Polynomials of several variable is integral domain Let $k$ be a field and $k[X_1,...,X_n]$ a ring of polynomials in $n$ variables. In which case is $k[X_1,...,X_n]$ an integral domain? 
 A: Always.
For a given ring $R$ we have: $R[X_1,\ldots,X_n]$ is an integral domain if and only if $R$ is. And in your case $R=k$ is a field, which is an integral domain.
Proof. Step 1. We will show that $R[X]$ is an integral domain if and only if $R$ is.
"$\Rightarrow$" Note that $R$ is a subring of $R[X]$. And subrings of integral domains are integral domains.
"$\Leftarrow$" Assume that $W=\sum w_nX^n$ and $U=\sum u_nX^n$ and $WU=0$. Assume that both $W,U\neq 0$ and note that the coefficient of $WU$ at $\deg(WU)$ is $w_{\deg{W}}\cdot u_{\deg{U}}$. But since $WU=0$ then $w_{\deg{W}}\cdot u_{\deg{U}}=0$ and since $R$ is an integral domain then $w_{\deg{W}}=0=u_{\deg{u}}$ which contradicts the definition of $\deg$ (which is the highest index with nonzero coefficient).
Step 2. $R[X_1,\ldots,X_n]$ is isomorphic to $R[X_1][X_2]\cdots[X_n]$ and therfore by Step 1 and induction on $n$ we conclude that $R[X_1,\ldots, X_n]$ is an integral domain if and only if $R$ is.
A: A proof that $R[X_1,\dots,X_n]$ is an integral domain if $R$ is this:
By a trivial induction, it is enough to show it for a single indeterminate. Consider two polynomials
$$p(X)=a_0+a_1X+\dots+a_n X^d\enspace(a_d\ne 0),\quad q(X)=b_0+b_1X+\dots+b_e X^e\enspace(b_e\ne 0).$$
The leading term of $p(X)q(X)$ is $a_db_e X^{d+e}$, by the definition of multiplication in $R[X]$, and it is non-zero since $a_db_e\ne 0$.
