= Solution
The <Krull dimension> $\dim R$ is the supremum of lengths $n$ of strict chains
$$
\mathfrak p_0\subsetneq\cdots\subsetneq\mathfrak p_n
$$
of prime ideals. The transcendence degree $\operatorname{trdeg}_kL$ is the cardinality of a transcendence basis of $L/k$.
By <Noether normalization lemma>, there are algebraically independent $y_1,\ldots,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\ldots,y_d]$. Their fraction field has transcendence degree $d$, and $L$ is algebraic over it, so $\operatorname{trdeg}_kL=d$. Going up and incomparability show that an integral extension preserves Krull dimension, while a polynomial ring in $d$ variables over a field has dimension $d$. Hence
$$
\boxed{\dim A=\operatorname{trdeg}_kL.}
$$
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