# Week 1 ## Exercise 1.3H: Right Exactness of Tensoring Show that the following endofunctor \` ` \begin{align*} \begin{align*} F: {\mathsf{R}{\hbox{-}}\mathsf{Mod}}&\to {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\\ X &\mapsto X\otimes_R N \\ (X\xrightarrow{f} Y) &\mapsto (X\otimes_R N \xrightarrow{f \otimes\operatorname{id}_N} Y\otimes_R N) \end{align*} \end{align*} `{=html}is exact. Solution: > Note: to make sense of the functor, we may need to show that there is an isomorphism \` ` \begin{align*} \begin{align*}\hom_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(X, Y) \otimes_R \hom_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(A, B) \to \hom_{\mathsf{R}{\hbox{-}}\mathsf{Mod}}(X\otimes_R A, Y\otimes_R B).\end{align*} \end{align*} `{=html}This is what makes taking $f:X\to Y$ and $g:A\to B$ and forming $f\otimes g: X\otimes A \to Y\otimes B$ well-defined? Let $A\xrightarrow{f} B \xrightarrow{g} C \to 0$ be an exact sequence, so - $\operatorname{im}f = \ker g$ by exactness at $B$ - $\operatorname{im}g = C$ by exactness at $C$. Applying the above $F$ yields \` ` \begin{align*} \begin{align*} A\otimes_R N \xrightarrow{f\otimes\operatorname{id}_N} B\otimes_R N \xrightarrow{g\otimes\operatorname{id}_N} C\otimes_R N \to 0 \end{align*} \end{align*} `{=html} We thus need to show 1. Exactness as $C\otimes_R N$: $\operatorname{im}(g\otimes\operatorname{id}_N) = C\otimes_R N$, i.e. this is surjective. 2. Exactness at $B\otimes_R N$: $\operatorname{im}(f\otimes\operatorname{id}_N) = \ker(g\otimes id_N)$. We'll use the fact that every element in a tensor product is a finite sum of elementary tensors. - Claim: $\operatorname{im}(g\otimes\operatorname{id}_N) \subseteq C\otimes_R N$. - Let $b\otimes n \in B\otimes_R N$ be an elementary tensor - Then $(g\otimes\operatorname{id}_N)(b\otimes n) \coloneqq g(b) \otimes\operatorname{id}_N (n) = g(b) \otimes n$ - Since $\operatorname{im}(g) = C$, there exists a $c\in C$ such that $g(b) = c$, so $g(b) \otimes n = c \otimes n \in C\otimes_R N$ - Extend by linearity: ` \begin{align*}\begin{align*} \qty{g\otimes_R \operatorname{id}_N}\qty{\sum_{i=1}^m r_i \cdot b_i \otimes n_i} =\sum_{i=1}^m (g\otimes\operatorname{id}_N)(r_i\cdot b_i \otimes n_i) \coloneqq\sum_{i=1}^m g(r_i\cdot b_i) \otimes\operatorname{id}_N(n_i) =_H \sum_{i=1}^m r_i\cdot c_i \otimes n_i \in C\otimes_R N \end{align*}\end{align*} `{=html} where we've used bilinearity for the first equality, and the equality marked with $H$ uses above the proof for elementary tensors, and noted that we can pull ring scalars $r_i\in R$ through ${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$ morphisms. - Claim: $C \otimes_R N \subseteq \operatorname{im}(g\otimes\operatorname{id}_N)$. - Let $c\otimes n \in C\otimes_R N$ be an elementary tensor. - Then $c\in C = \operatorname{im}(g)$ implies $c = g(b)$ for some $b\in B$. - So $c\otimes n = g(b) \otimes n = (g\otimes\operatorname{id}_N)(b\otimes n) \in B\otimes_R N$. - Extend by linearity: ` \begin{align*}\begin{align*} \sum_{i=1}^m r_i\cdot c_i \otimes n_i =_H \sum_{i=1}^m g(r_i\cdot b_i) \otimes n_i = \sum_{i=1}^m (g\otimes\operatorname{id}_N)(r_i\cdot b_i \otimes n_i) = (g\otimes\operatorname{id}_N)\qty{\sum_{i=1}^m r_i\cdot b_i \otimes n_i} .\end{align*}\end{align*} `{=html} This proves (1). - Claim: $\operatorname{im}(f\otimes\operatorname{id}_N) \subseteq \ker(g\otimes\operatorname{id}_N)$. - Let $b\otimes n \in \operatorname{im}(f\otimes\operatorname{id}_N)$, we want to show $(g\otimes\operatorname{id}_N)(b\otimes n) = 0 \in C\otimes_R N$. - Then $b\otimes n = f(a)\otimes n$ for some $a\in A$. - By exactness of the original sequence, $\operatorname{im}f \subseteq \ker g$, so $g(f(a)) = 0 \in C$ - Then ` \begin{align*}\begin{align*} (g\otimes\operatorname{id}_N)\qty{ b \otimes n} = (g\otimes\operatorname{id}_N)(f(a)\otimes n) \coloneqq g(f(a)) \otimes n = 0\otimes n = 0\in C\otimes_R N \end{align*}\end{align*} `{=html} where we've used the fact that $0\otimes x = 0$ in any tensor product. - Extend by linearity. - Claim (**nontrivial part**): $\ker(g\otimes\operatorname{id}_N) \subseteq \operatorname{im}(f\otimes\operatorname{id}_N)$. > Note: the problem is that ` \begin{align*}\begin{align*} x\in \ker(g\otimes\operatorname{id}_N) \implies x = \sum_{i=1}^m r_i\cdot b_i \otimes n_i \implies (g\otimes\operatorname{id}_N)\qty{\sum_{i=1}^m r_i\cdot b_i \otimes n_i} = \sum_{i=1}^m r_i\cdot g(b_i) \otimes n_i = 0\in C\otimes_R N \end{align*}\end{align*} `{=html} **but** this does not imply that $g(b_i) = 0\in C$ for all $i$, which is what you would need to use $\operatorname{im}f = \ker g$ to write $g(b_i) = 0\implies \exists a_i, f(a_i) = b_i$ and pull everything back to $A\otimes_R N$. - Strategy: use the first claim and the first isomorphism theorem to obtain this situation: ` \begin{align*} \begin{center} \begin{tikzcd} {B\otimes_R N \over \operatorname{im}(f\otimes_R \operatorname{id}_N)} \ar[r, hook, "i"]\ar[rrr, bend left, dotted, "\alpha"] & {B\otimes_R N \over \ker(g \otimes_R \operatorname{id}_N)} \ar[r, "\cong"] & \operatorname{im}(g\otimes_R \operatorname{id}_N) \ar[equal]{r} & C\otimes_R N \end{tikzcd} \end{center} \end{align*} `{=html} - The first injection $i$ will exist because $\operatorname{im}(g\otimes_R \operatorname{id}_N) \subseteq \ker(g\otimes_R \operatorname{id}_N)$ by the first claim. - The middle isomorphism is the first isomorphism theorem. - The RHS equality follows from surjectivity of $g\otimes_R \operatorname{id}_N$ - We then apply a strengthened version of the 1st isomorphism theorem for modules: > Hungerford Ch.4 Thm 1.7: If $f:A\to B$ is a $R{\hbox{-}}$module morphism and $C\leq \ker f$ then there is a unique map $\tilde f: A/C\to B$ which is an isomorphism iff $f$ is an epimorphism and $C = \ker f$. > > Following Hungerford Ch.4 Prop. 5.4, p.210. - Since $\operatorname{im}(g\otimes_R \operatorname{id}_N)\subseteq \ker(g\otimes_R \operatorname{id}_N)$, by the theorem the map $\alpha$ exists and satisfies the same formula, i.e. $\alpha = \tilde g \otimes\tilde{ \operatorname{id}_N}$ where the tilde denotes the induced map on quotients, so $\alpha([b\otimes n]) = g(b)\otimes n$. - We will show it is an isomorphism, which forces $\operatorname{im}(g\otimes_R \operatorname{id}_N) \cong \ker(g\otimes_R \operatorname{id}_N)$ by the above theorem. - Constructing the inverse map: define ` \begin{align*}\begin{align*} \tilde \alpha^{-1}: C\times N &\to {B\otimes_R N \over \operatorname{im}(g\otimes_R \operatorname{id}_N) } \\ (c, n) &\mapsto (b \otimes n) \operatorname{mod}\operatorname{im}(g\otimes_R \operatorname{id}_N) {\quad \operatorname{where} \quad} b \in g^{-1}(c) ,\end{align*}\end{align*} `{=html} which we will show well-defined (i.e. independent of choice of $b$) and $R{\hbox{-}}$linear, lifting to a map $\alpha^{-1}$ out of the tensor product by the universal property which is a two-sided inverse for $\alpha$. - Well-defined: - $g^{-1}(b)$ exists because $g$ is surjective. - If $b\neq b'$ and $g(b') = 0$, then $0 = g(b) - g(b') = g(b-b')$ so $b-b' \in \ker g$. - By the original exactness, $b-b' \in \operatorname{im}f$ so $b-b' = f(a)$ for some $a\in A$. - Then $f(a) \otimes n \in \operatorname{im}(f\otimes\operatorname{id})$ implies $f(a)\otimes n \equiv 0 \operatorname{mod}\operatorname{im}(f\otimes\operatorname{id})$. - Then noting that $b-b' = f(a) \implies b = f(a) + b'$, working mod $\operatorname{im}(g\otimes_R \operatorname{id}_N)$ we have ` \begin{align*}\begin{align*} b \otimes n \equiv (f(a) + b') \otimes n \equiv \qty{f(a) \otimes n} + \qty{b' \otimes n} \equiv b'\otimes n .\end{align*}\end{align*} `{=html} - $R{\hbox{-}}$linear: - ? - Two-sided identity: - $(\alpha \circ \alpha^{-1})(c\otimes n) = \alpha(b\otimes n) = g(b)\otimes n = c\otimes n$, so $\alpha\circ \alpha^{-1}= \operatorname{id}$. - $(\alpha^{-1}\circ \alpha)([b\otimes n]) = \alpha^{-1}(g(b) \otimes n) = [b'\otimes n]$ where $b'\in g^{-1}(g(b))$ implies $b'=b$, so $\alpha\circ\alpha^{-1}= \operatorname{id}$.