Question

Let D be a division ring, V =(infinite direct sum)eiD, and E = End(VD). Show that the ring E has exactly three ideals: 0, E, and the ideal consisting of endomorphisms of finite rank.

Answer 1

We specialize following result for case of finite cardinal, and obtain our statement.

For any infinite cardinal $\beta \leq \alpha$ , let $E_{\beta} = \{f \in E : \operatorname{rank}(f) < \beta\}$ , where $\operatorname{rank}(f)$ denotes the cardinal number $\dim_D f(V)$ . Since $\operatorname{rank}(g'fg) \leq \operatorname{rank}(f)$ , $E_{\beta}$ is an ideal of $E$ . We claim that the ideals of $E$ are $(0)$ , $E$ and the $E_{\beta}$ 's. For this, we need the following crucial fact. (*) If $f, h \in E$ are such that $\operatorname{rank}(h) \leq \operatorname{rank}(f)$ , then $h \in EfE$ .

Indeed, write $V = \ker(h) \circ V_1 = \ker(f) \circ V_2$ . Fix a basis $\{u_i : i \in I\}$ for $V_1$ , and a basis $\{v_j : j \in J\}$ for $V_2$ . We have $|I| = \operatorname{rank}(h) \leq \operatorname{rank}(f) = |J|$ , so let us assume, for convenience, that $I \subseteq J$ . Define $g \in E$ such that $g(\ker(h)) = 0$ and $g(u_i) = v_i$ for all $i \in I$ . Noting that $\{f(v_j) : j \in J\}$ are linearly independent, we can also find $g' \in E$ such that $g'(f(v_i)) = h(u_i)$ for all $i \in I$ . Then $h = g'fg$ . In fact, both sides are zero on $\ker(h)$ , and on $u_i (i \in I)$ we have $g'fg(u_i) = g'(f(v_i)) = h(u_i)$ . This proves (*). Now consider any ideal $A < > 0, E$ . Then, for any $f \in A$ , $\operatorname{rank}(f) < \alpha$ . For, if $\operatorname{rank}(f) = \alpha = \operatorname{rank}(\operatorname{Id})$ , then (*) implies $\operatorname{Id} \in EfE \subseteq A$ , a contradiction. Since the class of cardinal numbers is well-ordered, there exists a cardinal $\beta \leq \alpha$ which is least among cardinals larger than $\operatorname{rank}(f)$ for every $f \in A$ . We leave to the reader the easy task of verifying that $\beta$ is an infinite cardinal. Clearly, $A \subseteq E_\beta$ by the definition of $E_\beta$ . We finish by showing that $E_\beta \subseteq A$ . Let $h \in E_\beta$ , so $\operatorname{rank}(h) < \beta$ . By the choice of $\beta$ , we must have $\operatorname{rank}(h) \leq \operatorname{rank}(f)$ for some $f \in A$ . But then, by (*), $h \in EfE \subseteq A$ , as desired. Thus, ring $E$ has exactly three ideals: $0, E$ , and the ideal consisting of endomorphisms of finite rank.