All algebras below are associative, and not assumed unital, and, to fix ideas, over the complex numbers.


An algebra $A$ is *supercommutative-gradable* if it admits a grading $A=A\_0\oplus A\_1$ in $\mathbf{Z}/2\mathbf{Z}$ ($A\_iA\_j\subset A\_{i+j}$ for $i,j\in\mathbf{Z}/2\mathbf{Z}$) that makes it supercommutative: for $a,b$ homogeneous $ab=ba$ if either $a,b$ has even degree, and $ab=-ba$ for $a,b$ of odd degree.


I insist that by supercommutative-gradable, I assume that such a grading exists, but do not endow $A$ with it: I still view $A$ as a bare algebra, with no fixed grading.



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> What are the polynomial identities satisfied by supercommutative-gradable algebras? More precisely, in universal algebra terms: what is the variety generated by supercommutative-gradable algebras? [In particular, is it finitely generated? (Edit: [Yes!](https://mathoverflow.net/a/361760/14094))]
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(For readers not familiar with universal algebra or polynomial identities, see addendum below to make the question precise.)


For instance, the class of supercommutative-gradable algebras satisfies the identities $(xy-yx)z-z(xy-yx)$ and $x^2y^2-2xyxy+2yxyx-y^2x^2$, ~~and none of these two follows from the other one~~. (The identity $(xy-yx)z-z(xy-yx)$ holds because $xy-yx$ always has even degree, hence is central.)




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Note: (about the above convention above for the meaning of supercommutative gradable: $\mathbf{Z}$-gradings vs $\mathbf{Z}/2\mathbf{Z}$-gradings)


Let $\mathcal{A}$ be the class of supercommutative-gradable algebras. Some subclasses of $\mathcal{A}$ could compete for being called "supercommutative-gradable algebras", namely the class $\mathcal{A}\_{\mathbf{Z}}$ (resp. $\mathcal{A}\_{\mathbf{N}}$, resp. $\mathcal{A}\_{\mathbf{N}\_{>0}}$), those algebras admitting an algebra grading in $\mathbf{Z}$ (resp...) satisfying the supercommutativity rule. Also we have smaller classes $\mathcal{A}^1\_{\mathbf{Z}}$, $\mathcal{A}^1\_{\mathbf{N}}$ in which we assume the algebra unital with unit of degree $0$. All the obvious inclusions between these classes are strict. However, the question is not sensitive to the choice of class: indeed, if $A\in\mathcal{A}$, then it is quotient of an algebra in $\mathcal{A}\_{\mathbf{N}\_{>0}}$, which itself (adding a unit) is subalgebra of an algebra in $\mathcal{A}^1\_{\mathbf{N}}$. For the former quotient assertion: write $A=A\_1\oplus A\_2$ (writing $A\_2$ rather than $A\_0$) and consider the free $\mathbf{Z}$-graded supercommutative algebra $\tilde{A}$ over the vector space $A\_1\oplus A\_2$ with $A\_1,A\_2$ of degree $1,2$: then $A$ is canonically quotient of $\tilde{A}$.




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Addendum (basic definitions of identities in algebras, varieties) 


Fix the associative (non-unital) free $\mathbf{C}$-algebra $\mathbb{F}=\mathbf{C}\langle X\_n:n\in\mathbf{N}\rangle$. An element $P\in \mathbb{F}$ is a *polynomial identity* of a class $\mathcal{C}$ of algebras if $P$ vanishes in every $A\in\mathcal{C}$, that is, if $P$ belongs to the kernel of every homomorphism $\mathbb{F}\to A$ for every $A\in\mathcal{C}$. 


The set of polynomial identities of $\mathcal{C}$ forms a 2-sided ideal $I\_\mathcal{C}$ of $F$ satisfying strong conditions: it is fully invariant (=stable under all endomorphisms); it is strongly graded, in the sense that it is a graded ideal for the unique algebra grading of $\mathbb{F}$ in the free abelian group $\mathbf{Z}^{(\mathbf{N})}$ (with basis $(e\_n)$) for which $X\_n$ has degree $e\_n$ for every $n$ (for instance $x\_1x\_2x\_1^4x\_2-x\_2^2x\_1^5$ has degree $5e\_1+2e\_2$, while $x\_1^2+x\_2^2$ is not strongly homogeneous). Describing polynomial identities of $\mathcal{C}$, in practice, means exhibiting generators of $I\_\mathcal{C}$ as a fully invariant 2-sided ideal.


For instance, for $\mathcal{C}$ the class of commutative algebras: the polynomial identities of $\mathcal{C}$ are generated by $X\_0X\_1-X\_1X\_0$. 


The *variety* generated by $\mathcal{C}$ is the class of all algebras in which all $P\in I\_{\mathcal{C}}$ are polynomial identities. It is also the smallest class of algebras containing $\mathcal{C}\cup\{\{0\}\}$ and stable under taking quotients, subalgebras, and arbitrary (unrestricted) direct products. The mapping $\mathcal{V}\mapsto I\_\mathcal{V}$ is a canonical bijection between the "set" of varieties (of associative algebras) and fully invariant 2-sided ideals of $\mathbb{F}$. [It's not properly a set: to make it a set, cheat by fixing a set $X$ of cardinal $2^{\aleph\_0}$ and consider $\mathbf{C}$-algebra structures with underlying set $X$.]


A variety of associative algebras $\mathcal{V}$ is *finitely based* if the ideal $I\_\mathcal{V}$ is finitely generated as fully invariant ideal (~~it's not always the case~~). To my surprise it's always the case (I expected the contrary, by analogy with groups or Lie algebras in finite characteristic).



Polynomial identities of supercommutative-gradable algebras
I believe that the identity $(xy-yx)z-z(xy-yz)$ generates everything (in characteristic 0, at least). To show that no further identities are needed, it is enough to exhibit one algebra that has no further identities. It follows from  [an old theorem of Krakowski and Regev](https://www.jstor.org/stable/1996643?seq=1#metadata_info_tab_contents) that the Grassmann algebra of a countably dimensional vector space works for that purpose. 


