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
$$
\boxed{\dim A=\operatorname{trdeg}_kL.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 5 / b / i / Solution

{c}
{wiki=Noether_normalization_lemma}

If $A$ is a finitely generated algebra over a field $k$, then there are algebraically independent elements $y_1,\ldots,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\ldots,y_d]$.


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 Incoming links: Noether normalization lemma