Is a zero-dimensional algebra over a field a finite-dimensional vector space? I'm probably dumb, but I can't answer to the question in the title, that I state here again: let $A$ be a zero-dimensional algebra over a field $K$ ( - here I mean that the algebra has Krull dimension $=0$). Is it true that $\dim_{K} A < \infty$?
If the algebra is affine the answer is yes (cf. Theorem 5.11 of Kemper's A Course in Commutative Algebra). Can you give me an hint?
Thanks in advance!
 A: If the algebra $A$ is finitely generated over $K$, then $\dim A = 0$ always implies $\dim_K A < \infty$. To get an insight as to why this is the case, suppose for simplicity that $A: = K[x_0,\dots,x_n]/I$ is homogeneous. Then the maximal homogeneous ideal $\mathfrak{m}=(x_0,\dots,x_n)$ is the unique minimal prime over $I$, and so $\mathfrak{m}^\ell \subset I$ for some integer $\ell$. Since $A$ is finitely generated as a $K$-algebra, let $d$ be the degree of the largest homogeneous generator. Then for degrees $s>d$ we will have $A_s = \mathfrak{m}_{s-d} A_d$, and for $s-d \ge \ell$ we have
$A_s = \mathfrak{m}_{s-d} A_d \subset I_{s-d}k[x_0,\dots,x_n] / I_s=0$. This means that $A = K \oplus A_1 \oplus A_2 \oplus \cdots \oplus A_{\ell+d-1}$. This together with the fact that each of the $A_j$ is a finite-dimensional vector space give that $\dim_K A < \infty$. If $A$ is nonhomogeneous, use homogenization to reduce to the homogeneous case.
If $A$ is infinitely generated then the statement is not true: Let $T$ be a countably infinite set and consider the ring $R=K\left[\left\{x_i\right\}_{i\in T}\right]$ and the ideal $I$ generated by all $x_i^2, \, i \in T$. Then $R/I$ has zero Krull dimension and its homogeneous part of degree  $1$ is an infinite dimensional vector space.
A: No, it is not true that $\operatorname {dim}_KA\lt\infty $: take for $A$ the field of rational functions $A=K(X)$  over $K$. Then $$\operatorname {dim}_{\operatorname {Krull }}A=0 \quad\text {but} \quad \operatorname {dim}_KA=\infty $$
A: Take the algebra $R=k[x_i]/(x_i^i)$. Then any prime ideal contains $0=x_i^i$, and so contains $x_i$, thus $(x_i)$ is is the unique prime ideal and the ring is zero dimensional, yet $x_i$ is a countable linearly independent set.
