为何含I的ℂ[x₁,…,xₙ]极大理想与ℂ[x₁,…,xₙ]/I极大理想对应？

Great question! This is a common sticking point when first working with quotient rings and the correspondence theorem—let's unpack it step by step, no jargon overload.

First, Recap the Core of the Correspondence Theorem #

Let's restate the theorem clearly, since it's the foundation here:

For a ring ( R ) and an ideal ( I \subseteq R ), there’s a one-to-one bijection between:

1. Ideals of ( R ) that contain ( I ), and
2. Ideals of the quotient ring ( R/I ).

Most importantly, this bijection preserves inclusion: if ( J_1 ) and ( J_2 ) are ideals of ( R ) containing ( I ), then ( J_1 \subseteq J_2 ) if and only if ( J_1/I \subseteq J_2/I ). Strict inclusions (( \subsetneq )) are preserved too.

Why This Extends Straight to Maximal Ideals #

Remember what a maximal ideal is: an ideal ( M \subseteq R ) is maximal if there are no ideals ( J ) of ( R ) such that ( M \subsetneq J \subsetneq R ) (no "middle" ideals between ( M ) and the whole ring). Since maximality is defined entirely by these inclusion relationships, the correspondence theorem's preservation of inclusions lets us carry this property over to the quotient ring.

Let's prove both directions simply:

1. If ( M ) is a maximal ideal of ( \mathbb{C}[x_1,\dots,x_n] ) containing ( I ), then ( M/I ) is maximal in ( \mathbb{C}[x_1,\dots,x_n]/I ) #

Suppose ( M/I ) wasn't maximal. That would mean there's some ideal ( N/I ) in the quotient ring where ( M/I \subsetneq N/I \subsetneq (\mathbb{C}[x_1,\dots,x_n]/I) ). By the correspondence theorem's inclusion rule, this translates back to ( M \subsetneq N \subsetneq \mathbb{C}[x_1,\dots,x_n] ) in the original ring—but that contradicts ( M ) being maximal. So ( M/I ) has to be maximal.

2. If ( N/I ) is a maximal ideal of ( \mathbb{C}[x_1,\dots,x_n]/I ), then ( N ) is a maximal ideal of ( \mathbb{C}[x_1,\dots,x_n] ) containing ( I ) #

Flip the logic: suppose ( N ) wasn't maximal. Then there's an ideal ( M ) in ( \mathbb{C}[x_1,\dots,x_n] ) with ( N \subsetneq M \subsetneq \mathbb{C}[x_1,\dots,x_n] ). Since ( N ) contains ( I ), ( M ) must too (because ( I \subseteq N \subsetneq M )). By the correspondence theorem, this gives ( N/I \subsetneq M/I \subsetneq (\mathbb{C}[x_1,\dots,x_n]/I) )—which contradicts ( N/I ) being maximal. So ( N ) has to be maximal.

Why Artin Links This to the Correspondence Theorem #

Artin’s point is that you don’t need a separate, new theorem for maximal ideals—this correspondence is just a direct, obvious corollary of the original theorem. Maximality depends entirely on the "no middle ideals" rule, and since the bijection preserves all the inclusion relationships that define that rule, it automatically preserves maximality too.

内容的提问来源于stack exchange，提问作者user5826