Let \(R\) be a commutative ring containing a field \(k\) such that \(\dim_k(R) = d < \infty\). Let \(a \in R\). (1) Show that there exist \(n \in \mathbb{N}\) with \(1 \le n \le d\) and coefficients \(c_0, c_1, \dots, c_{n-1} \in k\) such that: \[a^n + c_{n-1}a^{n-1} + \cdots + c_1 a + c_0 = 0.\] (2) Suppose such a relation holds. Show that if \(c_0 \ne 0\), then \(a\) is a unit in \(R\). (3) Show that if \(a\) is not a zero-divisor in \(R\), then \(a\) is invertible in \(R\).
Solution
Since \(\dim_kR=d\), the \(d+1\) elements \[ 1,a,a^2,\dots,a^d \] are linearly dependent over \(k\). Thus there is a nonzero polynomial of degree at most \(d\) annihilating \(a\). Choosing one of minimal degree and dividing by its leading coefficient gives \[ a^n+c_{n-1}a^{n-1}+\cdots+c_0=0, \qquad 1\le n\le d, \] with \(c_i\in k\).
If \(c_0\ne0\), then \[ a\bigl(a^{n-1}+c_{n-1}a^{n-2}+\cdots+c_1\bigr)=-c_0, \] so \[ a^{-1}=-c_0^{-1}\bigl(a^{n-1}+c_{n-1}a^{n-2}+\cdots+c_1\bigr). \] Hence \(a\) is a unit.
Finally suppose \(a\) is not a zero-divisor. Multiplication by \(a\), \[ m_a:R\to R,\qquad x\mapsto ax, \] is an injective \(k\)-linear endomorphism of the finite-dimensional vector space \(R\). Therefore it is surjective. In particular there exists \(b\in R\) with \(ab=1\), so \(a\) is invertible.