Is the formal power series ring a graded ring?

Question

Let $k$ be a field, let $k[[t]]$ be the formal power series ring over $k$ in one variable. Does there exist a $\mathbb{Z}$-grading on $k[[t]]$? In other words, does there exist a direct sum decomposition $k[[t]] \simeq \bigoplus_{n \in \mathbb{Z}} A_{n}$ as $k$-vector spaces respecting the algebra structure, i.e. such that $A_{n} \cdot A_{m} \subset A_{m+n}$?

Answer 1

The only grading on $k[[t]]$ is the trivial grading $A_0=k[[t]]$. To prove this, note that if $A$ is any $\mathbb{Z}$-graded domain and $a\in A$ is a unit, then $a$ must be homogeneous (if $a$ had two parts of different degrees, that would remain true after multiplying by any nonzero element, since $A$ is a domain). In the case of $k[[t]]$, this means that any element with nonzero constant term is homogeneous. Now every element of $k[[t]]$ is a sum of at most two elements with nonzero constant terms, so this would mean every element of $k[[t]]$ has at most two nonzero homogeneous parts. Now if some element $x\in k[[t]]$ were nonzero and homogeneous of nonzero degree, then $x+x^2+x^3$ would have three nonzero homogeneous parts. Thus all nonzero homogeneous elements have degree $0$, and so all of $k[[t]]$ has degree $0$.

More generally, a similar argument shows no local domain can have a nontrivial $\mathbb{Z}$-grading.