How do I prove that, if $$k$$ is an algebraic closed field, the maximal ideals of $$R:=k[x_1,\dots, x_n]$$ are of the form $$(x_1-a_1,\dots,x_n-a_n)$$, where $$a_1,\dots, a_n\in k$$? I must prove it from the point of view of commutative algebra (the last lesson we saw Zariski's lemma and some equivalent forms of the weak Nullstellensatz). I reasoned like this: if I have a maximal ideal $$I\subseteq R$$, by weak Nullstellensatz there must be $$b_1,\dots, b_n\in k$$ such that $$f(b_1,\dots, b_n)=0$$ for every $$f\in I$$. Equivalently, $$I\subseteq (x_1-b_1,\dots, x_n-b_n)$$, so $$I=(x_1-b_1,\dots, x_n-b_n)$$. Is this proof correct? Are there more elegant or standard proofs? Thank you
Question: "How do I prove that, if $$k$$ is an algebraic closed field, the maximal ideals of $$R:=k[x_1,…,x_n]$$ are of the form $$(x_1−a_1,…,x_n−a_n)$$, where $$a_1,…,a_n∈k$$?"
Answer: If $$I⊆A:=k[x_1,..,x_n]$$ is maximal, it follows $$A/I$$ is a finite extension of $$k$$ (by the Zariski lemma) hence $$A/I≅k$$ and hence $$p(x_i)=a_i∈k$$ where $$p:A→A/I$$ is the canonical map. It follows $$I=(x_i−a_i)$$ is an equality of ideals. Hence any maximal ideal $$I \subseteq A$$ is on the form
$$I=(x_1-a_1,..,x_n-a_n)$$
with $$a_i \in k$$.