 Solution

For the polynomial ring, use the Euclidean function $\delta (f)=\mathrm{deg}f$ on nonzero polynomials. If $a,b\in F[X]$ and $b\ne 0$, subtract the multiple of $b$ that cancels the leading term of $a$. The leading coefficient of $b$ is invertible in the field, and every subtraction strictly lowers the degree of the remainder. This process terminates with $a=qb+r$ and either $r=0$ or $\mathrm{deg}r<\mathrm{deg}b$. Thus $\boxed{F[X]\text{ is a Euclidean domain}}$.

For the Gauss lemma for polynomials, call an integer polynomial primitive when its coefficient gcd is one. If two primitive polynomials had a nonprimitive product, a prime $p$ would divide every product coefficient. Reducing modulo $p$ would give a zero product of two nonzero polynomials in the integral domain ${\mathbb{F}}_p[X]$, a contradiction. Therefore the product of primitive integer polynomials is primitive.

To transfer a rational factorization, write each nonzero rational factor as a rational scalar times a primitive integer polynomial by clearing denominators and removing the coefficient gcd. Thus $f=qAB$ with primitive $A,B\in \mathbb{Z}[X]$ and $q\in \mathbb{Q}$. Their product is primitive by the lemma. If $q=m/n$ in lowest terms, integrality of every coefficient of $f$ forces $n$ to divide every coefficient of $AB$, so $n=1$. Hence $q\in \mathbb{Z}$; absorbing $q$ into either factor produces the required integer factorization with the same positive factor degrees. In particular irreducibility of a primitive polynomial over $\mathbb{Z}$ and over $\mathbb{Q}$ is equivalent.

Finally, let $a/b$ in lowest terms, $b>0$, be a rational root of a monic degree-$d$ integer polynomial. Multiplying the equation by $b^d$ shows $b\mid a^d$. Coprimality forces $b=1$. Thus $\boxed{\text{every rational root of a monic integer polynomial is an integer}}$.