Let $R = \mathbb{C}[x_1,...,x_n]/I$ be a quotient of a polynomial ring over $\mathbb{C}$, and let $M$ be a maximal ideal of $R$. How do I show that quotient ring $R/M$ is isomorphic to $\mathbb{C}$?

So I use the fact that $M$ is a maximal ideal of $R$ if and only if $R/M$ is a field. Obviously $\mathbb{C}$ is a field.

How would I use this theorem? Hilbert's Nullstellensatz: the maximal ideals of the polynomial ring $\mathbb{C}[x_1,...x_n]$ are in bijective correspondence with points of complex n-dimensional plane. A point a in $\mathbb{C^n}$ corresponds to the kernel of a substitution map which sends f(x) in $\mathbb{C}[x_1,...x_n]$ to f(a). the kernel of this map is the ideal generated by linear polynomials with roots consisting of the components of a

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    $\begingroup$ Claim: $R/M$ is a finite field extension of $\mathbb C$. If you can prove this, you're done. $\endgroup$ – Ian Coley Oct 22 '13 at 19:26
  • $\begingroup$ what is a finite field extension? $\endgroup$ – sarah Oct 22 '13 at 19:27
  • $\begingroup$ Have you done any field theory? $\endgroup$ – Ian Coley Oct 22 '13 at 19:27
  • $\begingroup$ i know what a field is $\endgroup$ – sarah Oct 22 '13 at 19:28
  • $\begingroup$ If $k$ is a field, we say that $F/k$ is a finite field extension if $k\subset F$, $F$ is a field, and $F$ is finitely generated as a $k$-algebra. $\endgroup$ – Ian Coley Oct 22 '13 at 19:29

This follows from Zariski's lemma, and from the fact that $\mathbf C$ is algebraically closed.

Edit: Since your edit, I see that the intended solution was most likely the one provided by Sammy, below. Nevertheless, I think that this proof is the "correct" proof, because it is "independent of coordinates".


By the Lattice Isomorphism Theorem, a maximal ideal $M$ in the ring $R$ corresponds to a maximal ideal $M'$ in $\mathbb{C}[x_1, \ldots, x_n]$ that contains $I$. Specifically, $$ M = M'/I. $$

In particular, $R / M \cong \mathbb{C}[x_1, \ldots, x_n] / M' \cong \mathbb{C}$.

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    $\begingroup$ Dear Sammy: This works, if the OP knows the result for $R=\mathbf C[x_1, \dots, x_n]$. $\endgroup$ – Bruno Joyal Oct 22 '13 at 19:35
  • $\begingroup$ That's true, @Marie. $\endgroup$ – Sammy Black Oct 22 '13 at 19:36

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