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Let $g$ be the geometric genus of $X$. Choose an integer $N\ge 2g+1$. For each prescribed point $x_i$, both $N{x}_i$ and $(N-1)x_i$ have degree greater than $2g-2$, so their complementary canonical divisors have negative degree and their complementary Riemann-Roch spaces vanish. The Riemann-Roch theorem therefore givesChoose $f_i\in L(N{x}_i)\setminus L((N-1)x_i)$. This is possible because the two dimensions differ by one. Such a rational function has its only pole at $x_i$, and its pole order there is exactly $N$. Set $f={\sum }_{i=1}^m{f}_i$.

At $x_i$ all terms except $f_i$ are regular. In a local parameter, the term of order $-N$ in $f_i$ has a nonzero coefficient, and none of the other terms has a negative-power term. Hence that coefficient cannot cancel. Away from the prescribed points every summand is regular. We obtain the stronger conclusionThus $f$ is regular on $U$ and has a pole at every $x_i$. This rational functions with prescribed poles construction works in every characteristic and does not require a generic-coefficient choice. If the prescribed list is empty, a constant function meets the vacuous requirements.