# Monday February 17th Last time: $R$ is a ring, $M$ a finitely-generated $R\dash$modules, $J\normal R$. Theorem (Nakayama) : If $M = JM$, then there exists an $x\in R$ with $x = 1 \mod (J)$ such that $xM = (0)$. Theorem (Generalized Nakayama) : $M = JM \iff J + \ann M = R$. Proof : The reverse implication is immediate, the forward is by Nakayama. ## Injective Modules Exercise : Show that for $R= R_1 \cross R_2$ and $M = M_1 \cross M_2$, $M_i$ is an $R_i\dash$module. Recall that every $R\dash$module is free $\iff R$ is a field. *Question:* What is the analogous condition for every $R\dash$module to be projective? *Answer:* Every SES $$ 0\to M_1 \to M_2 \to M_3 \to 0 $$ of $R\dash$modules splits. Focusing on $M_2$: every submodule $M_1$ of $M_2$ is a direct summand. Theorem (Characterization of Semisimplicity) : For an $R\dash$module $M$, TFAE 1. Every submodule of $M$ is a direct summand. 2. $M$ is a direct sum of simple modules (semisimple). 3. $M$ is generated by its simple submodules. Definition (Simple Modules) : $M$ is **simple** iff $\exists 0 \subsetneq N \subsetneq_R M$. > In this case, $M \cong R/\ann M$ (i.e. cyclic, monogenic) and the $\ann M$ is maximal. Proof : Omitted, see chapter 8 of notes. Thus every $R\dash$module is projective iff every $R\dash$module is semisimple. Definition (Injective Modules) : Dually, now focusing on $M_1$, every SES starting with $M_1$ is split iff whenever $M_1 \leq M_2$, $M_1$ is a direct summand. In this case we say $M_1$ is *injective*. Proposition (Characterization of Semisimple Modules) : For $R$ a ring, TFAE 1. Every SES of $R\dash$modules splits 2. Every $R\dash$module is projective 3. Every $R\dash$module is semisimple 4. Every $R\dash$module is injective 5. (Claim) $R$ is itself a semisimple $R\dash$module. Proof : $3 \implies 5$ is clear, and we'll prove $5\implies 3$ shortly using *Baer's Criterion*. Definition (Semisimple Rings) : $R$ is **semisimple** iff for all $I\normal R$, there exists a $J\normal R$ such that $I\oplus J = R$. Moreover, $\ann(I) = J$ and $\ann(J) = I$. Exercise (easy) : If $R_i$ are semisimple, $R_1 \cross R_2$ is semisimple. Corollary : Fields are semisimple, so any finite product of fields is semisimple. In fact, the converse is true: Theorem (Semisimple Rings are Products of Fields) : If $R$ is semisimple, then $R$ is a product of fields. Note that everything works here for left modules over non-commutative rings. Let $R$ be a ring. Theorem (Wedderburn-Artin) : A ring[^ring_note] $R$ is semisimple iff $R\cong \prod_{i=1}^r M_{n_i}(D_i)$ a product of matrix rings over division rings. [^ring_note]: Potentially non-commutative, but reduces to previous theorem in commutative case. Let $0\to M_1 \mapsvia{\iota} M_2 \to M_3 \to 0$ be a SES. > Note that splitting is slightly stronger than $M_2 \cong M_1 \oplus M_3$. This sequence is split iff there exists a retraction $\pi: M_2 \to M_1$ such that $\iota\circ\pi = \id_{M_2}$. In this case, $M_2 \cong \iota(M_1) \oplus \ker \pi$. Definition (Injective Modules) : An $R\dash$module $E$ is *injective* iff every SES $0\to E \to M_2 \to M_3 \to 0$ admits a retraction $\pi: M_2 \to E$. Theorem (Characterization of Injective Modules) : For an $R\dash$module $E$, TFAE 1. $E$ is injective 2. Reversing arrows of projective condition, there exists a lift of the following form: \begin{center} \begin{tikzcd} & & & & E \\ & & & & \\ 0 \arrow[rr] & & M \arrow[rr, hook] \arrow[rruu, "\varphi"] & & N \arrow[uu, "\exists \Phi", dotted] \end{tikzcd} \end{center} 3. If $M\injects N$, then $\hom(N, E) \surjects \hom(M, E)$. 4. The contravariant functor $\hom(\wait, E)$ is exact. Note big difference: no analog of being a direct summand of a free module! Free modules are usually not injective. Example : $\ZZ$ is a free but not injective $\ZZ\dash$module. Take $0 \to \ZZ \mapsvia{\times 2} \ZZ \to \ZZ/2\ZZ \to 0$. If this splits, we would have $\ZZ \cong \ZZ \oplus \ZZ/2\ZZ$ as $\ZZ\dash$modules. Why isn't this true? $\ZZ$ is a domain, the LHS is torsionfree, and the RHS has torsion. Example : Suppose now $R$ is a domain and not a field, then let $a$ be a non-unit and run the same argument with multiplication by $a$. This would yield $R \cong R \oplus R/aR$, where the LHS is torsionfree and the RHS has torsion. So $R$ itself need not be an injective $R\dash$module. Definition (Self-Injective Modules) : A ring $R$ is *self-injective* iff $R$ is injective as an $R\dash$module. Example : A field or a semisimple ring. Claim : Let $\tilde R$ be a PID, $\pi$ a prime element, $n\in \ZZ^+$, then take $R\definedas \tilde R/(\pi^n)$. Then $R$ is self-injective. Example : Let $R = \ZZ/p^n\ZZ$, and let $M$ be a finite $p\dash$primary commutative group (i.e. a $p\dash$group). Then $\exp M = p^n \iff \ann M = (p^n)$. $M$ is a faithful $\ZZ/p^n\ZZ\dash$module, so there exists an element $x\in M$ such that $\#\generators{x} = p^n$. There is a SES of $\ZZ/p^n \ZZ\dash$modules \begin{align*} 0\to \generators{x} \to M \to M/\generators{x} \to 0 .\end{align*} Since $\generators{x} \cong \ZZ/p^n \ZZ$, which is self-injective, so there exists a module $N$ such that $M = \generators{x} \oplus N \cong \ZZ/p^n \ZZ \oplus N$. Continuing on $N$ yields a decomposition of $M$ into a sum of cyclic submodules. **Conclusion: a finitely generated torsion module over a PID is a direct sum of cyclic modules.** In general, to show a module is injective, we need to consider lifts over all pairs of modules $M\injects N$. How to do this in practice? Theorem (Baer's Criterion) : If suffices to check the lifting condition for $N = R$ and $M = I\normal R$. I.e. if there is a lift of the following form: \begin{center} \begin{tikzcd} & & & & E \\ & & & & \\ 0 \arrow[rr] & & I \arrow[rr, hook] \arrow[rruu, "\varphi"] & & R \arrow[uu, "\exists \Phi", dotted] \end{tikzcd} \end{center} then $E$ is injective. Proof : Omitted for time. Application : Let $R$ be a semisimple $R\dash$module and let $E$ be any $R\dash$module. Let $I\normal R$, and $f\in \hom(I, E)$. If $R$ is semisimple, then there exists a $J\normal R$ such that $R = I \oplus J$. So extend $f$ to $f \oplus 0$, which yields a lift: \begin{center} \begin{tikzcd} & & & & E \\ & & & & \\ 0 \arrow[rr] & & I \arrow[rr, hook] \arrow[rruu, "f"] & & R = I\oplus J \arrow[uu, "{(f, 0)}", dotted] \end{tikzcd} \end{center} Exercise : Prove the claim that $R$ is self-injective for $R = \tilde R/(\pi^n)$ above.