I want to solve the following problem from Bosch's "Algebraic Geometry and Commutative Algebra" Sec. 6.1, Ex. 8:
Let $A$ be an algebra of finite type over a field $F$. Show that $Spm(A)$ (the maximal spectrum of $A$) is Hausdorff w.r.t. the subspace Zariski topology if $F$ is a finite field.
I have no idea where the finitness of the field comes into play. Maybe someone can give me a hint. Thanks a lot!
Best Answer
This is false. For instance, if $A=F[x]$, then $A$ has infinitely many maximal ideals (there are infinitely many irreducible polynomials over any field, even a finite field), and any proper closed subset of the maximal spectrum is finite (any nonzero element of $F[x]$ has only finitely many irreducible factors). So, the maximal spectrum is an infinite set with the cofinite topology, which is not Hausdorff.