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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

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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 $a_i \in k$.

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