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Consider the equational identities of the algebraic structure $(\mathbb{N};+,\cdot,0,1)$. I believe that the following identities are a basis for it:

  1. The commutative properties, of both addition and multiplication.
  2. The associative properties, of both addition and multiplication.
  3. The distributive property connecting addition and multiplication.
  4. $x+0=x$
  5. $x\cdot 1=x$
  6. $x\cdot 0=0$

Is this true? Can all equational identities of that structure be generated from my set of identities? If so, what is the proof?

Thomas Andrews
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user107952
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  • What is your definition of "equational identity"? Is it different from "identities that follow from the axiomatic structure of the natural numbers"? – lulu Aug 16 '22 at 23:52
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    @lulu This is standard terminology in universal algebra, which is the subject of this question. An equation, equational identity, or identity (yeah, logic is like that with terminology) is an expression of the form $s=t$ for terms $s,t$; there's an implicit universal quantifier here which is usually not written out explicitly. – Noah Schweber Aug 17 '22 at 00:10
  • I suspect the absence of an induction axiom will make some identities unprovable, but I don't have an example. – aschepler Aug 17 '22 at 00:11
  • @NoahSchweber So...presumably the types of relations are heavily constrained, no? Is $\lfloor \sqrt 2\rfloor=1$ an equational identity? What about $\sum \frac 1{n^2}=\frac {\pi^2}6$? My point, to be clear, is that it's not obvious (to me) why this isn't either false or tautologically true. – lulu Aug 17 '22 at 00:13
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    @lulu Note the opening of the question, where a structure (namely the naturals with addition, multiplication, zero, and one) is specified. Only terms in the language of that structure are considered (so neither of your examples is an identity here). Again, this is standard background which is appropriate to assume in any universal algebra question. – Noah Schweber Aug 17 '22 at 00:14
  • @NoahSchweber So...is the answer to this question something other than "tautologically, yes."? Sorry if I am missing something obvious here. – lulu Aug 17 '22 at 00:19
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    @lulu Why should it be? Each of the identities listed is true of $\mathbb{N}$, but there could be additional identities true of $\mathbb{N}$ which are not a consequence of those listed. (In particular, note that there are finite structures whose equational theories have no finite axiomatization by equations!) – Noah Schweber Aug 17 '22 at 00:20
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    So what if we show we can equate any term with a standard polynomial form, with a well-defined order of monomials, and coefficients in the form $1+1+\cdots+1$? We know any multi-variable polynomials which are equal for any variable evaluation are identical. – aschepler Aug 17 '22 at 00:20
  • @NoahSchweber Interesting. One last thing and I'll leave you alone: have you got a link for one of those "pathological" finite structures? – lulu Aug 17 '22 at 00:26
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    @lulu See https://math.stackexchange.com/questions/4155200/what-is-the-smallest-possible-cardinality-of-a-non-finitely-based-magma. (You may also be interested in https://mathoverflow.net/a/154019/8133.) – Noah Schweber Aug 17 '22 at 00:37
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    @aschepler yes, I think that approach works. for any term $s(\bar{x})$ in our language, the theory OP has posted proves there is a term $t(\bar{x})$ of the form you indicate such that $s(\bar{x})=t(\bar{x})$ identically, so you can reduce to the case of terms of the form you indicate. then I think we should be done by your remark. you should write it up as an answer! :) – Atticus Stonestrom Aug 17 '22 at 00:37
  • @NoahSchweber Thanks! – lulu Aug 17 '22 at 00:38
  • @aschepler You should add that as an answer to move this off the unanswered queue. – Noah Schweber Aug 22 '22 at 03:07
  • @NoahSchweber or OP: I'm not especially familiar with this domain. Does the $\mathbb{N}$ in $(\mathbb{N}; +, \cdot, 0, 1)$ mean that each natural number is a ground term in the formula language, or only that $\mathbb{N}$ acts as the model for evaluating equations true or false? – aschepler Aug 22 '22 at 22:26
  • @aschepler Only the latter. But in fact every natural number $n$ also has a corresonding canonical term ("numeral") $\underline{n}$, usually defined as $$\underline{0}=0,\quad\underline{1}=1,\quad\underline{2}=1+1,\quad\underline{3}=(1+1)+1,\quad\underline{4}=((1+1)+1)+1,\quad...$$ and the equational theory given is enough to convert any variable-free term into a numeral. – Noah Schweber Aug 22 '22 at 22:28

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