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Let $S$ be a finite set. Let $R$ be a complex vector space with basis indexed by subsets of $S$. Define a product on $R$ by defining it on the basis elements as $1_A\cdot 1_B=1_{A\Delta B}$, where $A\Delta B$ is the symmetric difference of $A$ and $B$. This gives $R$ the structure of a commutative and associative $C$-algebra.

Is this a well-understood algebra? Does it have a name?

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It is the complex group algebra over $((\mathbb{Z}/2)^S,+)$. This may be also described as the tensor product of $S$ copies of $\mathbb{C}[\mathbb{Z}/2] = \mathbb{C} \times \mathbb{C}$.

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  • $\begingroup$ Note that this description is also valid if $S$ is infinite. $\endgroup$ – Martin Brandenburg Sep 30 '10 at 12:46
  • $\begingroup$ Is it OK to speak purely algebraically of an infinite- (even uncountable-) fold tensor product? I guess that one has to construct it "all at once" by an obvious analogue of the 2-fold construction, satisfying an obvious analogue of the 2-fold universal property, since one can't sneak up on it by tensoring only 2 modules at a time. $\endgroup$ – LSpice Oct 17 '18 at 16:11
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This is the group algebra $\mathbb{C}G$ where $G$ is an elementary Abelian group of order $2^n$ where $n=|S|$ (the elements of $G$ are your $1_A$). Every group ring over $\mathbb{C}$ is isomorphic to the direct product of matrix algebras over $\mathbb{C}$. In this case as $G$ is Abelian, $\mathbb{C} G$ is isomorphic to the product of $2^n$ copies of $\mathbb{C}$.

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This is the group algebra of the additive group $(\mathbb Z/2\mathbb Z)^S$, hence it is the product of $2^{|S|}$ copies of $\mathbb C$.

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