k-points of k[x_1, ..., x_n]/I are exactly k-algebra maps to k

Let $\psi :A\to k$ be a $k$-algebra map, and let ${\bar{x}}_i$ be the class of $x_i$ in $A$. Put $a_i=\psi ({\bar{x}}_i)$. The composite $k[x_1,\dots ,x_n]\to A\stackrel{\psi }{\to }k$ is a $k$-algebra map sending $x_i$ to $a_i$, so by [L1] it is evaluation at $a=(a_1,\dots ,a_n)$. Since every $h\in I$ maps to $0$ in $A$, we get $h(a)=0$.

Conversely, let $a\in k^n$ satisfy $h(a)=0$ for every $h\in I$. By [L1], evaluation at $a$ is a $k$-algebra map $k[x_1,\dots ,x_n]\to k$, and the hypothesis says that $I$ lies in its kernel. Therefore it factors uniquely through a $k$-algebra map $A\to k$.

The two constructions are inverse because both record the same coordinate images of the classes ${\bar{x}}_1,\dots ,{\bar{x}}_n$. Hence $k$-points of $k[x_1,\dots ,x_n]\mathrm{/}I$ are exactly its $k$-algebra maps to $k$.