Transcendence degree and Krull dimension of finitely generated algebras

Question

Let $K$ be a field, and let $a_1,\dots,a_{n+1}$ be $n+1$ elements of a finitely generated $K$-algebra $A$ of Krull dimension $n$.

Are the elements $a_1,\dots,a_{n+1}$ always algebraically dependent over $K$?

I.e: Are the monomials $(a_1)^{m_1}\cdots(a_{n+1})^{m_{n+1}}$ always $K$-linearly dependent?

The answer is Yes if $A$ is a domain. Indeed, in this case, $n$ is the transcendence degree of $L$ over $K$, where $L$ is the field of fractions of $A$.

[The $K$-algebra $A$ is assumed to be commutative and unital.]

Answer 1

Suppose that $A$ is a finitely generated algebra over $k$ of Krull dimension $n$. Assume that $k[x_1,...,x_n,x_{n+1}]$ is a polynomial subalgebra of $A$. Pick $S = k[x_1,...,x_n,x_{n+1}]\setminus \{0\}$. This is a multiplicative subset of $A$. Hence there exists a prime ideal $\mathfrak{p}$ of $A$ such that $$\mathfrak{p}\cap S = \emptyset$$ Now consider $B = A/\mathfrak{p}$. Then $B$ is a finitely generated domain over $k$ of Krull dimension $\leq n$. Moreover, $B$ contains polynomial $k$-subalgebra in $n+1$ variables (image of $k[x_1,...,x_{n+1}]$ under the surjection $A\rightarrow B$). This is a contradiction.