@@ 1,5 1,6 @@
\newcommand{\id}{\operatorname{id}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\N}{\mathbb{N}}
\newcommand{\set}[3]{\{#1 \in #2 ~|~ #3\}}
\newcommand{\colim}{\operatorname{colim}}
@@ 70,6 70,34 @@ of $L$ and we may define a map $m \colon
With this definition we have $\varphi = \hom(m, A)$ since the two maps
agree on a basis of $\hom(K, A)$.
+The following example shows that we cannot assume
+the unrestricted functor
+$\hom(\_, A)$ to be full, if $A$ is a general ring.
+
+* *Example.*
+We consider $\hom (\N, \Z / p \Z)$.
+Let $\mathfrak{a}$ be the ideal of all $c \in \hom (\N, \Z / p \Z)$ with
+$c^{-1} (0)$ cofinite.
+By [Krull's theorem](https://en.wikipedia.org/wiki/Krull's_theorem)
+$\hom (\N, \Z / p \Z)$ has a maximal ideal $\mathfrak{m}$ with
+$\mathfrak{a} \subset \mathfrak{m}$ and this gives
+a homomorphism of fields
+$i \colon \Z / p \Z \rightarrow \hom (\N, \Z / p \Z) / \mathfrak{m}$.
+We further have
+$[c]^p - [c] = [c^p - c] = 0$
+for all $[c] \in \hom (\N, \Z / p \Z) / \mathfrak{m}$ and
+as $X^p - X$ is a polynomial of degree $p$ it has at most $p$ roots in
+$\hom (\N, \Z / p \Z) / \mathfrak{m}$ and thus $i$ is a bijection.
+Now the canonical homomorphism from $\hom (\N, \Z / p \Z)$ to the quotient
+$\hom (\N, \Z / p \Z) / \mathfrak{m}$ yields a homomorphism
+$\varphi \colon \hom (\N, \Z / p \Z) \rightarrow
+\hom(\{1\}, \Z / p \Z) \cong \Z / p \Z$
+which is not in the image of $\hom(\_, \Z / p \Z)$,
+since for any map
+$m \colon \{1\} \rightarrow \N$ the element
+$1_{m(1)} \in \mathfrak{a} \subset \mathfrak{m}$ is mapped to
+$1 \in \Z / p \Z$ under $\hom(m, \Z / p \Z)$.
+
* **Lemma.**
$\hom(\_, A)$ is continuous as a functor from the opposed category of sets
to the category of $A$-algebras.