Let $f(t)$ be an irreducible polynomial in $k(x_1, \ldots, x_n)[t]$. Show that $f(t)$ is irreducible in $k[x_1,\ldots,x_n][t]$.
I guess it's clear that I am required to apply Gauss' Lemma. The statement in my course notes is:
Let $R$ be a UFD and $F$ its field of fractions. A non-constant polynomial in $R[X]$ is irreducible in $R[X]$ if and only if it is both irreducible in $F[X]$ and primitive in $R[X]$. And a polynomial $P$ in $R[X]$ is primitive if the only elements of $R$ that divide all coefficients of $P$ at once are the invertible elements of $R$.
I first cleared the denominators of the coefficients $a_i \in k(x_1,\ldots,x_n)$ of $f(t)$. But this modified polynomial is not monic anymore, so I ran into trouble showing that $f$ is primitive in $k[x_1,\ldots,x_n][t]$ - or at least, things get very messy.
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EDIT: Additional context added. I am trying to understand Proposition 4.9 in Hartshorne's Algebraic Geometry. The argument given goes as follows:
Let $K$ be a finitely generated extension field of $k$. Then $K$ is seperably generated over $k$ since we are working over an algebraically closed field. Hence we can find a transcendence base $x_1,\ldots,x_n \in K$ such that $K$ is a finite seperable extension of $k(x_1,\ldots,x_n)$. By the primitive element therem we can find one further element $y \in K$ such that $K=k(x_1,\ldots,x_n,y)$. Now $y$ is algebraic over $k(x_1,\ldots,x_n)$. So it satisfies a polynomial equation with coefficients which are rational functions in $x_1,\ldots,x_n$. Clearing denominators we get an irreducible polynomial, satisfying $f(x_1,\ldots,x_n,y)=0$
My original question aimed to clarify this last step. I assumed he used Gauss' Lemma.
EDIT 2: Reformulation of original question (related to comments on first post).
Let $f(t)$ be an irreducible polynomial in $k(x_1, \ldots, x_n)[t]$. Let $f'(t)$ be the polynomial where we have cleared up the denominators of $f(t)$. Show that $f'(t)$ is irreducible in $k[x_1,\ldots,x_n][t]$.