$\operatorname{Spec}A$ is a finite set if and only if $A$ is a finite dimensional vector space Let $k$ be an algebraically closed field, and $A$ be a finitely generated $k$-algebra with no nilpotents.
Show that $\operatorname{Spec}A$ is a finite set if and only if $A$ is a finite dimensional vector space over $k$.
 A: Suppose $A$ is a finitely generated algebra (not necessarily reduced) over a field $k$ (not necessarily algebraically closed).
Then Noether's normalization theorem says that there exist $n\geq 0$  elements $y_1,\cdots, y_n\in A$, algebraically independent over $k$ ,  such that $A$ is module-finite over its sub-algebra $k[y_1,\cdots,y_n]$.
Since $\text {Spec} (A)\to \text {Spec} (k[y_1,\cdots, y_n])=\mathbb A^n_k$ is surjective by finiteness and since $\mathbb A^n_k$ is an infinite set for $n\gt 0$, it follows that $\text {Spec} (A)$ is a finite set if and only if $n=0$ i.e. if and only if $A$ is finite dimensional over $k$.
A: As $A$ is a finitely generated $k$-algebra, we have a surjection $k[x_1,\ldots x_r] \to A$ and thus $A \cong k[x_1,\ldots ,x_r]/I$. Assume that $Spec A$ is finite, then there are only finitely many maximal ideals containing $I$ in $k[x_1, \ldots ,x_r]$, say $m_1, \ldots m_n$. As these are co-maximal, we have $\cap m_i = \prod m_i$ and thus this product contains $I$, hence we have a map: 
$k[x_1, \ldots, x_r]/I \to k[x_1, \ldots, x_r]/\prod m_i$. Using The Chinese remainder theorem ,the right hand side is equal to $\oplus k[x_1, \ldots, x_r]/m_i$, and using the nullstellensatz + that $k$ is algebraically closed, we obtain that this is again equal to $\oplus_{i=1}^n k$ which is a finite dimensional vector space over $k$. The map above is obviously surjective, we now check injectivity: If $f$ is mapped to $0$ then, $f$ is contained in $\prod m_i = \cap m_i$, but as $\cap m_i = \cap P = \sqrt I$ in $k[x_1,\ldots x_r]$ (here the intersections are taken over maximal ideals containing $I$ and prime ideals containing $I$ respectively), it follows that the image of $f$ in $k[x_1,\ldots, x_r]/I$ is contained in the nilradical , but this is isomorphic to $A$ which was assumed not to have any nilpotents, thus the image of $f = 0$ and the map is an injection aswell. Conversely: Assume now that $A = k[x_1, \ldots, x_r]/I$ is a finite dimensional vector space over $k$ and assume for the sake of contradiction that $spec A$ is infinite, then given any $n$ we can find $n$ different prime ideals each containing $I$ , say $P_1, \ldots P_n$ and we have a surjection $k[x_1, \ldots x_r]/I \to k[x_1, \ldots x_r]/\prod P_i = \oplus k[x_1,\ldots x_r]/P_i$. The right hand side has dimension over $k$ at least equal to $n$, and thus $A$ has dimension greater or equal to $n$ for arbitrary $n$. This is a contradiction, thus $Spec A$ must be finite. 
Added: The reason why we can use the CRT on $k[x_1,\ldots x_r]/\prod P_i$, is because as $A$ is a finite dimensional vector space over $k$ the descending chain condition holds for sub-k-modules (and hence also for sub-A-modules i.e. the ideals of $A$), thus $A$ is Artin, and in an Artin ring we have that every prime ideal is maximal.  
A: Let me try an alternative proof.
If Spec $A$ is finite, then there are only finitely many maximal ideals, say $m_1,\dots, m_n$. On the other hand, since $A$ is a f.g. $k$-algebra, it follows from Nullstellensatz that nil $A:=\sqrt{(0)}=\bigcap$ Specm $A$, where Specm $A$ denotes the set of maximal ideals of $A$. Thus, there exists $l\in \mathbb{N}$ such that $(m_1\dots m_n)^l=(0)$. Now it is easy to conclude that $A$ has finite lenght as $k$-module, that is, $\dim_k A<\infty$.
Conversely, $\dim_k A<\infty$ implies that $A$ is artinian, hence $\dim A=0$. But noetherian rings with dimension $0$ have discrete spectrum (because they have only finitely many minimal prime ideals, which must also be maximal by dimension). Finally, since Spec $A$ is quasi-compact, if Spec ${A}$ is discrete, then it is finite.
