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 $dim_k(A)$?
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For examples where the cardinality of $\mathrm{Spec}(A)$ is strictly less than $\dim_k(A)$, take any finite local $k$-algebra of $k$-vector space dimension greater than $1$, e.g., $A=k[x]/(x^n)$ with $n>1$. – Keenan Kidwell May 17 '12 at 22:40
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Here is an answer – Georges Elencwajg Mar 06 '14 at 13:04
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$A$ 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 $k[x]/(x^2)$.
Dylan Moreland
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Another (similar, but slightly different, not totally elementary) argument for why $\mathrm{Spec}(A)$ is finite: since $A$ is Noetherian, being a finite $k$-algebra, it has finitely many minimal primes, $\mathfrak{p}_1,\ldots,\mathfrak{p}_r$, all maximal. Now, $\mathrm{Spec}(A)$ is the union of its irreducible components, and any prime $\mathfrak{p}$ of $\mathrm{Spec}(A)$ lies in one of them, i.e, $\mathfrak{p}\in V(\mathfrak{p}_i)={\mathfrak{p}_i}$ for some $i$, so $\mathfrak{p}=\mathfrak{p}_i$. – Keenan Kidwell May 15 '12 at 18:03
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@KeenanKidwell Oh, that's great. I like that fact because it seems halfway geometric—do you mind if I add that in? – Dylan Moreland May 15 '12 at 19:12
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