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 $\{i \in \mathbb Z| f_i \neq 0\}.$
Here is the statement I want to prove:
$\deg(fg) = \deg(f) + \deg(g).$
Here is my trial:
Let $f = f_0 + f_1 + \dots +f_k \neq 0$ where $f_i \in B_i, 0 \leq i \leq k, f_k \neq 0, \deg(f) = k$ and let $g = g_0 + g_1 + \dots +g_l \neq 0$ where $g_j \in B_j, 0 \leq j \leq l, g_l \neq 0, \deg(g) = l.$
Now, $\deg(fg) = \deg ((\sum_{i = 0}^k f_i)(\sum_{j = 0}^l g_j)) = \deg (\sum_{n=0}^{l+k} \sum_{i + j = n, i \geq 0, j \geq 0} f_i g_i)$ and $f_lg_k \neq 0 \implies \deg (fg) = l + k = \deg(f) + \deg(g)$ this is because we are in an integral domain.
If either $f$ or $g$ is $0,$ then $fg = 0$ and so $\deg(fg) = -\infty = \deg(f) + \deg(g) = -\infty$
But I think that my trial has 2 mistakes which are:
1- I took $i,j$ starting from $0.$ even though according to the support I defined above they should be in $\mathbb Z.$ How can I correct the proof to include $i,j$ as integers?
2- I think I used that I have $\mathbb Z$-graded integral domain $B$ in the wrong place. Could anyone help me in correcting this please?
Best Answer
You are correct about $(1)$, this does not cover all cases. I don't think $(2)$ is right though, you are using that $B$ is a domain and $\mathbb{Z}$-graded correctly. You just have to adapt the proof to the general case.
To do that, prove (or use) that $$ \left(\sum_{i=m}^ka_i\right) \cdot \left(\sum_{j=n}^l b_j\right) = \sum_{s = m+n }^{k+l}\left(\sum_{i+j=s} a_ib_j\right) $$
for any $m \leq k, n \leq l$.