Here is my attempt at proving each of the following, any insights would greatly be appreciated.
Given f(x) = $\sum_{i=0}^{n}a_ix_i ∈ Z[x]$ we define the content as the greatest common divisor of the coefficients: $$c(f) = c(a_0 + a_1x + · · · + a_nx^ n ) := \gcd(a_0, a_1, . . . , a_n) ∈ N.$$
(a) Let $d = \gcd(a_0, a_1, . . . , a_n)$ with $a_i = da_i'$ for all $i$. Prove that $\gcd(a_0', a_1',..., a_n') = 1$.
Pf: By definition, we have $d$ divides $a_i$ for all $i$. Suppose there exists some $e ∈ N$ such that $e$ divides $a_i$ for all $a_i$. Then $e$ divides $d$ implies $e \leq d$ and it follows from there that $e$ divides $a_i$, which implies $e$ divides $da_i'$ which then implies $e$ divides $a_i'$. So let $a_i' = ea_i''$. Then we can write $a_i = dea_i''$. Since $de$ is a common divisor of $a_0,...,a_n$, $de \leq d$, which implies $e$ $\leq$ $1$.
Thus it follows that $\gcd(a_0, a_1,... a_n)$ = $1$.
(b) If $f(x) ∈ Q[x]$ is monic, prove that there exists an integer $k ∈ N$ with $kf(x) ∈ Z[x]$ and $c(kf) = 1$. [Hint: Choose any $n ∈ N$ such that $nf(x) ∈ Z[x]$ and let $d = c(nf)$.]
Pf: Consider the monic polynomial $f(x) ∈ Q[X]$. Now let $n ∈ N$ be a natural number such that $nf(x) ∈ Z[X]$ with $d = c(nf)$. It follows that $nf(x)$ is of the form $\frac{na_0 + na_1 + ... + na_nx^n}{d}$ = $\frac{n}{d}f(x)$ where $n/d$ = $k ∈ N$ with $d$ $\leq$ $n$. Since we divided $nf(x)$ by $d$ to obtain $kf(x)$, we effectively divided $c(nf)$ by $d$ as well => $c(nf/d)$ = $c(uf)$ = $d/d$ = $1$.
(c) For all $f(x)$, $g(x)$ ∈ $Z[x]$ prove that $c(f) = c(g) = 1$ implies $c(fg) = 1$. [Hint: For any prime $p$ we know that $f$ and $g$ each have a coefficient not divisible by $p$. Show that $fg$ also has a coefficient not divisible by $p$.]
Pf: Consider some prime $p ∈ Z$ and suppose for contradiction that $p$ divides all of the coefficients of $f(x)g(x)$. It follows that $f(x)g(x) = 0$ in $Z/pZ[x]$, which implies that $p$ divides all of the coefficients of $f(x)$ or $p$ divides the coefficients of $g(x)$ since $f(x)$ or $g(x)$ is $0$ in the integral domain $Z/pZ[x]$. But this contradicts our assumption that $c(f) = c(g) = 1$, which finally implies that $c(fg) = 1$.
(d) If $f(x), g(x) ∈ Q[x]$ are monic with $f(x)g(x) ∈ Z[x]$, prove that $f(x), g(x) ∈ Z[x]$.
Pf: Let $f(x), g(x) ∈ Q(x)$ be monic with $f(x)g(x) ∈ Z[x]$. From part (b) we know that for $k$, $l$ ∈ $N$ we have $kf(x)$, $lg(x)$ ∈ $Z[X]$ with $c(kf)$ and $c(lg)$ = $1$. From part (c), it follows that $c(klfg)$ = 1 and hence it follows that $klfg$ = $1x^n +$ Lower Terms $= klx^n + kl$(Lower Terms) $∈Z$. But this implies that $kelc(kefg) = 1$ which finally implies that $kl$ = $1$, as desired.
Note: For $k ∈ Z$ and $f(x) ∈ Z[x]$ monic with $c(f) = d$ we have $c(kf) = kd$.
In (d), the idea is right, but there are some typos. I would write it this way:
In one hand, since $c(kf)=c(lg)=1$, by (c) you have $c(klfg)=1$.
On the other hand, since $fg\in \mathbb{Z}[x]$ and monic, we have $fg=a_0+a_1x+\dots +1x^n$ with $a_i\in\mathbb{Z}$ Then, $klfg=kla_0+kla_1x+\dots +klx^n$. So, $c(klfg)=\gcd(kla_0,kla_1,\dots, kl)=kl$.
We conclude that $kl=1$. (Hence $k=l=1$ , which imply that $f,g\in\mathbb{Z}[x]$.)
Edit: In part (a) it seems that there are still some tipos. You can write it this way. If $\gcd(a_0',a_1',\dots,a_n')=e$, then we could write $a_i=ea_i''$ with $a_i''\in\mathbb{Z}$. But then, $a_i=dea_i''$. So, $de$ would be a common divisor of the $a_i$'s. Hence, $de\leq d$ which implies $e\leq 1$, so $e=1$.