primitive polynomials and their factorisation A polynomial with integer coefficients is called
primitive if its coefficients are relatively prime. For
example, $$3{x^2} + 7x + 9$$ is primitive while $$10{x^2} + 5x + 15$$
is not.
(a) Prove that the product of two primitive polynomials is primitive.
(b) Use this to prove Gauss's Lemma: If a polynomial with integer coefficients can be factored
into polynomials with rational coefficients, it
can also be factored into primitive polynomials
with integer coefficients
 A: For the first part let $f$ and $g$ be primitive polynomials in $\mathbb{Z}$ and assume their product isn't, i.e. there exists prime number $p$ that divides all coefficients of $f\cdot g$. Now define the following mapping:
$$\phi = \begin{cases} \mathbb{Z}[x] \to \mathbb{Z}_p[x] \\ a_nx^n + a_{n-1}x^{n-1} + ... a_0 \to [a_n]x^n + [a_{n-1}]x^{n-1} + ... + [a_0] \end{cases}$$
Prove that it's epimorphism and as $\mathbb{Z}_p[x]$ is an integral domain we have that $\phi(f)\phi(g) = \phi(f \cdot g) =  0$ implies that one of the polynomials isn't primitive.

For the second part let $f,g \in \mathbb{Q}[x]$. Let $b$ be the LCM of the denominators and $a$ be the GCD of the nominators of $f$ and similarly define $c,d$ for $g$. Now $f = \frac{a}{b} \cdot F$ and $g = \frac{c}{d} \cdot G$, where $F,G \in \mathbb{Z}[x]$ and they are primitive.
Then we have $h = \frac{a}{b} \cdot F \cdot \frac{c}{d} \cdot G \implies bd \cdot h = ac \cdot F \cdot G$. But $F\cdot G$ is primitive, as well as $h$, so we must have $ac = bd$. So: $$h = f \cdot g = F \cdot G$$

Here's a more elementary solution to the first part. Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ and $g(x) = b_tx^t + b_{t-1}x^{t-1} + ... + b_0$. Now assume that $p \mid f \cdot g$ for some prime number $p$. As both are primitive there exist coefficients $a_i$ and $b_j$ that aren't divisible by $p$. Let $a_i$ and $b_j$ be the coefficients such that $p \not \mid a_i, b_j$, but $p \mid a_s, b_u$ for $s>i$ and $u>j$. Now let's consider the coefficient in front of $x^{i+j}$ in $f \cdot g$. We have:
$$ c_{i+j} = a_0b_{i+j} + a_1b_{i+j-1} + ... + a_ib_j + ... + a_{i+j-1}b_1 + a_ib_0$$
Using the previously mentioned things we have that $0 \equiv c_{i+j} \equiv a_ib_j \pmod p$, an ovcious contradiction. Hence the claim is proven by contradiction.
A: Since you tagged it elementary number theory and not abstract algebra, here is an approach that does not require any knowledge of ring theory.
Write the polynomials as $\,A,B.\,$ Suppose $\,p\nmid A,B.\,$ So when reduced mod $\,p,\,$ both are $\not\equiv 0$ hence they have leading coef's $\,a,b\not\equiv 0\pmod p.\,$ By $\,p\,$ prime:  $\,p\nmid a,b\,\Rightarrow\, \color{#c00}{p\nmid ab},\,$ so 
$$ \begin{eqnarray} A &\equiv&\ \  a\ x^j + \:\cdots,\quad\ \ a\not\equiv 0\pmod p\\   B &\equiv&\ \   b\ x^k + \:\cdots,\quad\ \ b\not\equiv 0\pmod p\\ \Rightarrow\ AB &\equiv& \color{#c00}{ab}\ x^{j+k}\! + \:\cdots,\ \color{#c00}{ab\not\equiv 0}\pmod p\end{eqnarray}$$
So $\,p\nmid AB.\,$ Said contrapositively, $\ p\mid AB\,\Rightarrow\,p\mid A\,$ or $\,p\mid B,\, $ i.e. a prime $\,p\in \Bbb Z\,$ remains prime in $\,\Bbb Z[x] = $ polynomials with integer coefficients. This preservation of the prime divisor property in the leading coef's will become clearer when one studies ring theory, where it follows from general structural results [$\,p$ prime $\Rightarrow \Bbb Z_p$ domain $\Rightarrow$ $\,\Bbb Z_p[x]$ domain].
For part 2, factor out fractional contents, then clear denominators and apply the first part.
Remark  $ $ Note that when reduced mod $p$ the polynomial may have different leading coef, e.g. $\, A = 10x^2+ax\equiv  ax\pmod{5}.\,$ But $\,p\nmid A\,\Rightarrow\,A\not\equiv 0\,$ so its reduction has lead coef $\,a\not\equiv 0$
