Problem in understanding of If $R$ is a UFD, then $R[x]$ is also a UFD(Gauss's theorem)

Question

Gauss's Theorem:If R is a UFD then so is R[X], the ring of polynomials over R.

I am studying the the from this pdf. But I have problem in understanding.

Proof: Take a (non-zero) polynomial $f ∈ R[X]$. We can factorize it into irreducibles in $F[X]$ (as F[X] is a PID and so a UFD), and we may as well take the irreducibles to be primitive polynomials in $R[X]$. So we can write $$f(X) = \frac{r}{s}g_1(X)g_2(X). . .g_k(X)$$ where $g_i ∈ R[X]$ is primitive and irreducible in $F[X]$.

Why $g_i ∈ R[X]$ is irreducible in $F[X]?$ Any help is appreciated. Thanks.

Answer 1

The author is combining some steps here, and you'll probably find things easier to understand if you keep them separate.

First, given $f\in R[X]$, we can think of $f$ as an element of $F[X]$, because $R\subseteq F$. Since $F$ is field, $F[X]$ is a PID, so we can factor $f$ as a product of irreducible elements $h_i$ of $F[X]$.

Second, for each of those irreducible factors $h_i$, look at all its coefficients; being in $F$, they are fractions of elements of $R$. Multiply $h_i$ by a common multiple $s_i$ of the denominators of its coefficients. The result is a new polynomial $k_i$, which is in $R[X]$ (because you've cleared the denominators) and is irreducible in $F[X]$ (because $h_i$ was irreducible and you've just multiplied by a unit $s_i$ of $F$). Of course, unlike the $h_i$'s, the $k_i$'s are not usually a factorization of $f$; their product is $sf$, where $s$ is the product of all the $s_i$'s.

Third, divide each $k_i$ by the least common multiple $r_i$ in $R$ of its coefficients, to get a new polynomial $g_i$. The point of this division is that each $g_i$ is now a primitive polynomial in $R[X]$. The $g_i$'s are irreducible in $F[X]$ (because you've just divided the $k_i$'s by units $r_i$ of $F$). Their product is $sf/r$, where $r$ is the product of the $r_i$'s.

So now we have $f$ equal to $\frac rs$ times the product of the $g_i$'s, each of which is a primitive polynomial in $R[X]$ and is irreducible in $F[X]$. So we've arrived at the end of the passage you quoted in the question.