Integral Domain and Degree of Polynomials in $R[X]$ In an algebra book there is the following proposition: Given the polynomial ring $R[X]$ in one variable over the ring $R$, then for polynomials $f,g \in R[X]$ it follows: $$\deg(f+g)\leq \max(\deg f, \deg g) $$ $$ \deg (f\cdot g)\leq \deg f +\deg g.$$ Moreover, in the case $R$ is integral, it follows: $$\deg (f \cdot g)= \deg f +\deg g.$$ 
In the same page it is presented the divison with rest in the polynomial ring $R[X]:$ Let $g\in R[X]$, whose largest coefficient $a_d$ is a unity in $R.$ Then for every $f\in R[X]$ there exist unique polynomials $q,r \in R[X]$ with $$ f=qg + r, \,\,\,\,\deg r< d.$$
In the proof of the latter statement one reads: $\deg (qg)= g\deg q + \deg g$ for polynomials $q\in R[X],$ even if $R$ is not integral. Here is the point which makes the trouble. It looks like a contradiction because in the former statement it was said the contrary, i.e. the equality holds only if $R$ is integrable. Cany somebody shed some light on this, explaining what i am missing here. Thanks.
 A: The former statement did not say "the equality holds only if $R$ is integrable".
What the former statement did say is "the equality holds if $R$ is integrable".
Also, "integrable" is very unusual terminology; I think you are referring to what we usually call "integral".

Assuming that you do mean "integral", then the following is true:

The identity $\deg(fg) =\deg(f) + \deg(g)$ holds for every every pair of polynomials $f,g$ only if $R$ is integral.

So, if $R$ is not integrable the identity has to fail for some polynomials. But only for some. It can hold for other polynomials.
For example, in any polynomial ring,
$$\deg(x^5) = 5 = 3 + 2 = \deg(x^3) + \deg(x^2)$$

The proof of this statement is simple, simply by looking at the leading coefficients:
$$ (f_d x^d + f_{d-1} x^{d-1} + \cdots) \cdot (g_e x^e + g_{e-1} x^{e-1} + \cdots ) = f_d g_e x^{d+e} + \cdots $$
Assuming $f_d \neq 0$ and $g_e \neq 0$, we see that
$$\deg(fg) = d + e = \deg(f) + \deg(g)$$
if and only if $f_d g_e \neq 0$. Thus any failure of this identity implies the existence of a pair of zero divisors, and any pair of zero divisors allows you to construct counterexamples.
