Remark 2

Let A be a commutative ring. The set of nilpotent elements $a$ of A is an ideal $\mathfrak{N}(A)$ of A, called the nilradical of A; it is the intersection of the prime ideals of A (V, §15, No. 1, p. 118, Proposition 2). We have $\mathfrak{N}(A)\subset \mathfrak{R}(A)$; $*$we have equality if A is an Artinian ring (VIII, p. 173, Corollary 2) or a finitely generated commutative algebra over a commutative field (Comm. Alg., V, § 3, n$^{\circ}4$, Theorem $3$)$*$. We can certainly have $\mathfrak{N}(A)\not=\mathfrak{R}(A)$. This is the case when A is the ring K[[X]], where K is a commutative field: we then have $\mathfrak{N}(A) = 0$ and $\mathfrak{R}(A) =$ AX (VIII, p. 154, Example 2).