# Tuesday October 29th ## Exact Sequences **Lemma (Short Five):** Consider a diagram of the following form: ```{=latex} \begin{tikzcd} 0 \arrow[r] & M \arrow[dd, "\alpha"] \arrow[r, "f"] & N \arrow[dd, "\beta"] \arrow[r, "g"] & Q \arrow[dd, "\gamma"] \arrow[r] & 0 \\ & & & & \\ 0 \arrow[r] & M' \arrow[r, "f'"] & N' \arrow[r, "g'"] & Q' \arrow[r] & 0 \end{tikzcd} ``` 1. $\alpha, \gamma$ monomorphisms implies $\beta$ is a monomorphism. 2. $\alpha, \gamma$ epimorphisms implies $\beta$ is an epimorphism. 3. $\alpha, \gamma$ isomorphisms implies $\beta$ is an isomorphism. Moreover, (1) and (2) together imply (3). *Proof:* Exercise. *Example proof of (2)*: Suppose $\alpha, \gamma$ are monomorphisms. - Let $n\in N$ with $\beta(n) = 0$, then $g' \circ \beta(n) = 0$. - $\implies \gamma \circ g (n) = 0$. - $\implies g(n) = 0$ - $\implies \exists m\in M$ such that $f(m) = n$ - $\implies \beta \circ f (m) = \beta(n)$ - $\implies f' \alpha(m) = \beta (n) = 0$ - $\implies \alpha(m) = 0$ - $\implies f'$ is injective, so $m=0$ and $n=f(m) = 0$. $\qed$ **Definition:** Two exact sequences are *isomorphic* iff in the following diagram, $f,g,h$ are all isomorphisms: ```{=latex} \begin{center} \begin{tikzcd} 0 \arrow[r] & M \arrow[dd, "f"] \arrow[r] & N \arrow[dd, "g"] \arrow[r] & Q \arrow[dd, "h"] \arrow[r] & 0 \\ & & & & \\ 0 \arrow[r] & M \arrow[r] & N \arrow[r] & Q \arrow[r] & 0 \end{tikzcd} \end{center} ``` **Theorem:** Let $0 \to M_1 \mapsvia f M_2 \mapsvia g M_3 \to 0$ be a SES. Then TFAE: - There exists an $R\dash$module homomorphisms $h: M_3 \to M_2$ such that $g\circ h = \id_{M_3}$. - There exists an $R\dash$module homomorphisms $k: M_2 \to M_1$ such that $k\circ f = \id_{M_1}$. - The sequence is isomorphic to $0 \to M_1 \to M_1 \oplus M_3 \to M_3 \to 0$. *Proof:* Define $\phi: M_1 \oplus M_3 \to M_2$ by $\phi(m_1 + m_2) = f(m_1) + h(m_2)$. We need to show that the following diagram commutes: ```{=latex} \begin{center} \begin{tikzcd} 0 \arrow[r] & M_1 \arrow[dd, "\id", latex'-latex',double,thin] \arrow[r] & M_1 \oplus M_3 \arrow[r] & M_3 \arrow[dd, "\id", latex'-latex',double,thin] \arrow[r] & 0 \\ & & & & \\ 0 \arrow[r] & M_1 \arrow[r] & M_2 \arrow[uu, "\phi"'] \arrow[r] & M_3 \arrow[r] & 0 \end{tikzcd} \end{center} ``` We can check that $$ (g\circ \phi)(m_1 + m_2) = g( f(m_1)) + g(h(m_2)) = m_2 = \pi(m_1 + m_2).$$ This yields $1 \implies 3$, and $2 \implies 3$ is similar. To see that $3 \implies 1, 2$, we attempt to define $k, h$ in the following diagram: ```{=latex} \begin{center} \begin{tikzcd} 0 \arrow[r] & M_1 \arrow[r] \arrow[dd, "\id", latex'-latex',double,thin] & M_1 \oplus M_3 \arrow[r] \arrow[l, "\pi_1"', bend right] & M_3 \arrow[dd, "\id", latex'-latex',double,thin] \arrow[l, "\iota_2"', bend right] \arrow[r] & 0 \\ & & & & \\ 0 \arrow[r] & M_1 \arrow[r] & M_2 \arrow[r] \arrow[uu, "\phi"'] \arrow[l, "k", bend left] & M_3 \arrow[r] \arrow[l, "h", bend left] \arrow[r] & 0 \end{tikzcd} \end{center} ``` So define $k = \pi_1 \circ \phi\inv$ and $h = \phi \circ \iota_2$. It can then be checked that $$ g \circ h = g \circ \phi \circ \iota_2 = \pi_2 \circ \iota_2 = \id_{M_3} .$$ $\qed$ ## Free Modules > Moral: A *free module* is a module with a basis. **Definition:** A subset $X = \theset{x_i}$ is *linearly independent* iff $$ \sum r_i x_i = 0 \implies r_i = 0 ~\forall i .$$ **Definition:** A subset $X$ *spans* $M$ iff $$ m\in M \implies m = \sum_{i=1}^n r_i x_i \quad \text{ for some }r_i \in R,~x_i \in X .$$ **Definition:** A subset $X$ is a basis $\iff$ it is a linearly independent spanning set. *Example:* $\ZZ_6$ is an abelian group and thus a $\ZZ\dash$module, but not free because $3 \actson [2] = [6] = 0$, so there are torsion elements. This contradicts linear independence for any subset. **Theorem (Characterization of Free Modules):** Let $R$ be a unital ring and $M$ a unital $R\dash$module (so $1\actson m = m$). TFAE: - There exists a nonempty basis of $M$. - $M = \oplus_{i\in I} R$ for some index set $I$. - There exists a non-empty set $X$ and a map $\iota: X \injects M$ such that given $f: X \to N$ for $N$ any $R\dash$ module, $\exists! \tilde f: M \to N$ such that the following diagram commutes. ```{=latex} \begin{center} \begin{tikzcd} M \arrow[rrdd, "\exists! \tilde f", dotted] & & \\ & & \\ X \arrow[rr, "f"] \arrow[uu, "\iota", hook] & & N \end{tikzcd} \end{center} ``` **Definition:** An $R\dash$module is *free* iff any of 1,2, or 3 hold. *Proof of $1 \implies 2$:* Let $X$ be a basis for $M$, then define $M \to \oplus_{x\in X} Rx$ by $\phi(m) = \sum r_i x_i$. It can be checked that - This is an $R\dash$module homomorphism, - $\phi(m) = 0 \implies r_j = 0 ~\forall j \implies m = 0$, so $\phi$ is injective, - $\phi$ is surjective, since $X$ is a spanning set. So $M \cong \bigoplus_{x\in X} Rx$, so it only remains to show that $Rx \cong R$. We can define the map \begin{align*} \pi_x: R &\to Rx \\ r &\mapsto rx .\end{align*} Then $\pi_x$ is onto, and is injective exactly because $X$ is a linearly independent set. Thus $M \cong \oplus R$. $\qed$ *Proof of $1 \implies 3$:* Let $X$ be a basis, and suppose there are two maps $X \mapsvia{\iota} M$ and $X \mapsvia{f} M$. Then define \begin{align*} \tilde f: M &\to N \\ \sum_i r_i x_i &\mapsto \sum_i r_i f(x_i) .\end{align*} This is clearly an $R\dash$module homomorphism, and the diagram commutes because $(\tilde f \circ \iota)(x) = f(x)$. This is unique because $\tilde f$ is determined precisely by $f(X)$. $\qed$ *Proof of $3 \implies 2$:* We use the usual "2 diagram" trick to produce maps \begin{align*} \tilde f: M \to \bigoplus_{x\in X} R \\ \tilde g: \bigoplus_{x\in X}R \to M .\end{align*} Then commutativity forces $$ \tilde f \circ \tilde g = \tilde g \circ \tilde f = \id .$$ $\qed$ *Proof of $2 \implies 1$:* We have $M = \oplus_{i\in I} R$ by (2). So there exists a map $$ \psi: \oplus_{i\in I} R \to M ,$$ so let $X \definedas \theset{\psi(1_i) \mid i\in I}$, which we claim is a basis. To see that $X$ is a basis, suppose $\sum r_i \psi(1_i) = 0$. Then $\psi(\sum r_i 1_i) = 0$ and thus $\sum r_i 1_i = 0$ and $r_i = 0$ for all $i$. Checking that it's a spanning set: Exercise. $\qed$ **Corollary:** Every $R\dash$module is the homomorphic image of a free module. *Proof:* Let $M$ be an $R\dash$module, and let $X$ be any set of generators of $R$. Then we can make a map $$ M \to \bigoplus_{x\in X} R $$ and there is a map $X \injects M$, so the universal property provides a map $$ \tilde f: \bigoplus_{x\in X} R \to M .$$ Moreover, $\bigoplus_{x\in X} R$ is free. $\qed$ *Examples:* - $\ZZ_n$ is **not** a free $\ZZ\dash$module for any $n$. - If $V$ is a vector space over a field $k$, then $V$ is a free $k\dash$module (even if $V$ is infinite dimensional). - Every nonzero submodule of a free module over a PID is free. **Some facts:** Let $R = k$ be a field (or potentially a division ring). 1. Every maximal linearly independent subset is a basis for $V$. 2. Every vector space has a basis. 3. Every linearly independent set is contained in a basis 4. Every spanning set contains a basis. 5. Any two bases of a vector space have the same cardinality. **Theorem (Invariant Dimension):** Let $R$ be a commutative ring and $M$ a free $R\dash$module. If $X_1, X_2$ are bases for $R$, then $\abs{X_1} = \abs{X_2}$. Any ring satisfying this condition is said to have the **invariant dimension property**. > Note that it's difficult to say much more about generic modules. For example, even a finitely generated module may *not* have an invariant number of generators.