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Theorem: (Hilbert) If $k$ is a field, $A$ is a finitely generated $k$-algebra, and $M$ is a maximal ideal in $A$, then the factor $A/M$ is a finite extension of $k$. In particular if $k$ is algebraically closed then $A/M\simeq k$.

Problem: Let $k$ be a field and $A=k[x_1,\dots,x_n]$ is the polynomial ring of $n$ variables with coefficients in $k$. Prove that:

a) If $\alpha_1,\dots\alpha_n\in k$, then $M=(x_1-\alpha_1,\dots, x_n-\alpha_n)$ is a maximal ideal in $A$ and $A/M\simeq k$.

b) If $k$ is algebraically closed, for each maximal ideal $M$ in $A$ exist elements $\alpha_1,\dots\alpha_n\in k$, such that $M=(x_1-\alpha_1,\dots, x_n-\alpha_n)$.

Solution:

a) Let $\varphi:A\to k$ be defined by $\varphi(f(x_1,\dots,x_n))=f(\alpha_1,\dots,\alpha_n)$ Apparently $\varphi$ is a surjective homomorphism and $M\subset \ker(\varphi)$. I can't show that $\ker(\varphi)\subset M$.

b) If $M$ is maximal then by Hilbert’s theorem $A/M\simeq k$. Let $\pi$ be the natural homomorphism from $A$ to the factor $A/M=k$. Let $\alpha_i=\pi(x_i)$. I can't prove that $M=(x_1-\alpha_1,\dots, x_n-\alpha_n)$.

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    $\begingroup$ The maximal ideal $M'=(x_1-\alpha_1,\ldots,x_n-\alpha_n)$ is contained in $M$ $\endgroup$ – Diego Apr 23 '13 at 8:00
  • $\begingroup$ You could also find interesting this related question. $\endgroup$ – A.P. Apr 23 '13 at 9:06

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