I'm reading Kunz's proof of the Nullstellensatz (pg 16), and he seems to use the following fact: if $A$ is a field extension of $K$ obtained by adjoining finitely many elements to $K$, then $A \simeq K[x_1,\cdots,x_n]/M$ where $M$ is maximal in $K[x_1,\cdots,x_n]$.
If we are adjoining algebraic elements $\alpha_1,\cdots, \alpha_k$, then we can let $M$ be generated by the minimal polynomials $p_i$ of $\alpha_i$ and we are done. However, I don't know how to proceed if the $\alpha_i$ are transcendental. Any advice (or if I have misinterpreted the statement) would be appreciated.
Kunz writes the following (emphasis mine):
Really, this is one way of defining ring adjunction; generally one says that, for rings $R\subseteq S$, $S$ is the ring obtained by adjoining elements $s_1,\dots,s_n$ to $R$ if $S$ is the only subring of itself containing both $R$ and the $s_i$. This is equivalent to saying that the evaluation morphism $$f:R[x_1,\dots,x_n]\to S$$ given by $x_i\mapsto s_i$ is surjective. In that case, we will have $S\cong R[x_1,\dots,x_n]\big/\ker f$, and if $S$ is a field then $\ker f$ must be a maximal ideal, from which Kunz's comment follows. Another commonly used terminology for this situation is to say that $S$ is "generated as an $R$-algebra" by the $s_i$.
However, it should be noted that ring adjunction is fundamentally different from field adjunction. In particular, the circumstances that you describe in your problem post – ie of some of the $\alpha_i$ being transcendental over $K$ – can actually never occur. This is Zariski's lemma, which says that, if a field $A$ containing $K$ is finitely generated as a $K$-algebra, then $\dim_KA<\infty$. In particular, in that case, $A$ cannot contain any element transcendental over $K$. So, put differently, if $A$ is a transcendental extension of $K$, then $A$ is not isomorphic to a ring of form $K[x_1,\dots,x_n]\big/M$ for any $n\in\mathbb{N}$.