Does there exist a finite set $S$ with a single binary operation $*$, where the only equational identities that hold are of the form $t=t$ for some term $t$?
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Yes, for example the empty set. And there's one more! – diracdeltafunk May 18 '21 at 23:05
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6@diracdeltafunk That's wrong - every equation holds in the empty magma (vacuously). Same with the one-element magma. – Noah Schweber May 18 '21 at 23:06
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Whoops! Thanks for setting me straight – diracdeltafunk May 19 '21 at 20:40
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In fact there is no finite magma which is generic even for one-variable terms.
Suppose $M$ is a finite magma. There are infinitely many one-variable terms but only finitely many functions $M\rightarrow M$; thus, there are distinct one-variable terms $t,s$ such that $t(a)=s(a)$ for all $a\in M$.
Note that it's important that we think about functions on $M$ as a whole rather than just individual elements: for example, for any set of $\vert M\vert+1$ one-variable terms $T$ there will be some $t,s\in T$ and $a\in M$ such that $t(a)=s(a)$, but there need not be distinct terms in $T$ which agree on all of $M$.
An interesting question is how many (parameter-free) one-variable terms there can be in a magma of size $n$ up to equality-of-corresponding-operations, as a function of $n$; of course we can get a trivial upper bound of $n^n$ by counting functions, but it's not clear to me that this upper bound is achievable. (We can also obviously ask the same question for arbitrary arity terms, with trivial upper bound of $n^{(n^{arity})}$, but already the unary case seems nontrivial to me.)
Noah Schweber
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Nice question! I suspect arbitrary magmas might be weird enough to encode all $n^n$ unary functions. Not at all convincing evidence: I've checked it for all $n\leq 2$. Maybe you should post this as its own question. – Alex Kruckman May 19 '21 at 01:59
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