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Usually, a graded ring $R$ over a monoid $M$ is defined as a ring which decomposes as an inner direct sum $R = \bigoplus_{m ∈ M} R_m$ of abelian subgroups such that for all $m, m' ∈ M$, $R_mR_{m'} \subseteq R_{m·m'}$.

I dislike this definition because it demands the existence of a decomposition. To my understanding, this definition cannot be formalised for a common first-order theory for two reasons:

  1. The decomposition involves subsets of $R$ (rather than elements of $R$).
  2. There are two structures involved, the ring $R$ and the monoid $M$ (even for $M = ℤ$).

Since Wikipedia also doesn’t mention the theory of graded rings on its list of first-order theories, it seems like the theory of graded rings is not a first-order theory. (Or is it?)

However, there is many-sorted logic which allows for instance a first-order theory for vector spaces; there is also second-order logic for dealing with quantification over subsets rather than elements.


This gives me hope: I would love to see a definition for graded rings as structures of (many-sorted) signature $(+_R,×_R,0_R,1_R,·_M,1_M,\operatorname{deg})$ (or something like that) such that a certain set of first-order or second-order axioms holds.

  • Is there such a first-order definition?
  • If not: Is there such a second-order definition?
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    $\begingroup$ I don't know of a natural presentation of graded rings as the models of a (multisorted) first-order theory. However, it's fairly straightforward to give a multisorted first-order presentation of filtered rings (a closely related notion). $\endgroup$ – Alex Kruckman Mar 3 '17 at 20:45
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    $\begingroup$ Also, the fact that Wikipedia doesn't mention a particular class of algebraic objects on its list is not good evidence that that class isn't captured by a first-order theory! Lists on Wikipedia are finite, and there are infinitely many first-order theories... $\endgroup$ – Alex Kruckman Mar 3 '17 at 20:47
  • $\begingroup$ I don't think it is possible to express this in a first-order way, but not quite for the reasons you have described. To be precise, no matter how you put the gradation (or even just filtering) into the language, (if the monoid is infinite) in a saturated model you will have a) elements which are not finite sums of homogeneous elements and b) nonzero elements with all coordinates zero. $\endgroup$ – tomasz Mar 6 '17 at 14:18
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The presentation of $M$-graded rings in Qiaochu's answer might or might not satisfy you, depending on what you want to do with your first-order theory. The $M$-indexed sequence of abelian groups, given together with the multiplication maps, captures all the information about an $M$-graded ring, in the sense that you can recover the ring from this data. But as tomasz points out in the comments, you can't actually quantify over elements of the ring (unless $M$ is finite), since you only have access to the homogeneous elements, and an arbitrary ring element is a finite sum of homogeneous elements.

But if you're satisfied with the presentation in Qiaochu's answer, you can easily adjust it to be independent of the choice of monoid. Consider the language with two sorts, $A$ and $M$, an element $e\in M$, a binary function $\cdot:M^2\to M$, a unary relation $0\subseteq A$, a ternary relation $+\subseteq A^3$, a binary function $\times\colon A^2 \to A$, and a unary function $\text{deg}\colon A\to M$.

We make a graded ring into a structure in this language by interpreting the $M$ sort as the monoid (and interpreting $e$ and $\cdot$ as identity and multiplication in the monoid), and interpreting the sort $A$ as the disjoint union of all the abelian groups $(R_m)_{m\in M}$ of homogeneous elements. Then the relation $0$ picks out exaclty one element of each $R_m$, the relation $+$ is the graph of addition (we have to make it a ternary relation, since you can only add two homogeneous elements of the same degree), $\times$ is multiplication, and $\text{deg}$ sends $a\in R_m$ to $m$.

I'll leave it to you to write down the first-order axioms.

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  • $\begingroup$ There's a separate $0$ in each group, just like in your answer. $\endgroup$ – Alex Kruckman Mar 7 '17 at 4:49
  • $\begingroup$ Ah, sorry, I didn't realize $A$ was supposed to be the disjoint union. $\endgroup$ – Qiaochu Yuan Mar 7 '17 at 4:49
  • $\begingroup$ Yes, this is more or less what I was hoping for, thank you! It’s a bit disillusioning that $+$ has to be ternary and $0$ has to be a unary relation, but I nevertheless like the description! $\endgroup$ – k.stm Mar 7 '17 at 11:15
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The definition does not merely demand existence; the decomposition is a structure, not a property, and morphisms of graded rings have to respect that structure.

In any case, here's an alternative definition. An $M$-graded ring is a sequence $R_m, m \in M$ of abelian groups together with a sequence of maps

$$\times_{m, n} : R_m \times R_n \to R_{m+n}$$

satisfying all of the obvious axioms. This is a first-order $|M|$-sorted theory.

A more categorical way of putting it is the following. Given a monoid $M$ there is a category of $M$-graded abelian groups, which are just sequences $A_m, m \in M$ of abelian groups indexed by $M$. This category has a monoidal structure given by Day convolution, and $M$-graded rings are precisely monoids in this category with respect to this monoidal structure.

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  • $\begingroup$ Thanks. Do you know of a (similar) definition independent of the grading monoid $M$? $\endgroup$ – k.stm Mar 4 '17 at 11:18
  • $\begingroup$ @k.stm: I don't understand what you mean by that. Can you clarify? $\endgroup$ – Qiaochu Yuan Mar 4 '17 at 19:39
  • $\begingroup$ Is it possible to give a fix a set of sorts $S$ and define an $S$-sorted theory of graded rings over monoids (instead of fixing a monoid $M$, setting $S = \lvert M \rvert$ and define $M$-graded rings)? Or rather: Do you know of such a definition? $\endgroup$ – k.stm Mar 5 '17 at 8:41
  • $\begingroup$ Clearly you can contain all the information about the whole ring in the structure you have described, but I don't see how you would fit the ring itself (as in, the whole direct sum) into it. $\endgroup$ – tomasz Mar 6 '17 at 14:14
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    $\begingroup$ @QiaochuYuan: if all you care about is the equivalence of categories, then I guess you are right, but from my point of view, usually in this context we want to see the structure as an expansion of the original (in this case, ring) structure. Besides, using multi-sorted language in this way is kind of cheating. ;-) $\endgroup$ – tomasz Mar 7 '17 at 0:33

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