Let $K$ be a field and $A$ an affine $K$-algebra. Show that $A$ has (Krull) dimension zero (is artinian) if and only if it is finite dimensional over $K$.
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If your $k$ - algebra $A$ is finite dimensional as a vector space over $k$ then it has to be Artinian. This is just linear algebra that the dimension $d$ of a subspace of a finite dimensional vector space $V$ is such that $0 \leq d \leq \dim V$. – Mar 25 '13 at 12:21
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Related to Zero-dimensional ideals and finite-dimensional algebras. – May 13 '13 at 10:05
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Finite dimensional (as a vector space) algebras are zero dimensional: Using quotients, it is enough to prove that every finite dimensional domain is a field. But this is standard: The multiplication map by a nonzero element is injective, hence also surjective.
If $A$ is a zero dimensional finitely generated $k$-algebra, then it is artinian, and we may even assume that $A$ is local, say with maximal ideal $\mathfrak{m}$. Now try to prove that each $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is finite-dimensional. Since $\mathfrak{m}$ is nilpotent, this shows that $A$ is finite dimensional (as a $k$-vector space).
Martin Brandenburg
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2The second part of your hint is unclear to me: $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is finite-dimensional as an $k$-vector space or as an $A/\mathfrak m$-vector space? And is not the main point in the proof of the question that $A/\mathfrak m$ is a finite dimensional $k$-vector space? – May 05 '13 at 08:45
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1Over $k$. But also over $A/m$, doesn't matter. Yes, $A/m$ is finite-dimensional over $k$, I assume such basic facts ... – Martin Brandenburg May 07 '13 at 14:23