Understanding a proof of "If $R$ is an integral domain, then $R[x]$ is an integral domain as well". Theorem: If $R$ is an integral domain, then $R[x]$ is an integral domain as well.
Proof:
Let $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ and $g(x)=b_0+b_1x+b_2x^2+\cdots+b_mx^m$ be two elements of $R[x]$. Let $0\leq i\leq n$ and $0\leq j\leq m$ such that $i$ & $j$ are the smalles numbers such that $a_i\ne 0$ & $b_j\ne 0$.
$f(x)\cdot g(x)=\sum_{i=0}^{n+m}{c_ix^i}.$
$c_{i+j}=\overbrace{a_{i+j}b_0+a_{i+j-1}b_1+\cdots +a_{i+1}b_{j-1}}^{=0} +\overbrace{a_ib_j}^{\ne 0}+\overbrace{a_{i-1}b_{j+1}+\cdots+a_0b_{i+j}}^{=0}$
$=a_i+b_j\ne 0\Longrightarrow f(x)g(x)\ne 0$


I don't know why $a_{i+j}b_0+a_{i+j-1}b_1+\cdots +a_{i+1}b_{j-1}=0$ and $a_{i-1}b_{j+1}+\cdots+a_0b_{i+j}=0$.

I see where it's coming from, but that proof is too complicated in my opinion.  I would simply say: consider nonzero polynomials $f, g \in R[x]$ such that $\deg(f) = n$ and $\deg(g) = m$.  That is, $\displaystyle f(x) = \sum_{k=0}^n a_kx^k$ and $\displaystyle g(x) = \sum_{k=0}^m b_kx^k$ where the coefficients $a_j$ and $b_j$ come from $R$.
We know the leading coefficients on $f$ and $g$, namely $a_n$ and $b_m$ respectively, are nonzero.  Now consider $f(x) \cdot g(x)$.  The leading term of this product, $a_nb_mx^{n+m}$, cannot be zero because $a_n b_m \neq 0$ due to $a_n$ and $b_n$ being nonzero elements of an integral domain.
$\Longrightarrow$ $f(x) \cdot g(x) \neq 0$ for all nonzero $f, g \in R[x]$.  Thus, $R[x]$ is a ring with no zero divisors, so it is an integral domain.
