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 $a_{i-1}b_{j+1}+\cdots+a_0b_{i+j}=0$.
 A: 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.
A: You probably mean $a_ib_j$, not the sum. Anyways, this looks simpler to me: we may assume the polynomials are not constants. Suppose that the leading coefficient of $f$ is $a$ and the leading coefficient of $g$ is $b$, both nonzero. Then the leading coefficient of $fg$ is $ab$, and this is not zero.
A: Hint $\ \ c_{i+j}=\overbrace{\underbrace{a_{i+j}}_{\large \color{#c00}{=\,0}}b_0+\underbrace{a_{i+j-1}}_{\large \color{#c00}{=\,0}}b_1+\cdots +\underbrace{a_{i+1}}_{\large \color{#c00}{=\,0}}b_{j-1}}^{\large \color{#c00}{=\,0}} +\overbrace{\underbrace{{a_i}_{\phantom f}\!\!\!}_{\color{#c00}{\ne\, 0}}\underbrace{b_j}_{\color{#0a0}{\ne\, 0}}}^{\large \color{#c00}\ne\, \color{#0a0}0}+\overbrace{a_{i-1}\underbrace{b_{j+1}}_{\large \color{#0a0}{=\,0}}+\cdots+a_0 \underbrace{b_{i+j}}_{\large \color{#0a0}{=\,0}}}^{\large\color{#0a0}{=\,0}}$
Simpler $ $ omit leading zeros, so the lead coef of the product is the product of the nonzero lead coefs  so is nonzero, by the coef ring is a domain, i.e. $\,\ell (f),\ell(g)\ne 0\,\Rightarrow\, \ell(fg) = \ell(f)\ell(g)\ne 0.$
Remark $ $ Compare a proof of Gauss's Lemma, that a prime $\,p\in R\,$ persists as prime in $\,R[x]$
It is instructive to give a direct proof of $\,p\nmid A,B\,\Rightarrow\,p\nmid AB.\,$ Since $\,p\nmid A,B\,$ when reduced mod $p$ both have lead coefs $\,\color{#0a0}{a,b\not\equiv 0}\,$ so $AB$ has lead coef $\,\color{#c00}{ab\not\equiv 0}\,$ (by $p$ prime), hence $AB\not\equiv 0,\,$ i.e.
$\qquad\qquad{\rm mod}\ p\!: \ \ \ \begin{eqnarray}
&&\ 0\ \not\equiv\ A\ \equiv\, \color{#0a0}a\, x^j\! + \:\cdots,\quad\ \ \ \color{#0a0}{a\not\equiv 0}\\  
&&\ 0\ \not\equiv\ B\ \equiv\, \color{#0a0}b\, x^k\! + \:\cdots,\quad\ \ \ \color{#0a0}{b\not\equiv 0}\\ \Rightarrow\,\ &&0 \not\equiv AB \equiv \color{#c00}{ab}\ x^{j+k}\! + \:\cdots,\, \color{#c00}{ab\not\equiv 0}\end{eqnarray}$
i.e. primes $\:p\in R\:$ remain prime in $\:R[x]\:$  because the prime divisor property $\,\color{#0a0}{p\nmid a,b}\,\Rightarrow\ \color{#c00}{p\nmid ab}\,$ persists when multiplying leading coefficients. This is one form of Gauss's Lemma.
This is an elementwise view of the common structural proof that factors out $p$ to reduce to the trivial domain case (i.e. it is simply the common proof of $\,D\,$ domain $\Rightarrow\,D[x]\,$ domain, in the special case that $\,D \cong R/p \cong R\bmod p,\,$ for $p$ prime).
