Proposition: Suppose $k$ is an algebraically closed field and $A$ is a finitely generated $k$-algebra that is also an integral domain. Then there exists a (unital) $k$-algebra homomorphism $\varphi:A\rightarrow k$.
This proposition is true since it follows from Zariski’s lemma: take a maximal ideal $\mathfrak{m}$, so $A/\mathfrak{m}$ is a field extension of $k$ which is also a finitely generated $k$-algebra, and hence an algebraic extension, and hence equal to $k$. Also this proposition implies weak Nullstellensatz: suppose $(f_1,\dots,f_n)$ is a proper ideal in $k[X_1,\dots,X_m]$, then take some maximal ideal $\mathfrak{m}\supseteq(f_1,\dots,f_n)$ so $A=k[X_1,\dots,X_m]/\mathfrak{m}$ is a finitely generated $k$-algebra and an integral domain; since $f_1,\dots,f_n$ have a common root in $A$, they also have a common root in $k$.
Question 1: Does this proposition have a name?
Question 2: Is there a direct way to prove it (without appealing to Zariski’s lemma)?
Here is my attempt. Suppose $A$ is generated over $k$ by $a_1,\dots,a_l$. Let $K$ be the fraction field of $A$ and $(t_1,\dots,t_d)$ be a transcendental basis for $K$ over $k$, which might be thought of as indeterminates. There exist finitely many rational functions $f_1,\dots,f_k$ in $t_1,\dots,t_d$ such that each $a_i$ is integral over $k[f_1,\dots,f_k]$.
There exists a homomorphism from $k[f_1,\dots,f_k]$ to $k$: we can plug any numbers into the indeterminates as long as none of the denominators of $f_1,\dots,f_k$ become zero. Now it suffices to build a homomorphism from $k[f_1,\dots,f_k,a_1,\dots,a_l]$ to $k$, for which it suffices to use the following claim plus induction.
Claim: If $A$ is a $k$-alegbra and an integral domain, $\varphi:A\rightarrow k$ is a $k$-algebra homomorphism, and $a$ is integral over $A$, then $\varphi$ can be extended to $\varphi:A[a]\rightarrow k$.
Proof of claim: Let $f$ be some monic polynomial in $A$ for which $a$ is a root. $\varphi$ extends to $\varphi:A[X]\rightarrow k[X]$, which induces $\varphi:A[X]/(f)\rightarrow k[X]/(\varphi(f))$. The former is isomorphic to $A[a]$, and the latter is a direct sum of copies of $k$ since it is algebraically closed. Composing this with a projection we get $\varphi:A[a]\rightarrow k$.
Question 3: Is this proof correct? Is it basically a special case of the standard proof of Zariski’s lemma?
