Suppose $M$ is what some authors call a rig, i.e. a "ring without negatives", that is a set endowed with two laws of composition $+$ and $\times$ with neutral elements $e_{+}$ and $e_{\times}$, such that $+$ and $\times$ are associative, commutative and $\times$ distributive over $+$, making $(M,+)$ and $(M,\times)$ commutative monoids.

I was told years ago that such a rig $M$ could be canonically embedded in a field $K_{M}$, just as we go from $\mathbb{N}$ to $\mathbb{Q}$.

My question is: is it always true that ${\rm Aut}(M)$ as the group of permutations of $M$ preserving its rig structure is isomorphic to ${\rm Aut}(K_{M})$? Does an hypothesis of maximality of $M$ make $K_{M}$ a separable closure of some underlying field? If not, which are the cases for which it holds true?

  • 2
    $\begingroup$ This is certainly false (the embeddability) if the addition isn't cancellative. $\endgroup$ Commented Apr 11, 2020 at 17:11
  • $\begingroup$ Without any assumptions on $M$ it is certainly not always the case that it can be embedded in a field. For one, $M$ has to have no zero divisors. I believe it is even possible that $(M,+)$ does not embed into a group, which obviously also excluded existence of $K_M$. $\endgroup$
    – Wojowu
    Commented Apr 11, 2020 at 17:11
  • $\begingroup$ The same formal properties as for the ring of integers hold in the instance of $M$ I work with. $\endgroup$ Commented Apr 11, 2020 at 17:18

1 Answer 1


$\Bbb{Q}[x]$ has less automorphisms than $\Bbb{Q}(x)$

Not a full answer since there are many more ways for the tempting correspondence to fail, but if the two monoids are cancellative it should be the rule : in most cases $Aut(K_M)$ is larger than $Aut(M)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .