# HG changeset patch
# User Benedikt Fluhr <http://bfluhr.com>
# Date 1445636619 -7200
#      Fri Oct 23 23:43:39 2015 +0200
# Node ID f832c5223a7dfc0ff3b027a3144b03e615cd916a
# Parent  1ac0ccce2566c4671fd874a1cbb4202730fdf78f
Counterexample to hom(_, A) being full

diff --git a/00_00_00_macros.tex b/00_00_00_macros.tex
--- a/00_00_00_macros.tex
+++ b/00_00_00_macros.tex
@@ -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}}
diff --git a/06_fromSetsToAlgebras.md b/06_fromSetsToAlgebras.md
--- a/06_fromSetsToAlgebras.md
+++ b/06_fromSetsToAlgebras.md
@@ -70,6 +70,34 @@
 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.