Week 1

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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 \(\alpha\circ\alpha^{-1}= \operatorname{id}\).

\(\hfill\blacksquare\)

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