Given a commutative unital ring $R$ and an ideal $I$ of $R,$ one can define a new ring called the associated graded ring of $R$ with respect to $I$ $$\operatorname{gr}_I(R) \stackrel{\text{def}}{=} \frac R I \oplus \frac I {I^2} \oplus \cdots = \bigoplus_{n = 0}^\infty \frac{I^n}{I^{n + 1}}.$$ We define the multiplication in $\operatorname{gr}_I(R)$ as follows.
Given any homogeneous elements $\bar r$ and $\bar s$ of $\operatorname{gr}_I(R),$ we have that $\bar r$ is in $I^m / I^{m + 1}$ and $\bar s$ is in $I^n / I^{n + 1}$ for some non-negative integers $m$ and $n.$ Let $r$ and $s$ denote the representatives of $\bar r$ and $\bar s$ in $I^m$ and $I^n,$ respectively. We define $rs$ to be the equivalence class of $\bar r \bar s$ in $I^{m + n} / I^{m + n + 1}.$ Observe that this is well-defined modulo $I^{m + n + 1}.$ We may multiply inhomogeneous elements of $\operatorname{gr}_I(R)$ by using the distributive property.
Generally, there is not a ring homomorphism $R \to \operatorname{gr}_I(R);$ however, there is always a set-theoretic map $-^* : R \to \operatorname{gr}_I(R)$ defined as follows. Given an element $r$ of $R,$ consider the quantity $\operatorname{ord}(r) = \sup \{n \,|\, r \in I^n \}.$ We define $r^*$ to be the equivalence class of $r$ in $I^{\operatorname{ord}(r) + 1}$ whenever $\operatorname{ord}(r)$ is finite and $r^* = 0$ otherwise. We refer to $r^*$ as the initial form of $r.$ Every homogeneous element of $\operatorname{gr}_I(R)$ is of the form $r^*$ for some element $r$ in $R.$
Chapter 5, Section 1 of Eisenbud's Commutative Algebra discusses the associated graded ring of $R$ in some detail; however, I am not comfortable enough with the material to solve the exercises on my own. Particularly, I would like to solve Exercise 5.1, which states that either $r^* + s^* = (r + s)^*$ or $r^* + s^* = 0$ and either $r^* s^* = (rs)^*$ or $r^* s^* = 0.$
Ultimately, I would like to understand the following questions.
1.) What is the addition operation in $\operatorname{gr}_I(R)?$
2.) Why does the initial form map fail to be a ring homomorphism?
3.) Under what conditions is it guaranteed that $r^* + s^* = (r + s)^*?$
4.) If $\operatorname{gr}_I(R)$ is a domain, then by Eisenbud's exercise, we have that $r^* s^* = (rs)^*.$ Under what other conditions is it guaranteed that $r^* s^* = (rs)^*?$
Thanks in advance for your time and consideration.
Note that each $I^n/I^{n+1}$ is an abelian group (in fact an $R$-module). So $\bigoplus_{n \ge 0} I^n/I^{n+1}$ inherits an abelian group structure from these. In particular, if $a=\sum^{\infty}_{n=0} r_n$ and $b=\sum^{\infty}_{n=0} r'_n$ with $r_n,r'_n \in I^n/I^{n+1}$ and only finitely many $r_n$'s and $r'_n$'s nonzero, then $a+b=\sum^{\infty}_{n=0} (r_n+r_n')$.
The initial form map is not additive even in the nicest of circumstances. For example, take $R=k[\![x]\!]$ (or $R=k[x]$), and $I=(x)$. Then $\operatorname{gr}_I(R) \cong k[x]$. Let $a=x+x^2$ and $b=-x+x^3$. Then $a^*=x$ and $b^*=-x$. But $(a+b)^*=x^2+x^3$ while $a^*+b^*=0$. You should be careful, however, the exercise you mention asks you to prove something false. Indeed, we could also take $a=x$ and $b=x^2$. Of course, $a^*=x$ and $b^*=x^2$, but $(a+b)^*=(x+x^2)^*=x$. Perhaps the best amendment to the exercise is to add the assumption that $r^*$ and $s^*$ have the same degree.
In fact, additivity holds for all $r,s \in R$ if and only if $I=I^2$. Indeed, if $x \in I-I^2$, then we set $a=-1$ and $b=1+x$. We have $a^*=-1$ and $b^*=1$ so $a^*+b^*=0$ while $(a+b)^*=x^* \ne 0$. The case where $I=I^2$ is easy to verify. Of course, in this case, the associated graded ring is just $R/I$ and the initial form map the natural surjection $R \to R/I$, so it's not super interesting.
The takeaway from this is that the initial form map is never meaningfully a homomorphism.