# Equivalent forms of weak Nullstellensatz

I read that the following proposition, that is basically a corollary of Zariski's lemma, is a form of the weak Nullstellensatz:

Let $$k$$ be a field, $$A$$ a finitely generated $$k$$-algebra. Let $$m$$ be a maximal ideal of $$A$$. Then the residue field $$A/m$$ is a finite algebraic extension of $$k$$. In particular, if $$k$$ is algebraically closed, $$A/m \cong k$$.

My question is: if $$k$$ is algebraically closed, saying that $$A/m \cong k$$ is actually equivalent to saying that for every proper ideal $$I\subset k[x_1,\dots,x_n]$$, there are $$a_1,\dots ,a_n\in k$$ such that $$f(a_1,\dots ,a_n)=0$$ for every $$f\in I$$? I suppose yes, but I really can't see how. Thanks in advance

Well every such ideal $$I$$ will be contained in a maximal ideal $$\mathfrak{m}$$ so that $$A/\mathfrak{m}\simeq k$$.
Now take $$a_i\in k$$ such that $$x_i$$ corresponds to $$a_i$$ under this isomorphism. So $$\mathfrak{m}=(x_1-a_2,\dots, x_n-a_n).$$ Now $$I \subset \mathfrak{m}=(x_1-a_1,\dots, x_n-a_n)$$ says that for all $$f\in I$$, $$f(a_1,\dots, a_n)=0$$.