In AG codes, we evaluate a set of points $\mathcal{P}$ on an algebraic variety over a certain Riemann Roch space whose support set is a set of points $\mathcal{Q}$ on the variety such that $\mathcal{P}\cap\mathcal{Q}=\emptyset$.
My question is what happens when we replace $\mathcal{P}$ with $\mathcal{P}'$ where $\mathcal{P}'$ is a set of points that is not in the variety and $|\mathcal{P}|=|\mathcal{P}'|$?
What happens is that basically everything breaks down.
The most fundamental thing that goes wrong is that the functions on that Riemann-Roch space are not well-defined outside the variety $V$. For example, those functions may have poles in ${\cal P}'$. Also, the functions in the Riemann-Roch space properly are elements in the field of quotient of the coordinate ring $\Gamma(V)$ of your variety. The elements of the coordinate ring are cosets of polynomials modulo the ideal $I(V)$ of polynomials vanishing on all of $V$, so the elements of the coordinate ring simply do not have a value at a point outside of $V$.
In some cases (e.g. if ${\cal P}$ consists of affine points only, and the RR-space consists of polynomials only (say, $V$ is a plane curve with a single point $Q$ on the line at infinity) we would be able to decide on a preferred set of polynomials such that they form a set of representatives of the functions in the RR-space. Even then, you would not know anything about neither the dimension of the resulting code no about its minimum distance.
Let me illustrate these points with a toy example.
Let $V$ be the $x$-axis of the projective $xy$-plane, so $\Gamma(V)=k[x]$. Let further $Q$ be the point $x=\infty$ (the one with homogeneous coordinates $[1:0:0]$). Then the RR-space of the divisor $nQ$ consists of polynomials in $k[x]$ of degree $\le n$. But if we want to extend those functions outside the $x$-axis, we immdeiately face the problem that for the purpose of $V$ the function $x$ equals the function $x(1+y^\ell)$ for all integers $\ell$. So which of those would we use when we evaluate a function outside $V$? Or consider the function $x(1+y/(y-1))$. Again, from the point of view of the variety $V(y=0)$ is the same as $x$, but this one has a pole at all the points $(x,1)$.
Ok. Let's decide to be simple and agree that we just embed $k[x]\hookrightarrow k[x,y]$, and use the "same" polynomials. Assume that $k$ is a finite field, and ${\cal P}$ consists of the $k$-rational points on $V$. In other words, the case where the resulting AG-code is equal to a Reed-Solomon code. Now what if we select ${\cal P}'$ to be the set of $k$-rational points with $x=1$. Then by evaluating the polynomials in $x$ you only get constant functions on ${\cal P}'$. In other words, you only ever get a repetition code. No good! Except that the minimum distance was good. But that may also fail. What if we select ${\cal P}'$ to consist of the point $(0,1)$ and the points $(1,y), y\in k, y\neq0$? The polynomial $x-1$ takes a non-zero value only at the point $(0,1)\in {\cal P}'$. In other words we get codes with minimu Hamming distance equal to one.