In this proof that the cyclotomic polynomials have integer coefficients, it is mentioned in the answer that it follows from Gauss Lemma. To be more specific,
if $f,g$ and $h$ are polynomials in $\mathbb Q[X]$ with $h = fg$ and $h, f \in \mathbb Z[X]$ are monic, why does this imply that $g \in \mathbb Z[X]$ by using Gauss Lemma?
I understand the proof using the division algorithm in $\mathbb Z[X]$ which appears in the linked thread above, and also I see that if we write down the coefficients of the product $fg$ and suppose that they must be in $\mathbb Z$, then we can successively work from the highest (using that $f$ is monic) down and find that the ones of $g$ must be in $\mathbb Z$. Or, by seeing that in the successive steps of (polynomial) long division, namely if $h_i = a_n X^n + \dots +a_0$ is the result so far, and $f = X^m + \dots +b_0$ with $n > m$, we compute $h_i - (a_nX^{n-m})f$, so that the result is still a polynomial in $\mathbb Z[X]$.
But, I do not see in what sense Gauss Lemma yields this immediately, namely the formulation of Gauss Lemma I know is:
An non-constant polynomial in $\mathbb Z[X]$ is irreducible in $\mathbb Z[X]$ iff it is irreducible in $\mathbb Q[X]$ and primitive in $\mathbb Z[X]$.
or the ones in terms of the content, like here.
From the version of Gauss' Lemma that says $\mathrm{cont}(h)=\mathrm{cont}(f)\mathrm{cont}(g)$, it follows from $h, f$ monic in $\mathbb{Z}[x]$ that $\mathrm{cont}(f) = \mathrm{cont}(h) = 1$ and so $\mathrm{cont}(g) = 1$ as well. Then we must have $g \in \mathbb{Z}[x]$: from the definition of content, any denominator in a coefficient of $g$ would show up in the content as well.