Here is my unique factorization domain: $\mathbb{R}[x_{1}, x_{2},...x_{n}]$. My goal is to show that for all $1 \le i < j \le n$ the polynomial $x_{j} - x_{i}$ is irreducible. Since the ring is a UFD, it is enough to show that $x_{j} - x_{i}$ is prime, since {irreducibles} = {primes} in an UFD. Here is my attempt at a proof, but I am not sure if it is 100% strong:
Suppose $x_{j} - x_{i}$ divides $f(x_{1},...,x_{n})*g(x_{1},...,x_{n})$. Suppose $x_{j} - x_{i}$ does not divide $f(x_{1},...,x_{n})$ Then by division algorithm (with respect to the variable $x_{j}$) I get: $f(x_{1},...,x_{n}) = (x_{j} - x_{i})Q + R(x_{1},...x_{j-1},x_{j+1},...,x_{n})$. Suppose further that $x_{j} - x_{i}$ does not divide $g(x_{1},...,x_{n})$ Then by division algorithm (with respect to the variable $x_{j}$) I get: $g(x_{1},...,x_{n}) = (x_{j} - x_{i})Q_{2} + R_{2}(x_{1},...x_{j-1},x_{j+1},...,x_{n})$. Then $fg = R_{1}R_{2}+...other stuff$. Contradicts the fact that fg is divisible by $x_{j} - x_{i}$
It is not clear (to me) how the expression $$fg=R_1R_2+\ldots\text{"other stuff"},$$ contradicts the fact that $fg$ is divisible by $x_j-x_i$. How about this very direct approach instead:
If $x_j-x_i=fg$ for some $f,g\in\Bbb{R}[x_1,\ldots,x_n]$ then $$1=\deg(x_j-x_i)=\deg(fg)=\deg(f)+\deg(g),$$ so either $\deg(f)=0$ or $\deg(g)=0$, meaning that either $f$ or $g$ is a unit, so $x_j-x_i$ is prime.