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Recently I came upon the following result:

Result 1. Let $K$ be a $\mathbb{Z}$-graded field. Then either $K$ is trivially graded (i.e. $K_k=0$ for $k\in\mathbb{Z}\setminus\{0\}$ with $K_0$ a field or we have $K\cong K_0[x,x^{-1}]$ for $x$ an element of positive even degree.

However, I've found conflicting statements about this: Lurie's 24th set of lecture notes on chromatic homotopy theory claims the above result as is, and this MSE answer offers a proof, while Remark 1.3.10 in Nastasescu–Oystaeyen's Methods of Graded Rings and Exercise 1.1 in these lecture notes claim that $K$ must necessarily be trivially graded with $K_0$ a field, not allowing for the case $K\cong K_0[x,x^{-1}]$, i.e., they claim the following result:

Result 2. Let $K$ be a $\mathbb{Z}$-graded field. Then $K$ is trivially graded with $K_0$ a field.

To be precise (and since I suspect this might be an issue with differing definitions), let me precisely define $\mathbb{Z}$-graded fields below:

Definition 3. A $\mathbb{Z}$-graded field $K$ consists of

  • The Collection of Abelian Groups. A collection $\{K_k\}_{k\in\mathbb{Z}}$ of abelian groups;
  • The Multiplication. A collection of maps $$\mu^{k,\ell}_K\colon K_k\otimes_{\mathbb{Z}}K_{\ell}\to K_{k+\ell}$$ indexed by $(k,\ell)\in\mathbb{Z}\times\mathbb{Z}$, whose action we denote by $(a,b)\mapsto ab$;
  • The Unit. A map $$\eta_K\colon\mathbb{Z}\to K_0$$ picking an element $1_K$ of $K_0$.

satisfying the following conditions:

  1. Associativity. We have $(ab)c=a(bc)$ for all $a\in K_i$, $b\in K_j$, and $c\in K_k$.
  2. Unitality. We have $1_Ka=a1_K=a$ for each $k\in\mathbb{Z}$ and each $a\in K_k$.
  3. Commutativity. We have $ab=ba$ for all $a\in K_k$ and all $b\in K_\ell$.
  4. Invertibility of Nonzero Homogeneous Elements. For each $k\in\mathbb{Z}$ and each nonzero $a\in K_k$, there exists an element $a^{-1}$ of $K_{-k}$ such that $aa^{-1}=a^{-1}a=1_K$.

Question I. Which result is correct? Result I or Result II?


More generally, I'm also interested in what is the situation for graded semifields, graded skew fields, and graded skew semifields.

  • Semifields are defined in almost the same way as fields, except that we allow its underlying additive abelian group of it to be just a commutative monoid, i.e. addition can be non-invertible. A famous example is the tropical semiring $(\mathbb{R}\cup\{-\infty\},\max,+,-\infty,0)$.
  • Skew fields are also defined similarly to fields, the exception now being that we allow multiplication to be noncommutative. The quaternions are a famous example.
  • Lastly, skew semifields are the common generalisation of the above notions: addition can be non-invertible and multiplication noncommutative.

Finally, one defines graded semifields, graded skew fields, and graded skew semifields in a similar way to Definition 3, except that (in order): 1) We ask the $K_k$'s to be just commutative monoids and replace $\mathbb{Z}$ by $\mathbb{N}$ and $\otimes_{\mathbb{Z}}$ by $\otimes_{\mathbb{N}}$ everywhere except in the grading 2) We remove condition 3 (commutativity) and 3) we do both of the things in 1) and 2).

Question II. Do we have results classifying

  1. Graded semifields;
  2. Graded skew fields;
  3. Graded skew semifields;
  4. Graded semifields (resp. skew fields, skew semifields) which are graded commutative, i.e. satisfy $ab=(-1)^{\mathrm{deg}(a)\mathrm{deg}(b)}ba$;

in a similar vein to Results I/II?

Emily
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1 Answers1

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Basically there is some confusion about definitions. First, note that what you wrote is not the definition of a graded field, but of a (commutative) graded ring. Note also that you can replace $\mathbb{Z}$ by any commutative group (actually monoid) $\Gamma$ for the grading.

Then there is the elementary result: in a $\Gamma$-graded ring, if $\Gamma$ is torsion-free then all invertible elements must be homogeneous. It is easy to deduce that, still when $\Gamma$ is torsion-free, if a $\Gamma$-graded ring is a field (not as a graded ring, but simply as a ring, so every non-zero element is invertible), then it is trivially graded.

On the other hand, there is a notion of a graded field: it is a (commutative) graded ring, such that every non-zero homogeneous element is invertible. Then of course $K=K_0[x,x^{-1}]$ is a $\mathbb{Z}$-graded field. It is actually isomorphic to the group algebra $K_0[\mathbb{Z}]$, and generally speaking, for any $\Gamma$ the group algebra $K_0[\Gamma]$ is a $\Gamma$-graded field. In general the structure of $\Gamma$-graded fields can be rather rich, but in the case of $\mathbb{Z}$, the fact that it is projective as an abelian group makes the classification simple.

I don't know much about the general structure of graded semirings, but if you want to learn about graded skew field (also called graded division algebras), then I recommand to you the beginning of "Value Functions on Simple Algebras, and Associated Graded Rings" by Tignol and Wadsworth. Note that they will be a generalization of what I called a graded field. It is still true that a (non-commutative) graded ring over a torsion-free group which is a skew field is trivially graded. You again have to require the non-zero homogeneous elements to be invertible, not all non-zero elements.

Captain Lama
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  • Aaagh I forgot to include the field condition indeed! (I've edited the question now.) Thank you so much for your answer! – Emily Mar 24 '23 at 08:52
  • Just to confirm, in the second case, where only the homogeneous elements are required to be invertible, is it indeed the case that the only $\mathbb{Z}$-graded fields are of the form $K_0$ and $K_0[x,x^{-1}]$? – Emily Mar 24 '23 at 08:52
  • Also thank you so much for the pointer to Tignol–Wadsworth, it seems there's lots of great stuff there! – Emily Mar 24 '23 at 08:53
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    Yes those are the only ones, but in $K_0[x,x^{-1}]$, $x$ can have any degree, which changes the grading. (Basically the non-zero components are those with degree in $n\mathbb{Z}$ for some $n$, which you can take as the degree of $x$.) – Captain Lama Mar 24 '23 at 19:51
  • Thank you for confirming it! I was thinking about this last night and figured out why Lurie said even degree: in that context the $\mathbb{Z}$-graded ring comes with the graded-commutativity condition I mentioned in the question body, and then we have $xx^{-1}=(-1)^{\mathrm{deg}(x)\mathrm{deg}(x^{-1})}x^{-1}x$, which forces $\deg(x)$ to be even. As you said though in general $\deg(x)$ can be odd too :) – Emily Mar 24 '23 at 19:55