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/Part 1: General Tools/Chapter 1: Category Theory

1.3 Universal Constructions for Modules

Definition 1.3.1 (Product).label Let $R$ be a ring and $\seqi{A}$ be $R$-modules, then there exists $(A, \bracsn{\pi_i}_{i \in I})$ such that:

  1. (1)

    For each $i \in I$, $\pi_{i} \in \hom(A; A_{i})$.

  2. (U)

    For any $(B, \seqi{T})$ satisfying (1), there exists a unique $T \in \hom(B; A)$ such that the following diagram commutes:

    \[\xymatrix{ B \ar@{->}[rd]_{T_i} \ar@{->}[r]^{T} & A \ar@{->}[d]^{\pi_i} \\ & A_i }\]

The module $A = \prod_{i \in I}A_{i}$ is the product of $\seqi{A}$.

Definition 1.3.2 (Direct Sum).label Let $E$ be a ring and $\seqi{A}$ be $R$-modules, then there exists $(A, \bracsn{\iota_i}_{i \in I})$ such that:

  1. (1)

    For each $i \in I$, $\iota_{i} \in \hom(A_{i}; A)$.

  2. (U)

    For each $(B, \seqi{T})$ satisfying (1), there exists a unique $T \in \hom(A; B)$ such that the following diagram commutes

    \[\xymatrix{ A \ar@{->}[r]^{T} & B \\ A_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & }\]

The module $A = \bigoplus_{i \in I}A_{i}$ is the direct sum of $\seqi{A}$.

Proof. Let

\[A = \bracs{x \in \prod_{i \in I}A_i \bigg | x_i \ne 0 \quad \text{for finitely many}\ i \in I}\]

For each $i \in I$, let

\[\iota_{i}: A_{i} \to A \quad (\iota_{i}x)_{j} = \begin{cases}x &i = j \\ 0 &i \ne j\end{cases}\]

then $\iota_{i} \in \hom(A_{i}; A)$.

(U): Let

\[T: A \to B \quad x \mapsto \sum_{i \in I}T_{i}x_{i}\]

then $T \in \hom(A; B)$ and the diagram commutes. Since $\bigcup_{i \in I}\iota_{i}(A_{i})$ spans $A$, $T$ is the unique linear map making the diagram commute.$\square$

Proposition 1.3.3.label Let $R$ be a ring and $(\seqi{A}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of $R$-modules, then there exists $(A, \bracsn{T^i_A}_{i \in I})$ such that:

  1. (1)

    For each $i \in I$, $T^{i}_{A} \in \hom({A_i; A})$.

  2. (2)

    For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:

    \[\xymatrix{ A_i \ar@{->}[rd]_{T^i_A} \ar@{->}[r]^{T^i_j} & A_j \ar@{->}[d]^{T^j_A} \\ & A }\]

  3. (U)

    For any pair $(B, \bracsn{S^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom({A, B})$ such that the following diagram commutes

    \[\xymatrix{ A_i \ar@{->}[d]_{T^i_A} \ar@{->}[rd]^{S^i_B} & \\ A \ar@{->}[r]_{g} & B }\]

    for all $i \in I$.

Proof. Let $M = \bigoplus_{i \in I}A_{i}$. For any $i, j \in I$ with $i \lesssim j$ and $x \in A_{i}$, let $x_{i, j}\in M$ such that for any $k \in I$,

\[\pi_{k}(x_{i, j}) = \begin{cases}x&k = i \\ T^{i}_{j} x&k = j \\ 0&k \ne i, j\end{cases}\]

Let $N \subset M$ be the submodule generated by $\bracs{x_{i, j}|i, j \in I, i \lesssim j, x \in A_i}$, $A = M/N$, and $\pi: M \to M/N$ be the canonical map.

(1): For each $i \in I$, let

\[T^{i}_{M}: A_{i} \to M \quad \pi_{k} T^{i}_{M} x = \begin{cases}x &k = i \\ 0 &k \ne i\end{cases}\]

and $T^{i}_{A} = \pi \circ T^{i}_{M}$.

(2): Let $i, j \in I$ with $i \lesssim j$, then for any $x \in A_{i}$, $T^{i}_{M}x - T^{j}_{M} T^{i}_{j} x \in N$. Hence $T^{i}_{A}x = T^{j}_{A} T^{i}_{j}x$.

(U): Let

\[S_{0}: M \to B \quad x \mapsto \sum_{i \in I}S^{i}_{B} \pi_{i} x\]

then $S_{0}$ is the unique linear map such that $S_{0} \circ T^{i}_{M} = S^{i}_{B}$ for all $i \in I$. For any $i, j \in I$ with $i \lesssim J$, $S^{i}_{B} x = S^{j}_{B} T^{i}_{j} x$, so $\ker S_{0} \supset N$. By the first isomorphism theorem, there exists a unique $S \in \hom(A; B)$ such that $S_{0} = S \circ \pi$.$\square$

Proposition 1.3.4.label Let $R$ be a ring and $(\seqi{A}, \bracs{T^i_j|i, j \in I, i \lesssim j})$ be a downward-directed system of $R$-modules, then there exists $(A, \bracsn{T^A_i}_{i \in I})$ such that:

  1. (1)

    For each $i \in I$, $T^{A}_{i} \in \hom(A; A_{i})$.

  2. (2)

    For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:

    \[\xymatrix{ A_i \ar@{->}[r]^{T^i_j} & A_j \\ A \ar@{->}[u]^{T^A_i} \ar@{->}[ru]_{T^A_j} & }\]

  3. (U)

    For any pair $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom(B; A)$ such that the following diagram commutes

    \[\xymatrix{ & A_i \\ B \ar@{->}[r]_{S} \ar@{->}[ru]^{S^B_i} & A \ar@{->}[u]_{T^A_i} }\]

    for all $i \in I$.

Proof. Let

\[A = \bracs{x \in \prod_{i \in I}A_i \bigg | \pi_j(x) = T^i_j\pi_i(x) \forall i, j \in I, i \lesssim j}\]

For each $i \in I$, let $T^{A}_{i} = \pi_{i}$, then $(A, \bracsn{T^A_i}_{i \in I})$ satisfies (1) and (2) by definition of $A$.

(U): Let $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2). Let

\[S: B \to \prod_{i \in I}A_{i} \quad \pi_{i}(Sx) = S^{B}_{i}\]

then for any $x \in B$ and $i, j \in I$ with $i \lesssim j$,

\[\pi_{j} (Sx) = S^{B}_{j}x = T^{i}_{j} S^{B}_{i}x = T^{i}_{j} \pi_{i}(S x)\]

so $S \in \hom(B; A)$, and the diagram commutes. Since any map $f: B \to A$ is uniquely determined by its composition with the projections, $S$ is unique.$\square$

Definition 1.3.5 (Tensor Product).label Let $R$ be a commutative ring and $\seqf{E_j}$ be $R$ modules, then there exists a pair $(\bigotimes_{j = 1}^{n} E_{j}, \iota)$ such that:

  1. (1)

    $\bigotimes_{j = 1}^{n} E_{j}$ is an $R$-module.

  2. (2)

    $\iota: \prod_{j = 1}^{n} E_{j} \to \bigotimes_{j = 1}^{n} E_{j}$ is a $n$-linear map.

  3. (U)

    For any pair $(F, \lambda)$ satisfying (1) and (2), there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^{n} E_{j}; F)$ such that the following diagram commutes:

    \[\xymatrix{ \prod_{j = 1}^n E_j \ar@{->}[rd]_{\lambda} \ar@{->}[r]^{\iota} & \bigotimes_{j = 1}^n E_j \ar@{->}[d]^{\Lambda} \\ & F }\]

  4. (4)

    $\bigotimes_{j = 1}^{n} E_{j}$ is the linear span of $\iota(\prod_{j = 1}^{n} E_{j})$.

The module $\bigotimes_{j = 1}^{n} E_{j}$ is the tensor product of $\seqf{E_j}$, and $\iota: \prod_{j = 1}^{n} E_{j} \to \bigotimes_{j = 1}^{n} E_{j}$ is the canonical embedding. For any $(x_{1}, \cdots, d_{n}) \in \prod_{j = 1}^{n} E_{j}$, the image

\[x_{1} \otimes \cdots \otimes x_{n} = \iota(x_{1}, \cdots, x_{n})\]

is its tensor product.

Proof. Let $M$ be the free module generated by $\prod_{j = 1}^{n}E_{j}$, and $N \subset M$ be the submodule generated by elements of the following form:

  1. (1)

    For any $1 \le j \le n$, $(x_{1}, \cdots, x_{n}) \in \prod_{k = 1}^{n} E_{k}$, and $x_{j} \in E_{j}$,

    \[(x_{1}, \cdots, x_{j} + x_{j}', \cdots, x_{n}) - (x_{1}, \cdots, x_{j}, \cdots, x_{n}) - (x_{1}, \cdots, x_{j}', \cdots, x_{n})\]

  2. (2)

    For any $(x_{1}, \cdots, x_{n}) \in \prod_{k = 1}^{n} E_{k}$ and $\alpha \in R$,

    \[(x_{1}, \cdots, \alpha x_{j}, \cdots, x_{n}) - \alpha(x_{1}, \cdots, x_{n})\]

(1), (2): Let $\bigotimes_{j = 1}^{n} E_{j} = M/N$ and

\[\iota: \prod_{j = 1}^{n} E_{j} \to \bigotimes_{j = 1}^{n} E_{j} \quad (x_{1}, \cdots, x_{n}) \mapsto (x_{1}, \cdots, x_{n}) + N\]

then by definition of $N$, $\iota$ is $n$-linear.

(U): Let $(F, \lambda)$ be a pair satisfying (1) and (2), then $\lambda$ admits a unique extension to a linear map $\ol \lambda: M \to F$. Since $\lambda$ is $n$-linear, $\ker \ol \lambda \supset N$. By the first isomorphism theorem, there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^{n} E_{j}; F)$ such that the given diagram commutes.

(4): Since $M$ is the free module generated by $\prod_{j = 1}^{n} E_{j}$, $M/N$ is generated by $\iota(\prod_{j = 1}^{n} E_{j})$.$\square$

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