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I just want to check that my proof is correct. For a given variety $K$, I let $F_K(X)$ denote the free $K$-algebra with generator set $X$.

Lemma 1. Assume a variety $V$ has arbitrary (small) coproducts as a category, which we denote by $\sqcup$. Then for all set $X$, $F_V(X) \simeq \displaystyle\sqcup_{x\in X} F_V(\{x\})$

Indeed, let $f: X\to A$ be any function, where $A$ is the underlying set of the algebra $\mathfrak{A}$. Then we obtain restriction functions $f_x: \{x\} \to A$ for all $x\in X$, and thus we obtain a unique morphism $\overline{f}_x: F_V(\{x\}) \to \mathfrak{A}$ and thus by the universal property of coproducts we get a unique morphism $\displaystyle\sqcup_{x\in X} F_V(\{x\}) \to \mathfrak{A}$.

Thus $\displaystyle\sqcup_{x\in X} F_V(\{x\})$ has the universal property of the free algebra with generating set $X$, thus it is isomorphic to it.

Lemma 2. Let $V$ be the variety of boolean algebras. Then $V$ has arbitrary (small) coproducts.

(2nd EDIT: For those interested, it's in this proof that the error noted by Keith Kearnes lies. What I call $\mathfrak{C}$ is not a subalgebra of $\mathfrak{B} $ )

Let $(\mathfrak{B}_i)_{i\in I}$ be a family of boolean algebras. We let $\mathfrak{B}= \displaystyle\prod_{i\in I} \mathfrak{B}_i$, and $\mathfrak{C}$ is the subalgebra of $\mathfrak{B}$ whose elements are those with only a finite number of nonzero coordinates, or all of its coordinates are $1$. Obviously, $\mathfrak{C} \in \mathbf{SP}(V) = V$.

We want to show that $\mathfrak{C}$ is a coproduct of the $\mathfrak{B}_i$'s. Let $\iota_i : \mathfrak{B}_i \to \mathfrak{C}$ be the morphism that sends $b$ to the family consisting of all $0$'s + a $b$ in place $i$. Let $f_i : \mathfrak{B}_i \to \mathfrak{A}$ be algebra homomorphisms.

Assume we have a morphism $f: \mathfrak{C}\to \mathfrak{A}$ such that $f\circ \iota_i = f_i$. Then for $c=$ all zeroes except in positions $i_1,...,i_n$ where there is $b_1,...,b_n$, well $c= \iota_{i_1}(b_1)\lor ... \lor \iota_{i_n}(b_n)$ so that $f(c)= f\circ\iota_{i_1}(b_1)\lor...\lor f\circ\iota_{i_n}(b_n) = f_{i_1}(b_1)\lor...\lor f_{i_n}(b_n)$ and $f(1,...,1,...) = 1$. Therefore $f$ is unique.

Conversely, defining $f$ by the above formulas shows that $f$ exists : $\mathfrak{C}$ is a coproduct (and so, up to isomorphism, the coproduct) of the $\mathfrak{B}_i$'s.

Lemma 3. Let $V$ be the variety of boolean algebras. For all $x$, $F_V(\{x\}) \in \mathbf{HSP}(\mathbf{B}_2)$ where $\mathbf{B}_2$ is the $2$-element boolean algebra.

It suffices to show it for one $x$, since all free algebras on one generator are isomorphic. For this we thus describe the free boolean algebra on one element. Let $B= \{ 0, x, y 1\}$ where $x\neq 0, 1$.Defining $x\land x= x, x\lor x = x, x\land 1= x, x\lor 1 = 1, x\land 0 = 0, x\lor 0 = x$, similar clauses for $y$, and $x\lor y = 1, x\land y =0$, $x' = y$, $y' = x$, we get that the algebra $\mathfrak{B}$ is a boolean algebra, and it is obviously free with one generator. Moreover, $\mathfrak{B} \simeq \mathbf{B}_2 ^2$. Indeed, sending $x\to (1,0)$ for instance provides an isomorphism.

Conclusion: By Lemma 3, for all $x$, $F_V(\{x\}) \in \mathbf{HSP}(\mathbf{B}_2)$. By lemmas 1 and 2, for all $X$, $F_V(X) \in \mathbf{HSP}(\mathbf{B}_2)$. Since any boolean algebra is the homomorphic image of a free boolean algebra, it follows that for any $\mathfrak{B}\in V$, $\mathfrak{B} \in \mathbf{HSP}(\mathbf{B}_2)$. It therefore follows that $V= \mathbf{HSP}(\mathbf{B}_2)$

Thank you for reading so far. My questions are: is this proof valid ? Are there any easier proofs of this fact ? Or not necessarily simpler, but essentially different ?

EDIT: Hopefully this remark will save the proof (which is not correct according to an answer below): when I claim that $F_V(X) \in \mathbf{HSP}(\mathbf{B}_2)$, it is because the coproduct of boolean algebras is a subalgebra of their product, that is it belongs to the variety generated by these algebras. Maybe I should have made this clearer by stating lemma 2 as "Let $V$ be a variety of boolean algebras" instead of "Let $V$ be the variety of boolean algebras".

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  • $\begingroup$ In the proof of Lemma 2, the functions $\iota_i$ are not BA morphisms. $\endgroup$ Jun 28, 2017 at 11:28
  • $\begingroup$ @KeithKearnes : Yes actually it seems I completely forgot about the complements. $\endgroup$ Jun 28, 2017 at 11:34

1 Answer 1

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The proof is not correct.

Let $V$ be the variety of Boolean algebras (= the variety axiomatized by the identities of Boolean algebras). The 2-element BA belongs to $V$, so $HSP(B_2)\subseteq V$. You correctly claim that $F_V(\{x\})$ belongs to $HSP(B_2)$. You then incorrectly deduce that the coproduct $F_V(X)$ must also belong to $HSP(B_2)$ for any $X$, and hence that $V=HSP(B_2)$.

Given varieties $U\subseteq V$, the coproduct of $U$-algebras, computed in $V$, need not belong to $U$. (Here $U=HSP(B_2)$.) Otherwise we would have $V = HSP(F_V(\{x\}))$ for every variety. This fails for the varieties of groups, rings, lattices, ETC.


For a correct proof, might try these steps:

  1. Every variety is generated by is subdirectly irreducible members.
  2. $B_2$ is the only subdirectly irreducible member of the variety axiomatized by the identities of Boolean algebra.
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  • $\begingroup$ But here I show that the coproduct is the subalgebra of the product, so why not ? Si $HSP(B_2)$ is stable under these operations, and $F_V(\{x\}) \in HSP(B_2)$, then any subalgebra of any product of $F_V(\{x\})$'s must belong to $HSP(B_2)$, and so, since $F_V(X)$ is isomorphic to said coproduct, it must also be in $HSP(B_2)$. Where's the mistake in that ? Thank you for your answer ! $\endgroup$ Jun 28, 2017 at 11:00
  • $\begingroup$ If in lemma 2, I had written "a variety of boolean algebras", instead of "the variety", the proof would go similarly, would it not ? $\endgroup$ Jun 28, 2017 at 11:01
  • $\begingroup$ ``But here I show that the coproduct is the subalgebra of the product...'' The statement of Lemma 2 is correct, but its proof is not. (1) $\mathfrak C$ is not a subalgebra. (2) in general, the coproduct of two BA's is not embeddable in their product. $\endgroup$ Jun 28, 2017 at 11:11
  • $\begingroup$ Oh right, I had forgotten about the complements ! That's why $\mathfrak{C}$ is not a subalgebra, right ? (or I made yet another mistake ?) Thank you so much ! $\endgroup$ Jun 28, 2017 at 11:18
  • $\begingroup$ (and thank you for the correct proof, I can show both of the items and therefore I can conclude !) $\endgroup$ Jun 28, 2017 at 11:57

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