𝑥 (𝑎) = 0$ and thus $𝑝 (𝑎) = 0$ , a contradiction. For the converse, if $𝑚 𝑎 (𝑥) = \sum 𝑛 𝑖 = 0 𝛼 𝑖 𝑥 𝑛$ and $𝛼 0 \neq 0$ , then
$- 1 𝛼 0 (𝑛 - 1 \sum 𝑖 = 1 𝛼 𝑖 𝑎 𝑖) 𝑎 = 1$
so
$𝑎 - 1 = - 1 𝛼 0 (𝑛 - 1 \sum 𝑖 = 1 𝛼 𝑖 𝑎 𝑖)$
is the inverse.

tidy | en | SemBr

2008. Advanced Linear Algebra, §18, pp. 459–461 ↩

An algebraic element is invertible iff its minimal polynomial has a nonzero constant term

Graph View

Backlinks

An algebraic element is invertible iff its minimal polynomial has a nonzero constant term

Footnotes

Graph View

Backlinks