Exercise 1.5 of Hartshorne asks us to show (in one direction) that any affine coordinate ring $k[x_1,\dots,x_n]/I(Y)$ is a finitely-generated $k$-algebra with no nilpotents. The second part is quite simple - we know $I(Y)$ is radical by Nullstellensatz so the quotient is reduced. However, the second part is trickier for me - we're looking for $k\to k[x_1,\dots,x_n]/I(Y)$ that is finitely generated. Presumably this is somehow canonical, but I'm not sure how to find it.
In short, how do I construct this map?
The ring $k[x_1,\ldots,x_n]/I(Y)$ is a finitely-generated $k$-algebra because it is generated over $k$ by $\overline{x_1},\ldots,\overline{x_n}$, the images of the $x_i\in k[x_1,\ldots,x_n]$ in the quotient ring $k[x_1,\ldots,x_n]/I(Y)$.
If you prefer your formulation of looking for a map $k\to k[x_1,\ldots,x_n]/I(Y)$, the map is just the ordinary map that realizes it as a $k$-algebra, namely map $k$ to $k$, i.e., the composite of $$k\hookrightarrow k[x_1,\ldots,x_n]\twoheadrightarrow k[x_1,\ldots,x_n]/I(Y)$$