Exercise 2.21

Question

Let $(A_i)_{i \in I}$ be a family of rings indexed by a directed set $I$, and for each pair $i \le j$ in $I$ let $\alpha_{ij} : A_i 	o A_j$ be a ring homomorphism, satisfying the conditions $(1)$ and $(2)$ of Exercise $14$. Regarding each $A_i$ as a $\mathbb{Z}$-module we can then form the direct limit $A = \displaystyle\lim_{\longrightarrow} A_i$. Show that $A$ inherits a ring structure form the $A_i$ so that the mappings $A_i 	o A$ are ring homomorphisms. The ring $A$ is the \emph{direct limit} of the system $(A_i,\alpha_{ij})$.
  \par If $A=0$ prove that $A_i = 0$ for some $i \in I$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Letting $\alpha_i : A_i \hookrightarrow A$ be the canonical ring homomorphic embedding that exists by considering $A = \bigcup A_i$ from AM Exercise $2.17$, we have that if $\xi,\eta \in A$, there exists $i,j$ such that $\xi = \alpha_i(x), \eta = \alpha_j(y)$ for $x \in A_i, y \in A_j$ from AM Exercise $2.15$. Choosing $k \geqslant i,j$ we have $\xi = \alpha_k(\alpha_{ik}(x)), \eta = \alpha_k(\alpha_{jk}(y))$; define $\xi\cdot\eta = \alpha_k(\alpha_{ik}(x)\alpha_{jk}(y))$. Since the $\alpha$'s are already ring homomorphisms, we only have to show this product is well-defined.
  \par So, suppose $\ell \geqslant i,j$ and $m \geqslant \ell,k$. Then,
  \begin{align*}
    \alpha_k(\alpha_{ik}(x)\alpha_{jk}(y)) &= \alpha_m(\alpha_{km}(\alpha_{ik}(x)\alpha_{jk}(y)))\
    &= \alpha_m(\alpha_{km}(\alpha_{ik}(x))\alpha_{km}(\alpha_{jk}(y)))\
    &= \alpha_m(\alpha_{im}(x)\alpha_{jm}(y))\
    &= \alpha_m(\alpha_{\ell m}(\alpha_{i\ell}(x))\alpha_{\ell m}(\alpha_{j\ell}(y)))\
    &= \alpha_m(\alpha_{\ell m}(\alpha_{i\ell}(x)\alpha_{j\ell}(y)))\
    &= \alpha_\ell(\alpha_{i\ell}(x)\alpha_{j\ell}(y)),
  \end{align*}
  and so the product is independent of $k$.
  \par Moreover, we would like to show this is independent of our choice of $x,y$. This follows since if we had other $x' \in A_{i'}, y' \in A_{j'}$ and choose $k \geqslant i,j,i',j'$, we would have
  \begin{multline*}
    \alpha_{k}(\alpha_{ik}(x) - \alpha_{i'k}(x')) = 0 \land \alpha_{k}(\alpha_{jk}(y) - \alpha_{j'k}(y')) = 0\
    \implies  \alpha_{k\ell}(\alpha_{ik}(x) - \alpha_{i'k}(x')) = 0 \land \alpha_{k\ell}(\alpha_{jk}(y) - \alpha_{j'k}(y')) = 0
  \end{multline*}
  for some $\ell \geqslant k$ by AM Exercise $2.15$. But this is equivalent to
  \begin{equation*}
    \alpha_{i\ell}(x) = \alpha_{i'\ell}(x') \land \alpha_{j\ell}(y) = \alpha_{j'\ell}(y'),
  \end{equation*}
  and so
  \begin{equation*}
    \alpha_{\ell}(\alpha_{i\ell}(x)\alpha_{j\ell}(y)) = \alpha_{\ell}(\alpha_{i'\ell}(x)\alpha_{j'\ell}(y)),
  \end{equation*}
  and so the product is independent from choice of $x,y$.
  \par Now suppose $A=0$; this implies $1_A = 0$. Then, there must exist $i \in I$ such that $0 = 1_A = \alpha_i(1_{A_i})$ since ring homomorphisms are unital; there then exists $j \geqslant i$ such that $\alpha_{ij}(1_{A_i}) = 0$ by AM Exercise $2.15$. But then, $\alpha_{ij}(1_{A_i}) = 0 = 1_{A_j}$ since ring homomorphisms are unital; thus, $A_j = 0$.
\end{proof}

Exercise 2.21

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Exercise 2.21

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