Fix a field $k$. Let $\mathbb{G}_m$ be the multiplicative affine group scheme over $k$. A $k-$character $\chi$ of $\mathbb{G}_m$ is an endomorphism of affine group schemes $\mathbb{G}_m \to \mathbb{G}_m$, or, equivalently, the data of a group homomorphisms $\chi_R: R^\times \to R^\times$ for each $k-$algebra $R$ such that for each map of $k-$algebras $f: R \to S$, we have $f \circ \chi_R = \chi_S \circ f$.
Another approach, which uses the Hopf algebra representing $\mathbb{G}_m$, is to look for group-like elements $P \in k[\mathbb{G}_m]:=k[X,X^{-1}]$. These are invertible elements such that $\Delta(P)=P\otimes P$. Since $\Delta$ is a map of $k-$algebras satisfying $\Delta(X)=X \otimes X$, if we write $$P = \sum_{i=-m}^n a_iX^i,$$ the above condition translates to $$\Delta(P) = \sum_{i=-m}^n a_iX^i\otimes X^i \overset{!}{=} \sum_{i,j=-m}^n a_ia_jX^i\otimes X^j = P \otimes P.$$ By comparing coefficients, we see immediately that $P=X^i$ for some $i \in \mathbb{Z}$, which tells us that $\chi$ is exponentiation to some integer power. Hence, the characters of $\mathbb{G}_m$ are in bijection with $\mathbb{Z}$.
My question is: is this fact true also if $k$ is not a field? In particular: my proof uses the fact that $k$ is a field when comparing coefficients. In a general ring $k$, there can be zerodivisors and idempotents that are not $0$ or $1$.