Show that quotient ring of a $\Bbb C$-algebra by a maximal ideal is isomorphic to $\mathbb{C}$.

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

• Claim: $R/M$ is a finite field extension of $\mathbb C$. If you can prove this, you're done. – Ian Coley Oct 22 '13 at 19:26
• what is a finite field extension? – sarah Oct 22 '13 at 19:27
• Have you done any field theory? – Ian Coley Oct 22 '13 at 19:27
• i know what a field is – sarah Oct 22 '13 at 19:28
• 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. – Ian Coley Oct 22 '13 at 19:29

This follows from Zariski's lemma, and from the fact that $\mathbf C$ is algebraically closed.
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, $R / M \cong \mathbb{C}[x_1, \ldots, x_n] / M' \cong \mathbb{C}$.
• Dear Sammy: This works, if the OP knows the result for $R=\mathbf C[x_1, \dots, x_n]$. – Bruno Joyal Oct 22 '13 at 19:35