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Let $\Phi$ denote the set of all identities satisfied by $(\mathbb{N},0,1,+,\times,\mathrm{gcd},\mathrm{lcm}).$

Question. Is $\Phi$ finitely axiomatizable? If so, I'd like to see a list of identities.

Noteworthy elements of $\Phi$:

  • $(\mathbb{N},0,1,+,\times)$ is a commutative semiring
  • $(\mathbb{N},1,0,\mathrm{gcd},\mathrm{lcm})$ is a bounded distributive lattice (with bottom $1$ and top $0$)
  • $\mathrm{gcd}(a,b+a) = \mathrm{gcd}(a,b)$
  • $\gcd(a+b,\operatorname{lcm}(a,b))=\gcd(a,b)$
  • $\mathrm{gcd}(a,b)\mathrm{lcm}(a,b) = ab$

The second last identity above actually follows from the preceeding one's; see Bill's answer here.

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goblin GONE
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  • "$(\mathbb{N}, 1, 0, \gcd, $lcm$)$ is a bounded distributive lattice (with least element 1)". And, by the way, greatest element is $0$. – amrsa Jul 11 '16 at 11:46
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    @amrsa, of course it is. I just couldn't fit that onto one line :) – goblin GONE Jul 11 '16 at 11:53
  • You can fit it in if you say "with top $0$ and bottom $1$". – user21820 Jul 11 '16 at 13:25
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    If this is possible it will be just barely. If you add merely a $\min$ operator to the language, then the set of identities is definitely not recursively enumerable, due to Matiyasevich's theorem. – hmakholm left over Monica Jul 11 '16 at 13:44
  • ... combined with $$\min(1+\gcd(a+1,b+1),\operatorname{lcm}(a+1,b+1))=1+\gcd(a+1,b+1) \iff a\ne b$$ – hmakholm left over Monica Jul 11 '16 at 14:09
  • @user21820, good thinking! – goblin GONE Jul 12 '16 at 05:26
  • @HenningMakholm, woah! You know some stuff. Why is that equivalence true? – goblin GONE Jul 12 '16 at 05:27
  • @goblin: I constructed it to be, while attempting (in vain) to prove your actual problem impossible. If we set $P=\gcd(a+1,b+1)$ and $Q=\operatorname{lcm}(a+1,b+1)$, then we have $P\le Q$ with $P=Q$ exactly when $a=b$. And then $\min(1+P,Q)=1+P$ exactly when $P\ne Q$, that is, when $a\ne b$. (The inner $+1$s are there to avoid $0$ which is the sole exception to "divides" implying "$\le$"). – hmakholm left over Monica Jul 12 '16 at 08:57
  • If we have both max and min, then we could use $P=\min(a,b)$ and $Q=\max(a,b)$, and thereby get a proof that the true identities in $(\mathbb N,0,1,+,\times,\min,\max)$ are not recursively enumerable. It feels unlikely that it would be easier to deal with gcd and lcm than with max and min, but I haven't found a way to actually prove that. – hmakholm left over Monica Jul 12 '16 at 09:26
  • @MattF.: As I explained in in the subsequent comment, having min would allow us to express $a\ne b$ as an identity. So if all the true identities involving min (and gcd and lcm) were recursively enumerable, we could determine whether a polynomial $P(x,y,\ldots)$ has roots in the integers: Just look for a solution and in parallel enumerate all the identities and see if we come across the representation of $P(x,y,\ldots)\ne 0$. But Matiyasevich (et al) teaches us that this problem is undecidable, which is a contradiction. – hmakholm left over Monica Jan 10 '19 at 12:24
  • @HenningMakholm, I find this clearer with an example. To prove that $x^2-2y^2=0$ has no solutions, it suffices to prove $$\min((x^2-2y^2)^2,\ 1)=1$$ or equivalently $$\min(x^4+4y^4,\ 1+4x^2y^2)=1+4x^2y^2$$ So a finite axiomatization for the identities in the language $(1,+,\cdot,\min)$ would make solving any such problem recursive, and Matiyasevich has showed us that it is not. –  Jan 10 '19 at 16:26
  • @HenningMakholm, so I agree with your original comment, and we can there without $gcd$ or $lcm$. –  Jan 10 '19 at 16:41
  • @MattF.: Right, that is much simpler! Unfortunately that still doesn't tell us anything about the original question where we have gcd and lcm but not min or max. – hmakholm left over Monica Jan 10 '19 at 18:10
  • I posted an answer based on $\gcd(1+x,1+x^3)=1+x$, $\gcd(1+x,1+x^5)=1+x$, etc., but I now see that they all follow from $\gcd(a,b)=\gcd(a,b+ac)$. E.g. $\gcd(1+x,1+x^3)=\gcd(1+x,(1+x^3)+(1+x)(3x))=\gcd(1+x,(1+x)^3)=1+x$. –  Jan 11 '19 at 19:54

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