Math 830 Fall 2025: Homework 3 Solutions

Problem 1

Let $\phi : R^n\to R^m$ be an $R$-module homomorphism. First show that there is an $m\times n$ matrix $A$ such that $\phi(v) = Av$, for all $v\in R^n$. Here we are writing the elements of $R^n$ and $R^m$ as column vectors. Then show that the induced map $\phi^*:\text{Hom}_R(R^m, R)\to \text{Hom}_R(R^n, R)$ is multiplication by $A^t$, the transpose of $A$. Use these facts to calculate $\text{Ext}^n_R(R/I, R)$, for all $n\geq 0$, for $R := k[x, y, z]/(xy-z^2)$, and $I = (x,z)R$.

Solution:

The first part is just like in linear algebra: If $e_1, \ldots, e_n \in R^n$ is the standard basis and $\phi(e_i) = C_i$, a column vector in $R^m$, then if we let $A$ be the $m\times n$ matrix whose columns are $C_1, \ldots, C_n$, then $\phi(v) = Av$, for all column vectors $v\in R^n$. Let us write $A := (a_{ij})$.

For the second part, we let $e_1^*, \ldots, e_n^*$ be the dual basis for $\text{Hom}_R(R^n, R)$ and $f_1^*, \ldots, f_m^*$ be the dual basis for $\text{Hom}_R(R^m, R)$, where $f_1, \ldots, f_m \in R^m$ is the standard basis. Here the dual basis means the same as in linear algebra $e^*_i(e_j) = \delta_{ij}$, the Kronecker delta. Now consider $\phi^*(f_j^*)$. We need to write this in terms of $e_1^*, \ldots, e_n^*$. By definition, $\phi^*(f_j^*) = f_j^*\circ \phi$ as a map from $R^n$ to $R$. Then

f^*_j(\phi(e_i)) = f_j^*(a_{1i}f_1 +\cdots + a_{mi}f_m) = a_{ji}.

It follows that $f_j^*\circ \phi = a_{j1}e_1^*+\cdots + a_{jn}e_n^*$, since these functionals agree on $e_1, \ldots, e_n$. In other words, $\phi^*(f_j^*) = a_{j1}e_1^*+\cdots + a_{jn} e_n^*$. Therefore the matrix of $\phi^*$ with respect to the bases $\{e_1^*, \ldots, e_n^*\}$ and $\{f_1^*, \ldots, f_m^*\}$ has the $j$th row of $A$ as its $j$th column, which means this matrix is just $A^t$. When we identify $\text{Hom}_R(R^n, R)$ with $R^n$, we are identifying each $e_i^*$ with $e_i$, and similarly when we identify $\text{Hom}_R(R^m, R)$ with $R^m$, so that with these identifications, $\phi^*$ is multiplication by $A^t$.

For the final statement, it turns out that once we know the relations on $x,z$ as elements of $R$, all further relations in the resolution of $R/I$ and the computation of cohomology in its dual are easily identified. Suppose $a+bz \equiv 0$ in $R$, then $ax+bz = c(xy-z^2)$ in $S := k[x,y,z]$. Thus, $(a-cy)x+(b+cz)z = 0$ in $S$. Therefore $\begin{pmatrix} a-cy\\b+cz\end{pmatrix} = d\begin{pmatrix} -z\\x\end{pmatrix}$, so that $\begin{pmatrix} a\\b\end{pmatrix} = d\begin{pmatrix} -z\\x\end{pmatrix}+ c\begin{pmatrix} y\\-z\end{pmatrix}$. Thus the relations on $x,z$ are contained in the submodule generated by the columns of $\begin{pmatrix}-z & y\\x & -z\end{pmatrix}$. On the other hand, the columns of this matrix are clearly relations on $x$ and $z$, so the start of a free resolution of $R/I$ is

R^2\overset{\begin{pmatrix}-z & y\\x & -z\end{pmatrix}}\longrightarrow R^2\overset{\begin{pmatrix} x & z\end{pmatrix}}\longrightarrow R\to R/I\to 0.

To calculate the kernel of multiplication by $A := \begin{pmatrix}-z & y\\x & -z\end{pmatrix}$, we note that if $\begin{pmatrix}-z & y\\x & -z\end{pmatrix} \cdot \begin{pmatrix} a\\b\end{pmatrix} \equiv \begin{pmatrix} 0\\0\end{pmatrix}$, then $ax+(-b) z \equiv 0$ in $R$, so that $\begin{pmatrix} a\\-b\end{pmatrix}$ is in the column space of $\begin{pmatrix}-z & y\\x & -z\end{pmatrix}$. Thus, $\begin{pmatrix} a\\ b\end{pmatrix}$ is in the column space of $B := \begin{pmatrix} z & y\\x & z\end{pmatrix}$. It is easy to check that $AB = \begin{pmatrix} 0\\0\end{pmatrix}$, thus, the free resolution of $R/I$ continues as

\cdots \longrightarrow R^2\overset{B}\longrightarrow R^2 \overset{A}\longrightarrow R^2\overset{\begin{pmatrix} x & z\end{pmatrix}}\longrightarrow R\longrightarrow R/I\longrightarrow 0

Continuing in this way we see that we obtain a periodic resolution of $R/I$

\cdots \longrightarrow R^2\overset{B}\longrightarrow R^2\overset{A}\longrightarrow R^2\overset{B}\longrightarrow R^2 \overset{A}\longrightarrow R^2\overset{\begin{pmatrix} x & z\end{pmatrix}}\longrightarrow R\longrightarrow R/I\longrightarrow 0

Applying $\text{Hom}_R(-, R)$ and dropping $R/I$, we seek the homology of the complex

0\to R\overset{\begin{pmatrix} x\\z\end{pmatrix}}\longrightarrow R^2\overset{A^t}\longrightarrow R^2\overset{B^t}\longrightarrow R^2\overset{A^t}\longrightarrow R^2\overset{B^t}\longrightarrow \cdots

Since $R$ is an integral domain, and the first map (on the left) in the complex above takes $r$ to $\begin{pmatrix} rx\\rz\end{pmatrix}$, the kernel of this map is zero, so $\text{Ext}_R^0(R/I,R) = 0$. Calculating as above, it is easy to see that the kernel of multiplication by $A^t$ is the image of multiplication by $B^t$ and the kernel of multiplication by $B^t$ is the image of multiplication by $A^t$. This shows $\text{Ext}_R^n(R/I,R) = 0$, for $n\geq 2$. This also gives that $\text{Ext}^1_R(R/I,R)$ is the cyclic module $L := \langle \begin{pmatrix} z\\y\end{pmatrix}, \begin{pmatrix} x\\z\end{pmatrix}\rangle/\langle\begin{pmatrix} x\\z\end{pmatrix}\rangle$. Now $x\cdot \begin{pmatrix} z\\y\end{pmatrix} \equiv z\cdot \begin{pmatrix} x\\z\end{pmatrix}$ and $z\cdot \begin{pmatrix} z\\y\end{pmatrix} \equiv y\cdot\begin{pmatrix} x\\z\end{pmatrix}$, which shows that $I$ annihilates $L$. On the other hand, if $t\cdot L \equiv 0$, then $tz \equiv sx$, for some $s\in R$, and our relations on $x, z$ show that $t\in I$. Thus, $\text{ann}(L) = I$, giving $\text{Ext}^1_R(R/I, R) \cong R/I$. ∎

Problem 2

Prove the injective version of the Ext Lemma. Namely: Given an exact sequence $0\to N\overset{i}\to Q\overset{\pi}\to C\to 0$, with $Q$ injective, then for any $R$-modules $A$, there exists an exact sequence

0\to \text{Hom}_R(A, N)\overset{\tilde{i}}\to \text{Hom}_R(A,Q)\overset{\tilde{\pi}} \to \text{Hom}_R(A, C)\to \text{Ext}^1_R(A, N)\to 0.

Moreover, for all $n\geq 1$, $\text{Ext}^n_R(A, C)\cong \text{Ext}^{n+1}_R(A, N)$. You may assume the equivalence in Proposition-Definition from Wednesday, October 8 hold.

Solution:

Since $0\to \text{Hom}_R(A, N)\overset{\tilde{i}}\to \text{Hom}_R(A,Q)\overset{\tilde{\pi}} \to \text{Hom}_R(A, C)$ is exact, we just need a surjective $R$-module homomorphism $g: \text{Hom}_R(A, C)\to \text{Ext}^1_R(A, N)$ whose kernel is the image of $\tilde{\pi}$. Towards this end, let $0\to N\overset{i}\to Q\overset{\psi_1}\to Q_2\overset{\psi_2}\to Q_3$ be the start of an injective resolution of $N$, so that the image of $\psi_1$ (and hence the kernel of $\psi_2$) is $C$. Write $j: C\to Q_1$ for the natural inclusion. Now suppose $h\in \text{Hom}_R(A, C)$. Then, $jh \in \text{Hom}_R(A, Q_1)$. If we show $\psi_2jh = 0$, then $jh \in \text{ker}(\tilde{\psi}_2)$, and we can define $g(h) := [jh]$, the class of $jh$ in $\text{ker}(\psi_2)/\text{im}(\psi_1) = \text{Ext}^1_R(A, N)$. Suppose $a\in A$. Take $q\in Q$ such that $\psi_1(q) = h(a)$. Then $\psi_2jh(a) = \psi_2\psi_1(q) = 0$, which shows that $\psi_2jh = 0$, as required.

Now, $h\in \text{ker}(g)$ if and only if $[jh] \equiv 0$ in $\text{ker}(\tilde{\psi}_2)/\text{im}(\tilde{\psi}_1)$ if and only if $jh = \psi_1d$, for some $d\in \text{Hom}_R(A, Q)$. If we show that $jh = \psi_1 d$ for $d\in \text{Hom}_R(A, Q)$ if and only if $h = \pi d$, then we have $h\in \text{ker}(g)$ if and only if $h\in \text{im}(\tilde{\pi})$, which is what we want. But this is clear from the definition of $j$ and the fact that $\psi_1(q) = \pi(q)$, for all $q\in Q$.

Finally, to see that $g$ is surjective, take $[l]\in \text{ker}(\tilde{\psi_2})/\text{im}(\tilde{\psi_1})$, where $l\in \text{Hom}_R(A, Q_1)$ satisfies $\psi_2l = 0$. Thus, $\text{im}(l) \subseteq \text{im}(\psi_1) = C$, so $l$ takes its values in $C$. Thus, if we let $l'\in \text{Hom}_R(A, C)$ be the map defined by $l$, then $jl' = l$ and hence $[l] = [jl'] = g(l')$, showing that $g$ is surjective. ∎

Problem 3

Let $A, B, C$ be $R$-modules. Prove that $\text{Hom}_R(A\otimes_R B, C)\cong \text{Hom}_R(A, \text{Hom}_R(B, C))$, the so-called Hom-Tensor adjointness. Hint: If $f: A\otimes_RB\to C$, show there exists $g\in \text{Hom}_R(A, \text{Hom}_R(B, C))$ satisfying $g(a)(b) = f(a\otimes b)$, for all $a \in A$ and $b\in B$.

Solution:

Fix $f\in \text{Hom}_R(A\otimes B, C)$. For $a\in A$, define $g(a): B\to C$ as follows: $g(a)(b) := f(a\otimes b)$, for all $b\in B$. The bilinear properties of tensor product show that $g(a) \in \text{Hom}_R(B, C)$. We next show that the function $g: A\to \text{Hom}_R(B, C)$ given by $a\to g(a)$ is an $R$-module homomorphism. For $a_1, a_2\in A$, and $b\in B$ we have

g(a_1+a_2)(b) = f((a_1+a_2)\otimes b) = f((a_1\otimes b)+ (a_2\otimes b)) = f(a_1\otimes b) + f(a_2\otimes b) = g(a_1)(b) + g(a_2)(b) = (g(a_1)+g(a_2))(b),

showing $g(a_1+a_2) = g(a_1)+g(a_2)$. Now take $r\in R$, $a\in A$. Then for any $b\in B$,

g(ra)(b) = f((ra)\otimes b) = f(r(a\otimes b)) = r f(a\otimes b) = r g(a)(b),

showing that $g(ra) = rg(a)$. Therefore, $g\in \text{Hom}_R(A, \text{Hom}_R(B, C))$. We can now define $\phi : \text{Hom}_R(A\otimes B, C)\to \text{Hom}_R(A, \text{Hom}_R(B, C))$ by $\phi(f) = g$.

We now show that $\phi$ is an isomorphism. Let $f_1, f_2\in \text{Hom}_R(A\otimes B, C)$, and set $g_1 := \phi(f_1)$ and $g_2 := \phi(f_2)$. Then, for $a\in A$ and $b\in B$, we have, on the one hand

\phi(f_1+f_2)(a)(b) = (f_1+f_2)(a\otimes b) = f_1(a\otimes b) + f_2(a\otimes b) = (\phi(f_1)(a)+\phi(f_2)(a))(b).

Thus, $\phi(f_1+f_2)(a) = \phi(f_1)(a)+\phi(f_2)(a) = (\phi(f_1)+\phi(f_2))(a)$, for all $a\in A$, and thus, $\phi(f_1+f_2)= \phi(f_1)+\phi(f_2)$. Similarly, one can show that $\phi(rf) = r\phi(f)$, for all $r\in R$ and $f\in \text{Hom}_R(A\otimes B, C)$. Thus, $\phi$ is an $R$-module homomorphism. Suppose $\phi(f) = 0$. Then, $\phi(f)(a)(b) = 0$, for all $a \in A$ and $b\in B$. Thus, $f(a\otimes b) = 0$, for all $a\in A$ and $b\in B$, which implies that $f = 0$. Therefore $\phi$ is one-to-one.

Finally, suppose $g\in \text{Hom}_R(A, \text{Hom}_R(B, C))$. The bilinearity of the map $l(a, b) := g(a)(b)$ from $A\times B$ to $C$ shows there exists $f: A\otimes B\to C$ satisfying $f(a\otimes b) = g(a)(b)$. This shows that $\phi(f) = g$, so that $\phi$ is surjective. ∎

Problem 4

Prove that $R/I\otimes_R M \cong M/IM$, for $I\subseteq R$ an ideal and $M$ an $R$-module. Conclude that for ideals $I, J\subseteq R$, $(R/I)\otimes_R (R/J)$ is isomorphic to $R/(I+J)$ and in particular, $\mathbb{Z}_n\otimes_{\mathbb{Z}} \mathbb{Z}_m \cong \mathbb{Z}_d$, for $d:= \text{GCD}(n, m)$.

Solution:

The bilinearity of the map $g(r+I, m) = rm+IM$ from $R/I\times M$ to $M/IM$ shows there exists $f: R/I\otimes M \to M/IM$ having the property that $f((r+I)\otimes m) = rm +IM$. We now define $l: M/IM \to R/I\otimes M$ as follows: $l(m+IM) := (1+I)\otimes m$. To see that $l$ is well defined, suppose $m+IM = m'+IM$, so that $m-m' = \sum_j i_jy_j$, with $i_j \in I$ and $y_j \in M$. Then

\begin{align*} (1+I)\otimes m &= (1+I)\otimes (m'+\sum_ji_jy_j)\\ &= (1+I)\otimes m'+ (1+I)\otimes \sum_j i_jy_j\\ &= (1+I)\otimes m' + \sum_j ((1+I)i_j)\otimes y_j\\ &= (1+I)\otimes m' + 0 = (1+I)\otimes m', \end{align*}

showing that $l$ is well-defined. Taking $(r+I)\otimes m$ in $R/I\otimes M$, we have

lf((r+I)\otimes m) = l(rm+IM) = (1+I)\otimes rm = (1+I)r\otimes m = (r+I)\otimes m.

Conversely, for $m+IM \in M/IM$, we have

fl(m+IM) = f((1+I)\otimes m) = m+IM,

showing that $R/I\otimes M\cong M/IM$.

For the second statement, by the first statement we have

(R/I)\otimes (R/J) \cong (R/J)/I(R/J) \cong (R/J)/\{(I+J)/J\} \cong R/(I+J).

For the final statement, we apply the second statement to obtain

\mathbb{Z}_n\otimes \mathbb{Z}_m = (\mathbb{Z}/n\mathbb{Z})\otimes (\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z}/(n\mathbb{Z}+ m\mathbb{Z}) = \mathbb{Z}/\langle n, m\rangle \cong \mathbb{Z}/d\mathbb{Z},

since $\langle n, m\rangle = d\mathbb{Z}$, where $d = \text{GCD}(n, m)$. ∎

Problem 5

Prove that if $0\to A\overset{f}\to B\overset{g}\to C \to 0$ is a short exact sequence of $R$-modules, and $D$ is a projective $R$-module, then $0\to A\otimes D\overset{f\otimes 1_D}\to B\otimes D\overset{g\otimes 1_D}\to C\otimes D\to 0$ is exact.

Solution:

Since $D$ is projective there is a free $R$-module $F$ and an $R$-module $K$ such that $F \cong D\oplus K$. We first note that $0\to A\otimes F\to B\otimes F\to C\otimes F\to 0$ is exact, and for this we just need to show that $f\otimes 1_F: A\otimes F\to B\otimes F$ is injective. Without loss of generality, we assume that $D, K\subseteq F$. We may assume $F = \bigoplus_{i\in I} R$, a direct sum of $|I|$ copies of $R$. Then it is easy to check that $A\otimes F \cong \bigoplus_i A$ and $B\otimes F \cong \bigoplus_i B$.

The commutativity of the induced maps shows that $f\otimes 1_F$ corresponds to the injective map $j$ from $\bigoplus_i A$ to $\bigoplus_i B$ induced by $f$. It follows that $f\otimes 1_F$ is injective. Since $f\otimes 1_F$ is injective and the natural isomorphisms preserve the structure, we can project to show that $f\otimes 1_D: A\otimes D \to B\otimes D$ is injective, which gives what we want. ∎

Problem 6

Let $A, B$ be $R$-modules and $S\subseteq R$ a multiplicatively closed set. Prove that there is an isomorphism of $R_S$-modules $\text{Tor}_n^R(A, B)_S \cong \text{Tor}^{R_S}_n(A_S, B_S)$, for all $n\geq 1$. You may use the fact that $(A\otimes_R B)_S \cong A_S\otimes_{R_S} B_S$, as $R_S$-modules.

Solution:

Let $\mathcal{P}: \cdots \to P_2\overset{\phi_2}\longrightarrow P_1\overset{\phi_1}\longrightarrow P_0\overset{\pi}\longrightarrow A\to 0$ be a projective resolution of $A$. To calculate $\text{Tor}^R_n(A, B)$, we calculate the $n$th homology of $\mathcal{P}\otimes B$, after dropping the $A$ term. Thus, $\text{Tor}^R_n(A, B) = \text{ker}(\phi_n\otimes 1_B)/\text{im}(\phi_{n+1}\otimes 1_B)$. Therefore,

\begin{align*} \text{Tor}^R_n(A, B)_S &= \{\text{ker}(\phi_n\otimes 1_B)/\text{im}(\phi_{n+1}\otimes 1_B)\}_S\\ &= \text{ker}(\phi_n\otimes 1_B)_S/\text{im}(\phi_{n+1}\otimes 1_B)_S\\ &= \text{ker}((\phi_n)_S\otimes (1_B)_S)/\text{im}((\phi_{n+1})_S\otimes (1_B)_S). \end{align*}

Since $\mathcal{P}_S$ is a projective resolution of $A_S$, this gives what we want. ∎

Problem 7

Calculate $\text{Tor}_n^{\mathbb{Z}_{48}}(\mathbb{Z}_{12}, \mathbb{Z}_{16})$ in two ways, for all $n\geq 0$.

Solution:

We set $R := \mathbb{Z}_{48}$ and start with a projective resolution of $\mathbb{Z}_{12} = R/12R$:

\cdots \overset{\cdot 4}\longrightarrow R_3 \overset{\cdot 12}\longrightarrow R_2 \overset{\cdot 4}\longrightarrow R_1 \overset{\cdot 12}\longrightarrow R_0\overset{\pi}\longrightarrow R/12R \to 0,

where $R_i$ denotes $R$ in homological degree $i$. Using the fact that $R\otimes \mathbb{Z}_{16} \cong \mathbb{Z}_{16}$, if we drop $\mathbb{Z}_{12}$ and tensor with $\mathbb{Z}_{16}$ we obtain the complex

\cdots \overset{\cdot 4}\longrightarrow \mathbb{Z}_{16} \overset{\cdot 12}\longrightarrow \mathbb{Z}_{16} \overset{\cdot 4}\longrightarrow \mathbb{Z}_{16} \overset{\cdot 12}\longrightarrow \mathbb{Z}_{16}\to 0.

The kernel of the map $\mathbb{Z}_{16}\overset{\cdot 12}\longrightarrow \mathbb{Z}_{16}$ is $4\mathbb{Z}_{16}$, since if $12a = 16b$ in $\mathbb{Z}$, then $3a = 4b$, so $a$ is divisible by 4. Thus the complex above is exact in odd homological degrees. On the other hand, in even homological degrees of 2 or more, the kernel of the map $\mathbb{Z}_{16}\overset{\cdot 4}\to \mathbb{Z}_{16}$ is $4\mathbb{Z}_{16}$, while the image is $12\cdot \mathbb{Z}_{16} = (12\mathbb{Z}+16\mathbb{Z})/16\mathbb{Z} = 4\mathbb{Z}/16\mathbb{Z} = 4\cdot \mathbb{Z}_{16}$, so these homology modules are also 0. Thus, $\text{Tor}_j^R(\mathbb{Z}_{12},\mathbb{Z}_{16}) = 0$, for $j\geq 1$. On the other hand the degree zero homology module is just $\mathbb{Z}_{12}\otimes_{\mathbb{Z}_{48}} \mathbb{Z}_{16}$. To calculate this, we note that this is

(R/12R)\otimes_R (R/16R) \cong R/\langle 12, 16\rangle \cong R/4R\cong \mathbb{Z}_4.

We now take a projective resolution of $\mathbb{Z}_{16}$:

\cdots \overset{\cdot 3}\longrightarrow R_3 \overset{\cdot 16}\longrightarrow R_2 \overset{\cdot 3}\longrightarrow R_1 \overset{\cdot 16}\longrightarrow R_0\overset{\pi}\longrightarrow \mathbb{Z}_{16}\to 0.

Tensoring with $\mathbb{Z}_{12}$ and dropping the $\mathbb{Z}_{16}$ term we have the complex

\cdots \overset{\cdot 3}\longrightarrow \mathbb{Z}_{12} \overset{\cdot 16}\longrightarrow \mathbb{Z}_{12} \overset{\cdot 3}\longrightarrow \mathbb{Z}_{12} \overset{\cdot 16}\longrightarrow \mathbb{Z}_{12}\to 0.

The zeroth homology in the complex above is $\mathbb{Z}_{12}/16\mathbb{Z}_{12} = \mathbb{Z}_{12}/4\mathbb{Z}_{12} \cong \mathbb{Z}_4$, which agrees with our previous calculation. For $n > 0$ and $n$ odd, we have $\text{Tor}_n^R(\mathbb{Z}_{16}, \mathbb{Z}_{12}) = 3\mathbb{Z}_{12}/3\mathbb{Z}_{12} = 0$, since the kernel of the map $\mathbb{Z}_{12}\overset{\cdot 16}\to \mathbb{Z}_{12}$ is $3\mathbb{Z}_{12}$. For $n$ even and $n \geq 2$, $\text{Tor}_n^R(\mathbb{Z}_{16}, \mathbb{Z}_{12}) = 4\mathbb{Z}_{12}/16\mathbb{Z}_{12} = 4\mathbb{Z}_{12}/4\mathbb{Z}_{12} = 0$, since the kernel of the map $\mathbb{Z}_{12}\overset{\cdot 3}\to \mathbb{Z}_{12}$ is $4\mathbb{Z}_{12}$. ∎

Problem 8

Let $p\in \mathbb{Z}$ be prime and consider the sequence of $\mathbb{Z}$-modules and $\mathbb{Z}$-module homomorphisms $\mathbb{Z}_p\overset{\cdot p}\to \mathbb{Z}_{p^2}\overset{\cdot p}\to \mathbb{Z}_{p^3}\overset{\cdot p}\to \mathbb{Z}_{p^4}\overset{\cdot p}\to \cdots$. Show that this is a directed system of $\mathbb{Z}$-modules whose direct limit is isomorphic to $\mathbb{Z}_{p^{\infty}}$.

Solution:

For this we set $I := \{1, 2, 3, \ldots\}$, $M_i := \mathbb{Z}_{p^i}$ and $f_{ij}: M_i\to M_j$ to be multiplication by $p^{j-i}$, for $i\leq j$. Then we clearly have a direct system of $\mathbb{Z}$-modules and $\mathbb{Z}$-module homomorphisms. In each of the modules $\varinjlim M_i$, $\mathbb{Z}_{p^i}$ and $\mathbb{Z}_{p^{\infty}}$, the elements are all equivalence classes, so we denote $[x]$ for an element of $\varinjlim M_i$, $\overline{a}$ for an element of $M_i := \mathbb{Z}_{p^i}$ and $\frac{a}{p^t}$ for an element of $\mathbb{Z}_{p^{\infty}}$. For each $i$, define $g_i: M_i\to \mathbb{Z}_{p^{\infty}}$ by $g_i(\overline{a}) := \frac{a}{p^i}$. This is well defined if we take $a < p^i$. To see that $g_i$ is an $R$-module homomorphism, suppose $\overline{b}\in M_i$, with $b < p^i$. Write $\overline{a+b} = \overline{c}$, with $c < p^i$. Then

g_i(\overline{a}+\overline{b}) = g_i(\overline{c}) = \frac{c}{p^i} = \frac{a+b-kp^i}{p^i} = \frac{a}{p^i}+\frac{b}{p^i} = g_i(\overline{a})+g_i(\overline{b}),

for some integer $k$. The argument showing $g_i(r\cdot \overline{a}) = rg_i(\overline{a})$, for $r\in \mathbb{Z}$ is similar.

Now, for $i\leq j$,

g_jf_{ij}(\overline{a}) = g_j(p^{j-i}\overline{a}) = g_j(\overline{p^{j-i}a}) = \frac{p^{j-i}a}{p^j} = \frac{a}{p^i} = g_i(\overline{a}),

so $g_i = g_j\circ f_{ij}$, for all $i\leq j$. Thus, by the universal property of direct limits, there exists a (unique) $\mathbb{Z}$-module homomorphism $\phi : \varinjlim M_i\to \mathbb{Z}_{p^{\infty}}$ given by $\phi([\overline{a}]) = \frac{a}{p^i}$, if $\overline{a}\in M_i$. Since $a < p^i$, $\frac{a}{p^i} \neq 0$, unless $\overline{a} = \overline{0}$, so $\phi$ is one-to-one. Moreover, given $\frac{b}{p^j} \in \mathbb{Z}_{p^{\infty}}$, we may assume that $b< p^j$, so that if we take $\overline{b}\in M_j$, then $\phi([\overline{b}]) = \frac{b}{p^j}$, showing that $\phi$ is surjective and hence an isomorphism. ∎

Problem 9

Let $A$ be an $R$-module and suppose $x\in R$ is a non-zerodivisor on both $R$ and $A$. Suppose

\mathcal{F}:\quad \cdots \to F_2\overset{\phi_2}\to F_1\overset{\phi_1} \to F_0\overset{\pi} \to A\to 0

is a free resolution of $A$.

Show directly that the induced sequence $\mathcal{F}/x\mathcal{F}:\quad \cdots \to F_2/xF_2\overset{\phi_2'}\to F_1/xF_1\overset{\phi_1'} \to F_0/xF_0\overset{\pi'} \to A/xA\to 0$ is exact.
Conclude that $\text{Tor}_n^R(A, R/xR) = 0$, for all $n\geq 1$.
Now reprove (i) by recalculating $\text{Tor}_n^R(A, R/xR)$, for all $n\geq 1$.

Solution:

Part (i): Clearly $\mathcal{F}/x\mathcal{F}$ is a complex and $\pi'$ is surjective. Suppose $\pi'(v') = 0$, for $v'\in F_0/xF_0$, the image of $v \in F_0$. Then $\pi(v) = xa$, for some $a \in A$. Since $\pi$ is surjective, $a = \pi(w)$, for some $w\in F_0$. Thus, $\pi(v) = \pi(xw)$, so that $\pi(v-xw) = 0$. Thus, $v-xw = \phi_1(u)$, for some $u\in F_1$. Thus, $v-\phi_1(u) \in xF_0$, so that $\phi_1'(u') = v'$ in $F_0/xF_0$, where $u'$ is the image of $u$ in $F_1/xF_1$. Thus, exactness holds at $F_0/xF_0$ in $\mathcal{F}/x\mathcal{F}$.

Now suppose $\phi_1'(y') = 0$, for $y'\in F_1/xF_1$. Then $\phi_1(y) = xz$, for some $z\in F_0$. Therefore, $0 = \pi(xz) = x\pi(z)$. Since $x$ is a non-zerodivisor on $A$, $\pi(z) = 0$. Thus, $z\in \text{ker}(\pi) = \text{im}(\phi_1)$, so that $z = \phi_1(t)$, for $t\in F_1$. It follows that $\phi_1(y) = \phi_1(xt)$, so that $\phi_1(y-xt) = 0$. Thus, $y-xt = \phi_2(h)$, for $h\in F_2$. Thus, in $F_1/xF_1$, we have $y' = \phi_2'(h')$, so exactness in $\mathcal{F}/x\mathcal{F}$ holds at $F_1/xF_1$. The proof proceeds in similar fashion for exactness at each $F_n/xF_n$, only for $n\geq 2$, one uses that $x$ is a non-zerodivisor on each $F_n$ by virtue of being a non-zerodivisor on $R$.

Part (ii): This follows immediately from (i), since $\mathcal{F}/x\mathcal{F}$ is obtained by tensoring $\mathcal{F}$ with $R/xR$.

Part (iii): We may calculate $\text{Tor}_n^R(A, R/xR)$ by taking a projective (or free) resolution of $R/xR$ and tensoring with $A$. Since $0\to R\overset{\cdot x}\to R\to R/xR\to 0$ is a free resolution of $R/xR$, tensoring with $A$ and dropping the $A/xA$ term on the right shows that the Tor modules are the homology modules in the complex $0\to A\overset{\cdot x}\to A\to 0$. Since $x$ is a non-zerodivisor on $A$, it follows that this complex has zero homology in all homological degrees greater than zero, i.e., $\text{Tor}_n^R(A, R/xR) = 0$, for all $n\geq 1$. ∎

Problem 10

An $R$-module $F$ is said to be flat if whenever $0\to C\to D\to E\to 0$ is a short exact sequence of $R$-modules, then the induced sequence $0\to C\otimes F\to D\otimes F\to E\otimes F\to 0$ is exact. Let $A$ be an $R$-module. A flat resolution of $A$ is an exact sequence of the form

\mathcal{F}:\quad \cdots \to F_2\to F_1\to F_0\to A\to 0

such that each $F_j$ is a flat $R$-module. The flat dimension of $A$ is the least $n\geq 0$ (if it exists) such that $A$ has a flat resolution of length $n$. For example, a flat module has flat dimension zero. For an $R$-module $A$, prove that the following conditions are equivalent:

$A$ has flat dimension less than or equal to $n$.
$\text{Tor}^R_j(A, B) = 0$, for all $j\geq n+1$ and all $R$-modules $B$.
$\text{Tor}^R_{n+1}(A, B) = 0$, for all $R$-modules $B$.

Solution:

We begin with three observations. The first is that if $\mathcal{M}: \cdots \to M_2\overset{f_2}\to M_1\overset{f_1}\to M_0\to 0$ is an exact sequence of $R$-modules and $A$ is a flat $R$-module, then tensoring $\mathcal{M}$ with $A$ preserves exactness. To see this, let $K$ be the kernel of $f_1$, so that $0\to K\overset{i}\to M_1\overset{f_1}\to M_0\to 0$ is exact. Thus, $0\to K\otimes A\overset{i\otimes1_A}\to M_1\otimes A\overset{f_1\otimes 1_A}\to M_0\otimes A\to 0$ is exact. This shows that $K\otimes A$ is the kernel of $f_1\otimes 1_A$. On the other hand, since $f_2$ maps $M_2$ onto $K$, $f_2\otimes 1_A$ maps $M_2\otimes A$ onto $K\otimes A = \text{ker}(f_1\otimes 1_A)$. In other words, the sequence $M_2\otimes A\overset{f_2\otimes 1_A}\to M_1\otimes A\overset{f_1\otimes 1_A}\to M_0\otimes A\to 0$ is exact. This argument can be repeated inductively to show $\mathcal{M}\otimes A$ is exact. The second observation is that if in the flat resolution $\mathcal{F}$ above, $K$ is the kernel of the map from $F_0$ to $A$, then for $n \geq 1$, the flat dimension of $A$ is less than or equal to $n$ if and only if the flat dimension of $K$ is less than or equal to $n-1$. This follows immediately from the definition of flat dimension. For the third observation, let $K$ be as in the second observation, so that we have the exact sequence $0\to K\to F_0\to A\to 0$. Tensoring with $B$ yields the long exact sequence in Tor

\cdots \to \text{Tor}_{j+1}^R(F_0, B)\to \text{Tor}_{j+1}^R(A, B)\to \text{Tor}_j^R(K,B)\to \text{Tor}_j^R(F_0, B)\to \cdots

By the first observation, $\text{Tor}_j^R(F_0, B) = 0$, for all $j\geq 1$, so that $\text{Tor}_j^R(K, B) \cong \text{Tor}_{j+1}^R(A,B)$, for all $j\geq 1$, showing that dimension shifting holds for a flat presentation.

To prove the equivalence of (i)-(iii), we induct on $n$. Suppose $n = 0$, so that $A$ is a flat $R$-module. Then (ii) follows from (i) by the first observation above. Clearly (ii) implies (iii). Now suppose (iii) holds. Take $0 \to D\to C\to B\to 0$ any short exact sequence of $R$-modules. By the long exact Tor sequence, we have

\cdots \to \text{Tor}_1^R(A,B) \to D\otimes A\to C\otimes A\to B\otimes A\to 0,

which by (iii) shows that the sequence $0\to D\otimes A\to C\otimes A\to B\otimes A\to 0$ is exact, and thus (iii) implies (i).

Now suppose $n > 0$ and (i)-(iii) hold for $n-1$. Suppose $A$ has flat dimension less than or equal to $n$ so that by the second observation, $K$ has flat dimension less than or equal to $n-1$. Thus, by induction and the third observation, $0 = \text{Tor}_j^R(K,B) = \text{Tor}_{j+1}^R(A,B)$, for all $j\geq n$, which gives (ii). Since (ii) implies (iii) is trivial, assume (iii). Then, by dimension shifting, $\text{Tor}_n^R(K, B) = 0$, so by induction $K$ has flat dimension less than or equal to $n-1$, and thus $A$ has flat dimension less than or equal to $n$. ∎