Let $k$ be a field and $A$ be a finite $k$-algebra.
How does one quickly see that $Spec(A)$ is a finite set?
Further, is it true that the cardinality of $Spec(A)$ is equal to $dim_k(A)$?
Let $k$ be a field and $A$ be a finite $k$-algebra.
How does one quickly see that $Spec(A)$ is a finite set?
Further, is it true that the cardinality of $Spec(A)$ is equal to $dim_k(A)$?
$A$ is Artinian for reasons of dimension, and in Artinian rings all primes are maximal and these are finite in number. For proofs, which are quite short, see Section 15 of Milne's notes.
In this specific case, showing that $A$ has dimension zero should be easier: if $\mathfrak p$ is a prime of $A$ then $A/\mathfrak p$ is a domain finite over $k$, hence a field. Thus $\mathfrak p$ is maximal. I don't see a shortcut for the second step: find maximal ideals such that $\mathfrak m_1 \cap \cdots \cap \mathfrak m_r$ is minimal, and show by contradiction that there are no others.
For a counterexample to your last question, take $A$ to be the ring of dual numbers $k[x]/(x^2)$.
In general the cardinality of $\mathrm{Spec}(A)$ is not greater than $\mathrm{dim}_k A$. The key ingredient is Chinese remainder theorem. For a proof you can see my answer here.