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In his answer link to the question whether $a|m$ and $a+1|m$ implies $a(a+1)|m$, Bill Dubuque takes a detour to derive the equality $$ \gcd(a,b)=ab/\mathrm{lcm}(a,b) $$ from the universal property of $\gcd$ and $\mathrm{lcm}$. Since they have a universal property, the natural question is: is $\gcd$ the right adjoint of something and is $\mathrm{lcm}$ the left adjoint of something?

A note about the answers below

Originally, the question was meant as: does the functor $\gcd:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ has a left adjoint. This was answered by Qiaochu Yuan by observing that $\gcd$ is a categorical product.

Hurkyl had interpreted my question as: does the functor $\gcd(-,b):\mathbb{N}\to\mathbb{N}$ have a left adjoint for each $b\in\mathbb{N}$? This is not the case, but interestingly the functor $\mathrm{lcm}(-,b):\mathbb{N}\to\mathbb{N}$ has a left adjoint. This shows that if we reverse the ordering of divisibility, the category $\mathbb{N}$ becomes cartesian closed.

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Egbert
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2 Answers2

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Well, as far as category theory goes, it is a categorical product (and $\text{lcm}$ is a categorical coproduct) in the category whose objects are the natural numbers where there is a single arrow $a \to b$ if $a | b$.

All limits and colimits are adjoints. If $C$ is a category and $J$ a diagram category, then the diagonal inclusion $C \to C^J$ sending every object $c \in C$ to the constant functor $J \to C$ with value $c$ potentially has a left and right adjoint, and these are the limit and colimitcolimit and limit respectively when they exist.

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    It is indeed the other way around. The mnemonic is, right adjoints preserve limits – so the formation of colimits can't be a right adjoint, since those don't usually commute with limits! – Zhen Lin May 21 '12 at 21:20
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Here's an explicit left adjoint $L$ to lcm in the poset of natural numbers ordered by divisibility.

I define $L(a,b)$ in terms of its valuation at each prime: $$ v_p(L(a,b)) = \begin{cases} 0 & v_p(a) \leq v_p(b) \\ v_p(a) & v_p(b) < v_p(a) \end{cases}$$

To verify it is a left adjoint:

  • $L(a,b) \mid c$
  • iff $\forall p: v_p(a) \leq v_p(b) \vee v_p(a) \leq v_p(c)$
  • iff $a \mid \mathop{\mathrm{lcm}}(b,c)$

It's sort of a weird function, and I don't think I've seen it before, but there it is!

I computed it by fixing $a,b$ and looking at the requirement

  • $L(a,b) \mid c$ if and only if $a \mid \mathop{\mathrm{lcm}}(b,c)$

I picked the smallest value for $c$ that made the right hand side true, which gives an upper bound on $L(a,b)$. Then, I checked that actually defining that upper bound to be $L(a,b)$ works.

There isn't a right adjoint to $\mathop{\mathrm{gcd}}$; my intuition attributes that to the fact that there is no upper bound on things. The proof is short: the condition is

  • $\mathop{\mathrm{gcd}}(a,b) \mid c$ if and only if $a \mid R(b,c)$

For a fixed $b,c$, we can pick arbitrarily large values of $a$ relatively prime to $b$, which makes the left hand side true. This contradicts the fact that $R(b,c)$ would be an upper bound on the set of values for $a$, and so $R(b,c)$ cannot exist.

  • I think I don't fully understand the functions $v_p$ and how you construct $L$ in terms of it. So you proved that $\mathrm{lcm}(-,b)$ has a left adjoint; in other words, that the dual of $(\mathbb{N},|)$ is cartesian closed? – Egbert May 22 '12 at 06:53
  • Is $L(a,b)=\mathrm{lcm}(a,b)/b$? – Egbert May 22 '12 at 07:31
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    $v_p(x)$ is the number of times the prime $p$ divides $x$. So, $v_2(18) = 1$ and $v_3(18) = 2$. A sample value of $L$ is $L(12,18) = 4$. It has this value because $2$ is the only prime that divides $12$ more than it does $18$, and it does so twice. –  May 22 '12 at 08:50
  • Ok, I get it. But I believe that the assertion in the second item should read: iff $\forall p: v_p(a)\leq v_p(b)\vee v_p(a)\leq v_p(c)$. – Egbert May 22 '12 at 12:48
  • I've made the fix. –  May 22 '12 at 16:10
  • I still think that it should be disjunction, because in the conjunction case you'd get $L(a,b)=1$. Or I'm still misunderstanding it... – Egbert May 22 '12 at 16:30
  • Whoops, I had missed that I made two typos! :( –  May 22 '12 at 18:17
  • @Hurkyl Is there a standard name for $v_p$ ? – magma May 23 '12 at 08:39
  • @magma: $v_p(x)$ is the "valuation of $x$ at $p$". –  May 23 '12 at 08:49
  • @Hurkyl Thank you. In wikipedia I only found http://en.wikipedia.org/wiki/Valuation_%28algebra%29 . Does it relate to your valuation function? – magma May 23 '12 at 09:05
  • @magma Yes: it's the "P-adic valuation". But really, you should think of it in terms of "the exponent of $p$ in the factorization of $x$" -- one point of the notion of valuation is to generalize this simpler notion. –  May 23 '12 at 20:16