= Solution
For the <polynomial ring>, use the <Euclidean function> $\delta(f)=\deg f$ on nonzero <polynomials>. If $a,b\in F[X]$ and $b\ne0$, 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 $\deg r<\deg b$. Thus $\boxed{F[X]\text{ is a Euclidean domain}}$.
For the <Gauss lemma for polynomials>, call an <integer> <polynomial> <primitive polynomial>[primitive] when its <coefficient> gcd is one. If two <primitive polynomial>[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 \b[the product of <primitive polynomial>[primitive] <integer> <polynomials> is <primitive polynomial>[primitive]].
To transfer a rational factorization, write each nonzero rational factor as a rational scalar times a <primitive polynomial>[primitive] <integer> <polynomial> by clearing denominators and removing the <coefficient> gcd. Thus $f=qAB$ with <primitive polynomial>[primitive] $A,B\in\mathbb Z[X]$ and $q\in\mathbb Q$. Their product is <primitive polynomial>[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 $\boxed{\text{every rational root of a monic integer polynomial is an integer}}$.
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