Assume that $\mathbb{F}$ is a finite field of order $q$ and $R=\mathbb{F}[x_1,...,x_n]$. Say $I=(f_1,...,f_n)$ and define $J = I+(x_1^q-x_1,x_2^q-x_2,...,x_n^q-x_n)$. Now for any $p\in \mathbb{F}$, we know that $p^q-p=0$. Therefore the zero locus of $I$ and $J$ in $\mathbb{F}^n$ should be same.
For example for $q=2$, if we take $I=(xy)$, there are $3$ points in the zero locus, namely $(1,0),(0,0)$ and $(0,1)$. But since $\deg(xy)=2$, the multiplicity of $(0,0)$ is $2$. But when I look at $J=(x^2+x,xy,y^2+y)$, the zero locus again contains exactly three points but the footprint of $J$ is $\{1,x,y\}$ which contains $3$ points (which is not $4$). Therefore the variety does not change but the multiplicities seem to differ. Am I correct and if I am how can I see that all multiplicities in the ideal $J$ should be $1$?