When product of multivariable polynomials is a monomial I am considering $f$ and $g \in k[x_{1},\ldots,x_{n}]$ be multivariable polynomials over a field $k$. Suppose both of them have more than $2$ terms. I would like to know if it is possible that their product is a monomial.
$n=1$ is simple, but if $n>1$ you might find some cancellations. I currently don't have a good idea to approach general $n$ directly or conduct reduction/induction method.
Any comments are welcome.
 A: According to this answer, a ring $R$ is a UFD if and only if $R[x]$ is a UFD.
So in particular, we can show by induction on $n$ that if $R$ is a UFD, so is $R[x_1, ..., x_n]$. This is because $R[x_1, ..., x_{n + 1}] \cong R[x_1, ..., x_n][x_{n + 1}]$.
Another result is needed which characterises the units in a polynomial ring.
It is well-known that (for any ring $R$ at all) $P$ is a unit in $R[x]$ if and only if $P$ can be written as $a_0 + a_1 x + ... + a_n x^n$ where $a_0$ is a unit and $a_1, ..., a_n$ are nilpotent. To prove this, consider an arbitrary prime ideal $K$ of $R$. Then $P$ reduced mod $K$ is a unit in $(R / K)[x]$, which is an integral domain. Thus, $P$ reduced mod $K$ must be a constant polynomial which is a unit. Therefore, all coefficients of $P$ except $a_0$ must be in all prime ideals, hence must be nilpotent.
Since there are no nontrivial nilpotents in an integral domain, we see that whenever $R$ is an integral domain, the only units in $R[x]$ are constant polynomials.
Furthermore, if $R$ is an integral domain, then so is $R[x]$. Thus, there are no nontrivial nilpotents in $R[x]$.
Therefore, we see that by induction, if $R$ is an integral domain then so is $R[x_1, ..., x_n]$.
We also see by induction that the only units in $R[x_1, ..., x_n]$ are the constant polynomial units.
Combining these two results, we see that if $R$ is a UFD and we have two polynomials $P, Q$ such that $PQ$ is a monomial, then both $P$ and $Q$ must be a monomial. This is because monomials factor uniquely as $\prod\limits_{j = 1}^k x_{i_j}$, and the only associates of $x_{i_j}$ are $u x_{i_j}$ where $u \in R$ is a unit.
And clearly, a field is a UFD.
