In a $\mathbb Z$-graded integral domain $B = \bigoplus_{i \in \mathbb Z} B_i$ my definition for the degree function is as follows:
Given $f \in B,$ define $\deg : B \rightarrow \mathbb Z \cup \{-\infty \}$ by $$\deg(f) = \mathrm{maxSupp}{(f)} \text{ if } f \neq 0 \text{ and } \deg(0) = -\infty $$
my definition for the $\mathrm{Supp}$ of $f$ is as follows $\{i \in \mathbb Z| f_i \neq 0\}.$
I am trying to prove that $\deg(f) = - \infty$ iff $f=0.$ The backward direction is clear by the definition of the $\deg$ function given to me. The forward direction, I am trying to prove it by contra positive i.e. Assume that $f \neq 0$ and I want to prove that $\deg(f) \neq - \infty$, does that follows from the definition of the support I have because my $i \in \mathbb Z$ or from what? Does the $\mathrm{maxSupp}{(f)}$ always $\geq 0$ if $f \neq 0$? if so, why?
Any help will be appreciated!
Since $B = \bigoplus_{i \in \mathbb Z}B_i$, an element $f \in B$ can be written uniquely as a sum
$$ f = \sum_{i \in \mathbb Z}f_i, \quad f_i\in B_i $$
with finitely many $f_i$ non zero. In paritcular, the support of $f$ is always a finite set. If $f \neq 0$, then not all $f_i$ are zero, hence $\mathbf{supp} f \neq \emptyset$ and thus $\max \mathbf{supp} f \in \mathbb Z$.
Note that this does not mean that $\deg$ is always non-negative. For example, consider the ring $B = \mathbb{Z}[t,t^{-1}]$ of Laurent polynomials with the grading defined as $B_n = \mathbb{Z}t^n$ for each integer $n$. In this ring, we have $\deg t^{-1} = -1$.