Let $A$ be a commutative ring, with unit, and suppose $A$ is graded over a commutative monoid $M$. (In particular $1\in A_e$, where $e\in M$ is the identity.) If $x\in A$ is invertible and homogeneous, does $x^{-1}$ need to be homogeneous as well? I've been unable to find previous questions mentioning this on MSE. The thing is that the only example of a graded ring I've ever worked with is the polynomial ring in $n$ variables, so I don't know instances of graded rings to look for counterexamples.


2 Answers 2


The other answer only works when $M$ has the cancellation property.

Here is the general case: If $A=0$, the claim is trivial. So assume $A \neq 0$. Let $d = \deg(x)$. Write $x^{-1} = \sum_{m \in M} y_m$ with $y_m \in A_m$. Then $$1 = x \cdot x^{-1} = \sum_{m \in M} \underbrace{x \cdot y_m}_{\in A_{d+m}} = \sum_{n \in M}\, \underbrace{\sum_{m \in M,\, d+m=n} x \cdot y_m}_{\in A_n}.$$ The idea here is that, although the map $m \mapsto d+m$ might not be injective in general, we can still gather all preimages of a fixed $n$, in order to get a homogeneous decomposition.

It follows $\sum_{m \in M,\, d+m=n} x \cdot y_m = 0$ for $n \neq 0$ (which we do not need in the following) and $$\sum_{m \in M,\, d+m=0} x \cdot y_m = 1.$$ Since $1 \neq 0$, we have $d+m = 0$ for at least one $m \in M$, but then this is clearly uniquely determined with $m = -d$, and the equation simplifies to $$x \cdot y_{-d} = 1.$$ So $y_{-d}$ is homogeneous and inverse to $x$, hence it must be $x^{-1}$.

If $M$ and $A$ are not commutative, we can use a similar argument, but we also need to use $1 = x^{-1} x$ to conclude that $d$ is left and right invertible, hence has a unique inverse, and proceed as before.


Inverse element $x^{-1}$ is homogenous.

In fact we take homogenous decomposition $x^{-1}=y_1+\cdots+y_k$ where $y_i$ is homogenous element s.t. $\deg(y_i)\neq \deg (y_j)$.

Then $xy_1+\cdots+xy_k=xx^{-1}=1\in A_e$, but because of $\deg(xy_i)\neq \deg(xy_j)$ we get $k=1$. It means $x^{-1}$ is homogenous.

  • 4
    $\begingroup$ This argument is not completely correct. It becomes true when the grading monoid $M$ satisfies the cancellation property (ie $a+b=a+c$ implies $b=c$). This is the case in the vast majority of applications, though. $\endgroup$ Dec 11, 2022 at 16:39

You must log in to answer this question.

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