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I came across the following answer that proofs that every $n$-ary operation on a finite set is a finite composition of binary operations:

A proof is especially simple for operations on a finite set $\rm\:A\:.\:$ Namely, if $\rm\:|A| = n\:$ then we may encode $\rm\:A\:$ by $\rm\:\mathbb Z/n\:,\:$ the ring of integers $\rm\:mod\ n\:,\:$ allowing us to employ Lagrange interpolation to represent any finitary operation as a finite composition of the binary operations $\rm\: +,\ *\:,\:$ and $\rm\: \delta(a,b) = 1\ if\ a=b\ else\ 0\:,\:$ namely $$\rm f(x_1,\ldots,x_n)\ = \sum_{(a_1,\ldots,a_n)\ \in\ A^n}\ f(a_1,\ldots,a_n)\ \prod_{i\ =\ 1}^n\ \delta(x_i,a_i) $$

For me, it is unclear how $\delta(a,b)$ is decomposed into the binary operations $\rm\: +,\ *\:$ in the case of $n$ not being a prime. If $n$ is a prime, the ring $\rm\:\mathbb Z/n\:$ is a field and we can write $$\delta(x_i,a_i) = \prod_{a_j\ \in\ A\\a_j \neq a_i} \frac{x_i - a_j}{a_i - a_j} ,$$ as the inverse of $a_i - a_j$ exists. Am I missing a decomposition of $\delta(a, b)$ like the above that also holds true for rings? Or does Sierpinski's proof rely on the existence of the Kronecker Delta as a third "binary operation" $\delta: A \to \{0, 1\}$ that cannot be decomposed further for the cases where $n$ is not prime.

If I were to miss such a decomposition of $\delta(a, b)$, this proof actually states that every $n$-ary operation on a finite set can be re-interpreted as a polynomial over the ring $\rm\:\mathbb Z/n\:$. Is that reasoning correct?

If Sierpinski's proof relies on the existence of the Kronecker Delta as a third "binary operation", is there a proof that, for $n$ not prime, $n$-ary operations exist that cannot be re-interpreted as a polynomial over a ring?

The same proof is also stated here.

Niklas Netter
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    $\delta(x,y)$ is a already binary operation. It doesn't have to be decomposed into other binary operations. (Moreover, it is impossible to express $\delta$ in terms of $+$ and $$, since, for a prime divisor $p$ of $n$, $+$ and $$ preserve congruence modulo $p$ and $\delta$ does not.) – Keith Kearnes Apr 03 '23 at 23:21
  • A "binary operation" on a set $A$ means a map $A \times A \to A$. That is all it means. $+, \times, -$ are all binary operations, but they are not the only binary operations. If $|A| = n$, then there are $n^{(n^2)}$ distinct binary operations on $A$. $\delta$ is one of those other binary operations. – Paul Sinclair Apr 05 '23 at 00:12

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Let me convert my coment to an answer, so that I can expand on the 'Moreover' part of the comment.

If $n>1$ and $A=\mathbb Z_n$ has elements $\{0,1,\ldots,n-1\}$, then $\delta$ is a function $\delta\colon A\times A\to A$ where $\delta(a,a)=1$ and $\delta(a,b)=0$ if $a\neq b$. Thus, $\delta$ is a binary operation on $A$.

If $n=q$ is prime, then $\delta$ is a polynomial of the ring $\mathbb Z_n=\mathbb Z_q$, since every finitary operation on $\mathbb Z_q$ is a polynomial operation (by Lagrange interpolation).

But, if $n$ is not prime, then $\delta$ is not a polynomial operation of $\mathbb Z_n$ since, for any prime $p$ dividing $n$, all polynomial operations of $\mathbb Z_n$ preserve congruence modulo $p$ while $\delta$ does not. The statement that every polynomial operation preserves congruence modulo $p$ is the statement that if $f(x_1,\ldots,x_k)$ is a polynomial operation and $a_i\equiv b_i\pmod{p}$ for $i=1,\ldots,k$, then $f(a_1,\ldots,a_k)\equiv f(b_1,\ldots,b_k)\pmod{p}$. To see that $\delta$ does not preserve this relation, let $(a_1,b_1)=(0,0)$ and $(a_2,b_2)=(0,p)$. Then $a_1\equiv b_1\pmod{p}$ and $a_2\equiv b_2\pmod{p}$, but $$\delta(a_1,a_2)=\delta(0,0)=1\not\equiv 0=\delta(0,p)=\delta(b_1,b_2)\pmod{p}.$$

Keith Kearnes
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