# Does the inverse of an invertible homogeneous element need to be homogeneous?

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.

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.

• 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. Dec 11, 2022 at 16:39