Zero-dimensional ideals and finite-dimensional algebras [duplicate]

I encountered in the literature the term zero-dimensional ideal, however I can not find a relevant definition anywhere in Atiyah-MacDonald or Matsumura. In fact, I encounted the statement:

$I$ is a zero-dimensional ideal of $k[x_1,\dots,x_n]$ $\Leftrightarrow$ $k[x_1,\dots,x_n]/I$ is a finite-dimensional $k$-vector space, where we assume that $k$ is algebraically closed.

Question 1: What is the definition of the dimension of an ideal?

Question 2: How do we prove the above statement?

Question 3: Is $k$ being algebraically closed a necessary condition for the statement to be true?

If you want us to be sure that we are correctly interpreting terminology that you have encountered "in the literature", you had better identify the piece of literature in question: no one can be responsible for knowledgeable about all possible ways that a term has been used across all mathematical papers and books.

But based on the context it is quite plausible that a zero dimensional ideal $I$ in a commutative ring $R$ means that the quotient ring $R/I$ has Krull dimension zero: prime ideals are maximal. At least this definition renders true the statement

$k[x_1,\ldots,x_n]/I$ has Krull dimension zero $\iff$ it is finite-dimensional as a $k$-vector space.

This is true over any field $k$. Recall that a ring is Artinian iff it is Noetherian and dimension zero. The implication $\Longleftarrow$ is then immediate. For $\Longrightarrow$ use the fact that an Artinian ring is a finite product of local Artinian rings to reduce to the case in which $I$ is a power of a maximal ideal (c.f. Theorem 8.35 here) and then to the case in which $I$ is maximal. In this latter case the result is Zariski's Lemma: see e.g. $\S$ 11.1 of loc. cit..

Also let me note that when $k$ is algebraically closed, the given conditions are also equivalent to: the corresponding affine algebraic set $V(I)$ is finite. This equivalence does not hold over any non-algebraically closed field. A simple example is $I = \langle y^2+1 \rangle$ in $\mathbb{R}[x,y]$. The quotient ring is isomorphic to $\mathbb{C}[x]$, which has Krull dimension one, but there are no points $(a_1,a_2) \in \mathbb{R}^2$ with $a_2^2+1 = 0$.