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


1 Answer 1


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 $f\in I$, $f(a_1,\dots, a_n)=0$.


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