This is from my book, $k$ is a field.
"Suppose in additional that $k$ is algebraically closed. Then every maximal ideal of $A=k[X_1,....,X_n]$ is of the form $m=(X_1-a_1,...,X_n-a_n)$ for some $a_1,...,a_n \in k$. The map $k[X_1,....,X_n] \to k[X_1,....,X_n]/m =k$ is just the natural "evaluation" map $f(X_1,...,X_n) \mapsto f(a_1,...,a_n)$"
This is my first course in algebraic geometry and I am already having troubles the very first days, for instance why do we have $k[X_1,....,X_n]/m = k$, is it because they are isomorphic? I also wonder what the residue classes of $X_1,...,X_n$ means, is this the images of $X_1,...,X_n$ under this natural "evaluation" map? Please help me to bring some clarity to this.
It's not literally true that $k[X_1...X_n]/m=k$, rather, only that they're isomorphic, as you say. The $A\to A/(X_1-a_1...X_n-a_n)$ sending $X_i\mapsto [X_i]$ its residue class is a surjection onto a field isomorphic to $k$, because in the quotient ring $[X_i]=[a_i]$ so that we may uniquely rewrite each polynomial as a constant. (For instance $[X_1^2-2X_2]=[a_1^2-2a_2]$). After this rewriting the isomorphism is natural, namely, $[1]\mapsto 1$.