2003 April

Let $R$ be a ring, $I$ a right ideal and $J$ a left ideal of $R$. The expression $I+J$ stands for the Abelian group generated by $I\cup J$. Prove that $$\frac{R}{I}\otimes_R \frac{R}{J} \cong \frac{R}{I+J}.$$ Define your maps carefully with particular attention to welldefinedness.

Let $A_1$ and $A_2$ be modules over some ring, $K_1\subseteq A_1$ and $K_2\subseteq A_2$ be submodules, and $\gamma_1: A_1\rightarrow A_2$ and $\gamma_2: A_2\rightarrow A_1$ and be homomorphisms with the following properties.
i. $\gamma_1(K_1) \subseteq K_2$ and $\gamma_2(K_2) \subseteq K_1$,
ii. $(1\gamma_2\gamma_1)(A_1) \subseteq K_1$ and $(1\gamma_1\gamma_2)(A_2) \subseteq K_2$.
a. Show that the maps $f \colon A_1\oplus K_2 \rightarrow A_2\oplus K_1$ and $g \colon A_2\oplus K_1 \rightarrow A_1\oplus K_2$ defined by \begin{align*} f(x_1,y_2) &= (\gamma_1(x_1)  y_2, x_1 + \gamma_2\gamma_1(x_1)  \gamma_2(y_2)), \text{ and }\\ g(x_2,y_1) &= (\gamma_2(x_2)  y_1, x_2 + \gamma_1\gamma_2(x_2)  \gamma_1(y_1)) \end{align*} are welldefined homomorphisms and that $g = f^{1}$.
b. Let $P_1$ and $P_2$ be projective modules and $M$ a module over some ring $R$. Suppose that $\beta_1: P_1\rightarrow M$ and $\beta_2: P_2 \rightarrow M$ are epimorphisms. Show that $P_1\oplus \ker(\beta_2) \cong P_2\oplus \ker(\beta_1)$. 
Let $M_n(R)$ be the ring of matrices with coefficients in a ring $R$ (with 1), where $n$ is a natural number $> 1$. For a (twosided) ideal $I$ of $R$, let $M_n(I) = \{[a_{ij}]\in M_n(R) : a_{ij}\in I\}$.
a. Prove that $M_n(I)$ is an ideal of $M_n(R)$.
b. Prove that every ideal of $M_n(R)$ is of the form $M_n(I)$ for a suitable ideal $I$ of $R$. 
Recall that a module is simple if it is nontrivial and has no proper submodules.
a. Prove that the endomorphism ring of a simple module is a division ring (= skew field).
b. Prove that a ring $R$ with identity 1 contains a left ideal $I$ such that $R/I$ is a simple left $R$module.
c. Let $R$ be a ring with identity 1 and suppose that $M = _RM$ is a left $R$module with the property that every submodule is a direct summand (“everybody splits”). Show that every nonzero submodule $K$ of $M$ contains a simple submodule.