# 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*} is exact.

Solution:

Note: to make sense of the functor, we may need to show that there is an isomorphism \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*} 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*} 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*}

We thus need to show

• Exactness as $$C\otimes_R N$$: $$\operatorname{im}(g\otimes\operatorname{id}_N) = C\otimes_R N$$, i.e. this is surjective.
• 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*}
\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*}

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*}
\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*}

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*}
(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*}
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*}
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*}
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{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}

• 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*}
\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*}
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*}
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*}
• $$R{\hbox{-}}$$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 $$\alpha\circ\alpha^{-1}= \operatorname{id}$$.

$$\hfill\blacksquare$$