Irreducible Laurent polynomial in multiple variables: Ideal intersection

Question

In a larger proof of some theorem in algebraic geometry, an argument is used which I don't really understand. Giving the full proof is clearly over the top, so I tried to minimize the information needed (although I'm not sure if more details are needed). I think it boils down to the following:

Let $f,g \in K[x_1^{\pm1},\dots,x_n^{\pm1}]$ be Laurent polynomials in multiple variables, where $K$ is a field and $f$ is irreducible. Suppose that $\langle f,g \rangle \cap K[x_1^{\pm1},\dots,x_{n-1}^{\pm1}] = \{0\}$. Then, since $f$ is irreducible, $g$ is a multiple of $f$.

Here, "$\langle f,g \rangle$" denotes the ideal generated by $f$ and $g$. I don't clearly see the argument behind it. Can somebody tell me why this needs to hold?

Thanks in advance!

EDIT: Another piece of information that might be helpful is that $f$ has the following special form: When regarding $f$ as polynomial in $x_n$ with coefficients in $K[x_1^{\pm1},\dots,x_{n-1}^{\pm1}]$, the coefficients are all monomials of the form $\displaystyle cx_1^{u_1} \cdots x_{n-1}^{u_{n-1}}$ for $c \in K$ and $u_1, \dots, u_{n-1} \in \mathbb Z$.

Answer 1

After multiplying by an appropriate power of $x_{n}$, we may assume that $f,g \in K[x_{1}^{\pm 1} , \dotsc , x_{n-1}^{\pm 1} , x_{n}]$ and that $f$ has nonzero constant term when viewed as a polynomial in $x_{n}$ with coefficients in $K[x_{1}^{\pm 1} , \dotsc , x_{n-1}^{\pm 1}]$ (this implies $f$ is irreducible in $K[x_{1}^{\pm 1} , \dotsc , x_{n-1}^{\pm 1} , x_{n}]$). Now we can apply the following lemma to the case $A = K[x_{1}^{\pm 1} , \dotsc , x_{n-1}^{\pm 1}]$:

Let $A$ be a UFD. Let $f,g \in A[x]$ be polynomials such that $f$ is irreducible in $A[x]$ and $(f,g) \cap A = (0)$. Then $f$ divides $g$ in $A[x]$.

Proof: We may assume that $g \not\in A$. Suppose $f$ does not divide $g$ in $A[x]$. Set $F := \mathrm{Frac}(A)$. Then $f$ is irreducible in $F[x]$ as well by Gauss' lemma. Moreover $f$ does not divide $g$ in $F[x]$ (since $A[x]$ is a UFD and irreducibles in $A$ remain irreducible in $A[x]$). Since $F[x]$ is a PID, there exist $a,b \in F[x]$ such that $af+bg = 1$. Clearing the denominators of $a,b$ gives a nonzero element of $(f,g) \cap A$, contradiction.