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Let $K$ be a field and $K\{X\}$ be the free non-associative algebra, freely generated by the countably infinite set $X$. We consider elements of $K\{X\}$ as (non-associative) polynomials in the variables of $X$.

Given $\mathfrak{F}\subseteq K\{X\}$, one can consider the variety of algebras defined by $\mathfrak{F}$. This is the class of all algebras which satisfy all identities of $\mathfrak{F}$, i.e., an algebra $A$ is an algebra in such variety if given $p\in \mathfrak{F}$, for any $a_i\in A$, we have $p(a_1,\dots,a_n)=0$.

For example, if one consider the polynomial $p=x(yz)-(xy)z$, the variety of algebras defined by $\{p\}$ is the variety of associative algebras.

My question is the following:

Is the variety of algebras defined by some $\mathfrak{F}\subseteq K\{X\}$, a set?

A simpler question: Is a variety of associative algebras a set?

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I don't understand the question. If $A$ is an algebra in the "variety" deined by $\mathfrak{F}$, then each $A \times \{ \alpha \}$ is in the same variety, for all $\alpha \in \mathsf{Ord}$. – Qfwfq Feb 8 '13 at 14:37
Every nontrivial variety $V$ has members of arbitrarily large cardinality, and therefore is a proper class: in particular, if $A\in V$ has at least two elements, then $A^\kappa\in V$ for every cardinal $\kappa$, and $|A^\kappa|\ge2^\kappa$. – Emil Jeřábek Feb 8 '13 at 15:02
That is, as Qfwfq says, every variety is a proper class for the simple reason it contains a proper class of isomorphic copies of any its algebra. What I meant is that a nontrivial variety does not even have a set of representatives up to isomorphism. – Emil Jeřábek Feb 8 '13 at 15:18
up vote 3 down vote accepted

Summarizing the above, the answer is no.

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