CAlg(R).sql

INSERT INTO category_property_assignments (
	category_id,
	property_id,
	is_satisfied,
	reason
)
VALUES
(
	'CAlg(R)',
	'locally small',
	TRUE,
	'There is a forgetful functor $\mathbf{CAlg(R)} \to \mathbf{Set}$ and $\mathbf{Set}$ is locally small.'
),
(
	'CAlg(R)',
	'finitary algebraic',
	TRUE,
	'Take the algebraic theory of a commutative ring.'
),
(
	'CAlg(R)',
	'strict terminal object',
	TRUE,
	'If $f : 0 \to R$ is a homomorphism, then $R$ satisfies $1=f(1)=f(0)=0$, so that $R=0$.'
),
(
	'CAlg(R)',
	'Malcev',
	TRUE,
	'This follows in the same way as for groups, see also Example 2.2.5 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.'
),
(
	'CAlg(R)',
	'coextensive',
	TRUE,
	'One can use the same proof as for $\mathbf{CRing}$.'
),
(
	'CAlg(R)',
	'balanced',
	FALSE,
	'Take a prime ideal $P \subseteq R$ and consider the commutative $R$-algebra $A := R/P$ (which is an integral domain). Then the inclusion $A \hookrightarrow Q(A)$ is a counterexample.'
),
(
	'CAlg(R)',
	'cogenerating set',
	FALSE,
	'We apply <a href="/lemma/missing_cogenerating_sets">this lemma</a> to the collection of commutative $R$-algebras which are fields: If $F$ is a commutative $R$-algebra that is also a field and $A$ is a non-trivial commutative $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).'
),
(
	'CAlg(R)',
	'skeletal',
	FALSE,
	'This is trivial.'
),
(
	'CAlg(R)',
	'countably codistributive',
	FALSE,
	'The canonical homomorphism $A \otimes_R R^{\mathbb{N}} \to A^{\mathbb{N}}$ is given by $a \otimes (r_n)_n \mapsto (r_n a)_n$ and does not have to be surjective: Since $R \neq 0$, there is a commutative $R$-algebra $K$ which is a field. Now take $A := K[X]$ and consider the sequence $(X^n)_{n} \in A^{\mathbb{N}}$.'
),
(
	'CAlg(R)',
	'semi-strongly connected',
	FALSE,
	'Choose a maximal ideal $\mathfrak{m}$ of $R$, so $K := R/\mathfrak{m}$ is a field. If $\mathbf{CAlg}(R)$ is semi-strongly connected, then also $\mathbf{CAlg}(K)$ is semi-strongly connected. This has been disproven in <a href="https://math.stackexchange.com/questions/5129689" target="_blank">MSE/5129689</a>.'
),
(
	'CAlg(R)',
	'coregular',
	FALSE,
	'See <a href="https://math.stackexchange.com/questions/3745302" target="_blank">MSE/3745302</a>.'
),
(
	'CAlg(R)',
	'co-Malcev',
	FALSE,
	'See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{CAlg}(R) \to \mathbf{Set}$ and the relation $S \subseteq U^2$ defined by $S(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $R[X]$ and $S$ by $R[X,Y] / \langle XY-X^2 \rangle$. It is clear that $S$ is reflexive, but not symmetric.'
),
(
	'CAlg(R)',
	'regular quotient object classifier',
	FALSE,
	'The strategy is similar to the one for $\mathbf{CRing}$: Assume that $P \to R$ is a regular quotient object classifier. If $J$ denotes the kernel of $P \to R$, every ideal $I \subseteq A$ of any commutative $R$-algebra has the form $I = \langle \varphi(J) \rangle$ for a unique homomorphism $\varphi : P \to A$. If $\sigma : A \to A$ is an automorphism with $\sigma(I)=I$, then uniqueness gives us $\sigma \circ \varphi = \varphi$, which means that $\varphi(J)$ lies in $A^{\sigma}$, the fixed algebra of $\sigma$. But then $I$ is generated by elements in $A^{\sigma} \cap I$. If $K$ is a residue field of $R$, this fails for $A = K[X,Y]$, $I = \langle X,Y \rangle$, $\sigma(X)=Y$, $\sigma(Y)=X$. The fixed algebra is the subalgebra of symmetric polynomials, which is $K[X+Y,XY]$. So $\langle X,Y \rangle$ is generated by symmetric polynomials without constant term, which implies $\langle X,Y \rangle \subseteq \langle X+Y,XY \rangle$ in $K[X,Y]$. But reducing an equation like $X = a(X,Y) \cdot (X+Y) + b(X,Y) \cdot (XY)$ modulo $\langle X^2,Y^2,XY \rangle$ yields a contradiction.'
);