Now is a very good time for a quick foray into the ideal-theoretic version of Sun-Ze (better known as the Chinese Remainder Theorem). Let $R$ be a commutative ring with $1$ and $I,J\triangleleft R$ coprime ideals, i.e. ideals such that $I+J=R$. Then
$$\frac{R}{I\cap J}\cong\frac{R}{I}\times\frac{R}{J}.$$
First let's recover the usual understanding of SZ from this statement, then we'll prove it. Thanks to Bezout's identity, $(n)+(m)={\bf Z}$ iff $\gcd(n,m)=1$, so the hypothesis is clearly analogous. Plus we have $(n)\cap(m)=({\rm lcm}(m,n))$. As $nm=\gcd(n,m){\rm lcm}(n,m)$, if $n,m$ are coprime then compute the intersection $(n)\cap(m)=(nm)$. Thus we have ${\bf Z}/(nm)\cong{\bf Z}/(n)\times{\bf Z}/(m)$. Clearly induction and the fundamental theorem of arithmetic (unique factorization) give the general algebraic version of SZ, the decomposition ${\bf Z}/\prod p_i^{e_i}{\bf Z}\cong\prod{\bf Z}/p_i^{e_i}{\bf Z}$.
(How this algebraic version of SZ relates to the elementary-number-theoretic version involving existence and uniqueness of solutions to systems of congruences I will not cover.)
Without coprimality, there are counterexamples though. For instance, if $p\in\bf Z$ is prime, then the finite rings ${\bf Z}/p^2{\bf Z}$ and ${\bf F}_p\times{\bf F}_p$ (where ${\bf F}_p:={\bf Z}/p{\bf Z}$) are not isomorphic, in particular not even as additive groups (the product is not a cyclic group under addition).
Now here's the proof. Define the map $R\to R/I\times R/J$ by $r\mapsto (r+I,r+J)$. The kernel of this map is clearly $I\cap J$. It suffices to prove this map is surjective in order to establish the claim. We know that $1=i+j$ for some $i\in I$, $j\in J$ since $I+J=R$, and so we further know that $1=i$ mod $J$ and $1=j$ mod $I$, so $i\mapsto(I,1+J)$ and $j\mapsto(1+I,J)$, but these latter two elements generate all of $R/I\times R/J$ as an $R$-module so the image must be the whole codomain.
Now let's work with ${\cal O}={\bf Z}[i]$, the ring of integers of ${\bf Q}(i)$, aka the Gaussian integers. Here you have found that $(2)=(1+i)(1-i)=(1+i)^2$ (since $1-i=-i(1+i)$ and $-i$ is a unit), that the ideal $(3)$ is prime, and that $(5)=(1+2i)(1-2i)$. Furthermore $(1+i)$ is obviously not coprime to itself, while $(1+2i),(1-2i)$ are coprime since $1=i(1+2i)+(1+i)(1-2i)$ is contained in $(1+2i)+(1-2i)$. Alternatively, $(1+2i)$ is prime and so is $(1-2i)$ but they are not equal so they are coprime. Anyway, you have
- ${\bf Z}[i]/(3)$ is a field and
- ${\bf Z}[i]/(5)\cong{\bf Z}[i]/(1+2i)\times{\bf Z}[i]/(1-2i)$ is a product of fields.
Go ahead and count the number of elements to see which fields they are. However, ${\bf Z}[i]/(2)={\bf Z}[i]/(1+i)^2$ is not a field or product of fields, although the fact that its characteristic is prime (two) may throw one off the chase. In ${\bf Z}[i]/(1+i)^2$, the element $1+i$ is nilpotent. Since this ring has order four, it is not difficult to check that it is isomorphic to ${\bf F}_2[\varepsilon]/(\varepsilon^2)$, which is not a product of fields since $\epsilon\leftrightarrow 1+i$ is nilpotent and products of fields contain no nonzero nilpotents.
Notice the first construction you gave can also be described as the semigroup ring $R[\Bbb N ^k]$, and makes a polynomial ring with $k$ commuting indeterminates. Moreover, the coefficients commute with indeterminates, but not always with each other.
Of course, you can let go of both these properties and generate a ring of "generalized polynomials," which look like finite sums of finite products of coefficients interleaved with indeterminates.
I'm not sure that these rings have significant usage (definitely nothing compared to polynomials over commutative rings); however, its key advantage is that evaluation maps work properly.
For example, in the ordinary polynomial ring $\Bbb H[x]$ over quaternions, $xi-ix=0$, since coefficients commute with indeterminates. But this equation doesn't remain true if you try to evaluate it at quaternion values: $j$ would yield $0\neq 2ij$. But in the ring of generalized polynomials, evaluation works just fine.
Naturally, things are messier than normal for solutions to such generalized polynomials. For example, $ix-xi +1=0$ doesn't have any solutions over $\Bbb H$, but $x^2+1=0$ has uncountably many solutions.
Best Answer
This should answer both of your questions.
Characteristic $p > 0$ means (informally, see the post by Thomas Andrews for a formal definition) that the multiple $$ p \cdot 1 = \underbrace{1+ \dots + 1}_{\text{$p$ times}} $$ is zero. It follows $p \cdot a = 0$ for all $a$ in the ring. This is because $$p \cdot a = \underbrace{a + \dots + a}_{\text{$p$ times}} = (\underbrace{1+ \dots + 1}_{\text{$p$ times}}) a = (p \cdot 1) a = 0 a = 0.$$
Also, if $p$ divides the integer $n$, so that $n = m p$ for some $m$, then $$n \cdot a = (m p) \cdot a = m \cdot (p \cdot a) = m \cdot 0 = 0.$$
Now the binomial theorem in a commutative ring is $$ (x + y)^{n} = \sum_{i=0}^{n} \binom{n}{i} \cdot x^{i} y^{n-i}, $$ where note that the binomal coefficients act as multiples.