[proofplan] We reduce both $\dim(A)$ and $\operatorname{trdeg}_k(A)$ to the same integer $d$ using Noether Normalisation. By the Noether Normalisation Lemma, there exists an integral injective $k$-algebra homomorphism $k[T_1, \ldots, T_d] \hookrightarrow A$ for some $d \ge 0$. The [Integral Extensions Preserve Dimension](/theorems/2896) theorem equates $\dim(A)$ with $\dim(k[T_1, \ldots, T_d]) = d$, and the same integral extension equates $\operatorname{trdeg}_k(A)$ with $\operatorname{trdeg}_k(k[T_1, \ldots, T_d]) = d$. [/proofplan]

[step:Apply Noether Normalisation to obtain a polynomial subring over which $A$ is integral]Since $A$ is a finitely generated $k$-algebra that is an integral domain, the Noether Normalisation Lemma provides an integer $d \ge 0$ and an injective $k$-algebra homomorphism \begin{align*} \iota: k[T_1, \ldots, T_d] \hookrightarrow A \end{align*} such that $A$ is integral over the image $\iota(k[T_1, \ldots, T_d])$. Identifying $k[T_1, \ldots, T_d]$ with its image, we have an integral extension $k[T_1, \ldots, T_d] \subset A$.[/step]

[guided]The Noether Normalisation Lemma states: if $A$ is a finitely generated $k$-algebra and $k$ is a field, then there exist algebraically independent elements $y_1, \ldots, y_d \in A$ such that $A$ is integral over $k[y_1, \ldots, y_d]$. Equivalently, there is an injective $k$-algebra homomorphism $\iota: k[T_1, \ldots, T_d] \hookrightarrow A$ making $A$ integral over $\iota(k[T_1, \ldots, T_d])$. We verify the hypotheses: $A$ is a finitely generated $k$-algebra (given) and $k$ is a field (given). The conclusion gives us an integer $d \ge 0$ and an integral extension $k[T_1, \ldots, T_d] \subset A$, where we identify $k[T_1, \ldots, T_d]$ with its image under $\iota$. The integer $d$ is the number of algebraically independent generators needed — it will turn out to be both $\dim(A)$ and $\operatorname{trdeg}_k(A)$.[/guided]

[step:Equate Krull dimensions via the integral extension]By [Integral Extensions Preserve Dimension](/theorems/2896) (part 1), since $k[T_1, \ldots, T_d] \subset A$ is an integral extension: \begin{align*} \dim(A) = \dim(k[T_1, \ldots, T_d]). \end{align*} The Krull dimension of a polynomial ring in $d$ variables over a field is $d$ (the chain $(0) \subsetneq (T_1) \subsetneq (T_1, T_2) \subsetneq \cdots \subsetneq (T_1, \ldots, T_d)$ has length $d$, and no longer chain exists). Therefore $\dim(A) = d$.[/step]

[guided]The [Integral Extensions Preserve Dimension](/theorems/2896) theorem states: if $A \subset B$ is an integral extension of rings, then $\dim(A) = \dim(B)$. We apply it with the roles of $A$ and $B$ played by $k[T_1, \ldots, T_d]$ and $A$ respectively. The hypothesis — that $A$ is integral over $k[T_1, \ldots, T_d]$ — was established in the previous step. The conclusion gives $\dim(A) = \dim(k[T_1, \ldots, T_d])$. It remains to compute $\dim(k[T_1, \ldots, T_d])$. The polynomial ring $k[T_1, \ldots, T_d]$ over a field $k$ has Krull dimension exactly $d$. One direction: the chain of prime ideals \begin{align*} (0) \subsetneq (T_1) \subsetneq (T_1, T_2) \subsetneq \cdots \subsetneq (T_1, T_2, \ldots, T_d) \end{align*} has length $d$, so $\dim(k[T_1, \ldots, T_d]) \ge d$. The other direction follows from the fact that $k[T_1, \ldots, T_d]$ is a Noetherian ring and every maximal ideal has height $d$ (a consequence of the weak Nullstellensatz and induction on $d$). Hence $\dim(A) = d$.[/guided]

[step:Equate transcendence degrees via the integral extension]By [Integral Extensions Preserve Dimension](/theorems/2896) (part 2), since $k[T_1, \ldots, T_d] \subset A$ is an integral extension of $k$-algebras that are both integral domains: \begin{align*} \operatorname{trdeg}_k(A) = \operatorname{trdeg}_k(k[T_1, \ldots, T_d]). \end{align*} The elements $T_1, \ldots, T_d$ are algebraically independent over $k$ and generate $k[T_1, \ldots, T_d]$, so $\operatorname{trdeg}_k(k[T_1, \ldots, T_d]) = d$. Therefore $\operatorname{trdeg}_k(A) = d$.[/step]

[guided]We now compute the transcendence degree. The [Integral Extensions Preserve Dimension](/theorems/2896) theorem (part 2) states: if $A \subset B$ is an integral extension of integral domains that are both $k$-algebras, then $\operatorname{trdeg}_k(A) = \operatorname{trdeg}_k(B)$. We verify the hypotheses: $k[T_1, \ldots, T_d]$ is an integral domain ($k$ is a field, so the polynomial ring is a domain), $A$ is an integral domain (given), both are $k$-algebras (the polynomial ring canonically, $A$ by hypothesis), and the extension is integral (by the Noether Normalisation step). The conclusion gives $\operatorname{trdeg}_k(A) = \operatorname{trdeg}_k(k[T_1, \ldots, T_d])$. Why is $\operatorname{trdeg}_k(k[T_1, \ldots, T_d]) = d$? The transcendence degree of the fraction field $k(T_1, \ldots, T_d)$ over $k$ is $d$, since $\{T_1, \ldots, T_d\}$ is a transcendence basis: the $T_i$ are algebraically independent over $k$ by definition, and every element of $k(T_1, \ldots, T_d)$ is algebraic over $k(T_1, \ldots, T_d)$ (trivially). Since $\operatorname{trdeg}_k$ of a domain means $\operatorname{trdeg}_k$ of its fraction field, we get $\operatorname{trdeg}_k(k[T_1, \ldots, T_d]) = d$.[/guided]

[step:Conclude $\dim(A) = \operatorname{trdeg}_k(A)$] Combining the two equalities: \begin{align*} \dim(A) = d = \operatorname{trdeg}_k(A). \end{align*} [/step]