By Hilbert Nullstelensatz we know that for any field $k$, every maximal ideal of $k[x_1, ..., x_n]$ has residue field a finite extension of $k$. I also did an exercise which goes: any integral domain $A$ which is a finite $k$- algebra is a field.
Then there is a remark after this exercise that says: it follows by Hilbert Nullstelensatz and this exercise that prime ideals of $k[x_1, ..., x_n]$ with finite residue ring are the same as maximal ideals of $k[x_1, ..., x_n]$.
Could someone please explain me how this works?
If $P$ is a prime ideal then the residue ring $k[x_1,...,x_n]$ is an integral domain and naturally a $k$-algebra (quotients of $k$-algebras by ideals are also $k$-algebraS). So if we assume the residue ring is finitely generated, then we have the hypotheses of the second so the residue ring is a field. It is then a classical result of algebra that in a ring $R$, $P$ an ideal is maximal iff $R/P$ is a field.