Proposition 5

a) The radical of a ring A is the intersection of the annihilators of simple A-modules; it is also the smallest annihilator of a semisimple A-module. It is, in particular, a two-sided ideal of A. If A is not reduced to 0, then the radical of A is distinct from A.

b) Let $\mathfrak{a}$ be a two-sided ideal of A. We have $(\mathfrak{R}(A) +\mathfrak{a})/\mathfrak{a}\subset \mathfrak{R}(A/\mathfrak{a})$. If $\mathfrak{a}$ is contained in $\mathfrak{R}(A)$, then we have $\mathfrak{R}(A/\mathfrak{a}) =\mathfrak{R}(A)/\mathfrak{a}$.

c) The ring $A/\mathfrak{R}(A)$ is without radical; conversely, every two-sided ideal $\mathfrak{a}$ of A such that $A/\mathfrak{a}$ is without radical contains $\mathfrak{R}(A)$.

d) The radical of A is contained in the intersection of the maximal two-sided ideals of A.

Let $x\in A$. Saying that $x$ belongs to the annihilator of every simple A-module is equivalent to saying that $x$ belongs to the annihilator of every element of every simple A-module, in other words (VIII, p. 46, Proposition 1), to every maximal left ideal of A.

Let M be a semisimple A-module. Its annihilator is the intersection of the annihilators of the simple submodules of M, so it contains $\mathfrak{R}(A)$. Furthermore, if $\mathscr{S}$ is the set of classes of simple A-modules (VIII, p. 51), then the direct sum $\oplus_{\lambda\in\mathscr{S}}\lambda$ is a semisimple A-module whose annihilator is $\mathfrak{R}(A)$. Suppose that A is not reduced to 0; the relation $\mathfrak{R}(A)\not= A$ follows from Proposition 2, a) of VIII, p. 153 applied to the A-module $A_s$. We have proved a).

Let $\mathfrak{a}$ be a two-sided ideal of A. The maximal left ideals of $A/\mathfrak{a}$ are the ideals of the form $\mathfrak{m}/\mathfrak{a}$, where $\mathfrak{m}$ is a maximal left ideal of A containing $\mathfrak{a}$. Consequently, the radical of the ring $A/\mathfrak{a}$ is equal to the radical of the A-module $A_s/\mathfrak{a}$. Assertions b) and c) therefore follow from Corollary 1 of VIII, p. 152.

Let $\mathfrak{a}$ be a maximal two-sided ideal of A. In the ring $A/\mathfrak{a}$, the only two-sided ideals are 0 and $A/\mathfrak{a}$. Since the ring $A/\mathfrak{a}$ is not reduced to 0, its radical is not equal to $A/\mathfrak{a}$. The ring $A/\mathfrak{a}$ is therefore without radical, and we have $\mathfrak{R}(A)\subset \mathfrak{a}$ by c). This proves d).

We say that a left (or right) ideal of A is a nil ideal if it consists of nilpotent elements. We say that a two-sided ideal $\mathfrak{a}$ of A is nilpotent if there exists an integer $n\geqslant 1$ such that $\mathfrak{a}^n= 0$, that is (I, §8, No. 9, p. 107), such that we have $x_1\cdots x_n= 0$ for every sequence $(x_1, . . . , x_n)$ of elements of $\mathfrak{a}$. Every nilpotent two-sided ideal is a nil ideal, but there can exist nil ideals that are not contained in a nilpotent two-sided ideal (VIII, p. 167, Exercise 9).