show that Ring and Alg(R) do not have cocartesian cofiltered limits

Author: ScriptRaccoon

database/data/004_property-assignments/Alg(R).sql

@@ -82,4 +82,12 @@ VALUES
 	'regular quotient object classifier',
 	FALSE,
 	'We may copy the proof for the <a href="/category/CAlg(R)">category of commutative algebras</a> (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\mathbf{Alg}(R)$ would produce one in $\mathbf{CAlg}(R)$ by <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).'
+),
+(
+	'Alg(R)',
+	'cocartesian cofiltered limits',
+	FALSE,
+	'Consider the ring $A = R[X]$ and the sequence of rings $B_n = R[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose colimit is $R[[Y]]$. Every element in the coproduct of rings $R[X] \sqcup R[[Y]]$ has a finite "free product" length. Now consider the elements
+	<br>$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n$.</br>
+	Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.'
 );

database/data/004_property-assignments/Ring.sql

@@ -82,4 +82,12 @@ VALUES
 	'regular quotient object classifier',
 	FALSE,
 	'We may copy the proof for the <a href="/category/CRing">category of commutative rings</a> (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\mathbf{Ring}$ would produce one in $\mathbf{CRing}$ by <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).'
+),
+(
+	'Ring',
+	'cocartesian cofiltered limits',
+	FALSE,
+	'Consider the ring $A = \mathbb{Z}[X]$ and the sequence of rings $B_n = \mathbb{Z}[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose colimit is $\mathbb{Z}[[Y]]$. Every element in the coproduct of rings $\mathbb{Z}[X] \sqcup \mathbb{Z}[[Y]]$ has a finite "free product" length. Now consider the elements
+	<br>$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n$.</br>
+	Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.'
 );