Example 4

Let K be an integral domain, I a nonempty set, and A the polynomial ring $K[X_i]_{i\in I}$. Let us prove that the ring A is without radical. The only invertible elements of A are the invertible elements of K (IV, §1, No. 5, p. 9, Corollary 2). Let $f\in \mathfrak{R}(A)$. Choose an element $i\in I$. Then $1 +fX_i$ is invertible (Theorem 1), which implies $f= 0$.

Note that when K is a commutative field, the ring $A = K[X_i]_{i\in I}$ is a subring of $B = K[[X_i]]_{i\in I}$ and we have $\mathfrak{R}(A) = 0$ and $A\cap \mathfrak{R}(B)\not= 0$ (cf. VIII, p. 154, Example 2).