Throughout R is a ring with 1 and all modules are left R-modules. In Definition 2 in this post, we defined Z(M), the singular submodule of a module M.

Problem 1. Let M be an R-module and suppose that N_1, \cdots, N_k are submodules of M. Prove that \bigcap_{i=1}^k N_i \subseteq_e M if and only if N_i \subseteq_e M for all i.

Solution. We only need to solve the problem for k = 2. If N_1 \cap N_2 \subseteq_e M, then N_1 \subseteq_e M and N_2 \subseteq_e M because both N_1 and N_2 contain N_1 \cap N_2. Conversely, let P be a nonzero submodule of M. Then N_1 \cap P \neq \{0\} because N_1 \subseteq_e M and therefore (N_1 \cap N_2) \cap P = N_2 \cap (N_1 \cap P) \neq \{0\} because N_2 \subseteq_e M. \ \Box

Problem 2. Prove that if M is an R-module, then Z(M) is a submodule of M and Z(R) is a proper two-sided ideal of R. In particular, if R is a simple ring, then Z(R)=\{0\}.

Solution. First note that 0 \in Z(M) because \text{ann}(0)=R \subseteq_e R. Now suppose that x_1,x_2 \in Z(M). Then \text{ann}(x_1+x_2) \supseteq \text{ann}(x_1) \cap \text{ann}(x_2) \subseteq_e M, by Problem 1. Therefore \text{ann}(x_1+x_2) \subseteq_e M and hence x_1+x_2 \in Z(M). Now let r \in R and x \in Z(M). We need to show that rx \in Z(M). Let J be a nonzero left ideal of R. Then Jr is also a left ideal of R. If Jr = \{0\}, then J \subseteq \text{ann}(rx) and thus \text{ann}(rx) \cap J = J \neq \{0 \}. If Jr \neq \{0\}, then \text{ann}(x) \cap Jr \neq \{0\} because x \in Z(M). So there exists s \in J such that sr \neq 0 and srx = 0. Hence 0 \neq s \in \text{ann}(rx) \cap J. So rx \in Z(M) and thus Z(M) is a submodule of M. Now, considering R as a left R-module, Z(R) is a left ideal of R, by what we have just proved. To see why Z(M) is a right ideal, let r \in R and x \in Z(R). Then \text{ann}(xr) \supseteq \text{ann}(x) \subseteq_e R and so \text{ann}(xr) \subseteq_e R, i.e. xr \in Z(R). Finally, Z(R) is proper because \text{ann}(1)=\{0\} and so 1 \notin Z(R). \ \Box

Problem 3. Prove that if M_i, \ i \in I, are R-modules, then Z(\bigoplus_{i \in I} M_i) = \bigoplus_{i \in I} Z(M_i). Conclude that if R is a semisimple ring, then Z(R)=\{0\}.

Solution. The first part is a trivial result of Problem 1 and this fact that if x = x_1 + \cdots + x_n, where the sum is direct, then \text{ann}(x) = \bigcap_{i=1}^n \text{ann}(x_i). The second now follows trivially from the first part, Problem 2 and the Wedderburn-Artin theorem. \Box

Problem 4. Suppose that R is commutative and let N(R) be the nilradical of R. Prove that

1) N(R) \subseteq Z(R);

2) it is possible to have N(R) \neq Z(R);

3) if Z(R) \neq \{0\}, then N(R) \subseteq_e Z(R), as R-modules or Z(R)-modules.

Solution. 1) Let a \in N(R). Then a^n = 0 for some integer n \geq 1. Now suppose that 0 \neq r \in R. Then ra^n=0. Let m \geq 1 be the smallest integer such that ra^m = 0. Then 0 \neq ra^{m-1} \in \text{ann}(a) \cap Rr and hence a \in Z(R).

2) Let R_i = \mathbb{Z}/2^i \mathbb{Z}, \ i \geq 1 and put R=\prod_{i=1}^{\infty}R_i. For every i, let a_i = 2 + 2^i \mathbb{Z} and consider a = (a_1,a_2, \cdots ) \in R. It is easy to see that a \in Z(R) \setminus N(R).

3) Let a \in Z(R) \setminus N(R). Then \text{ann}(a) \cap Ra \neq \{0\} and thus there exists r \in R such that ra \neq 0 and ra^2=0. Hence (ra)^2 = 0 and so ra \in N(R). Thus 0 \neq ra \in N(R) \cap Ra implying that N(R) is an essential R-submodule of Z(R). Now, we view Z(R) as a ring and we want to prove that N(R) as an essential ideal of Z(R). Again,  let a \in Z(R) \setminus N(R). Then \text{ann}(a) \cap Ra^2 \neq \{0\} and thus there exists r \in R such that ra^2 \neq 0 and ra^3 = 0. Let s = ra \in Z(R). Then (sa)^2=0 and thus 0 \neq sa \in N(R) \cap Z(R)a implying that N(R) is an essential ideal of Z(R). \ \Box

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