Remark 4

Let $R\in K(X)$ be a rational fraction and let $x\in A$ be an element which can be substituted in R, that is to say (A, IV, p. 20), that there exist P and $Q\in K[X]$ such that $R = P/Q$ and $Q(x)$ is invertible; one can then form the element $R(x) = P(x)\cdot Q(x)^{-1}= Q(x)^{-1}\cdot P(x)$ of A; it does not depend on the choice of P and Q. One has $0\notin$ Q(Sp($x$)), so that every element of Sp($x$) can be substituted in R.

One has R(Sp($x$))$\subset$ Sp(R($x$)). For let $\lambda \in$ Sp($x$); there exists a polynomial $P_1$ such that $R(\lambda )-R(X) = (\lambda -X)P_1(X)/Q(X)$; then, $R(\lambda )e-$ $R(x) = (\lambda e-x)(P_1(x)/Q(x))$, so that $R(\lambda )-R(x)$ is not invertible, hence $R(\lambda )\in$ Sp(R($x$)).

Conversely, suppose that the field K is algebraically closed. Suppose first that R is not constant and let us prove that one has Sp(R($x$)) $=$ R(Sp($x$)). Let $\mu\in$ Sp(R($x$)). Since R is not constant, $P-\mu Q$ is not the zero polynomial; let $\mu Q-P =\alpha \prod(\lambda_i-X)$ be a decomposition into factors of degree 1, so that $\mu e-R(x) =$ $\alpha \prod(\lambda_ie-x)Q(x)^{-1}$. Since $\mu e-R(x)$ is not invertible, there exists $i$ such that $\lambda_ie-x$ is not invertible, hence $\lambda_i\in$ Sp($x$), and then $R(\lambda_i) =\mu\in$ R(Sp($x$)).

When R is constant, the equality Sp(R($x$)) $=$ R(Sp($x$)) also holds, provided that Sp($x$) is nonempty.