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Steven Sam
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This is a bit late, but here is one example I like:

Theorem. A localization of a regular local ring at a prime ideal is still regular.

One way to prove this is to deduce it from

Theorem. Let $R$ be a local ring. Then the following are equivalent:

  1. $R$ is regular

  2. Every $R$-module has a finite length projective resolution

  3. The residue field has a finite length projective resolution.

(To use it, note that iflet $P$ be the prime ideal. Since $R$ is regular, then its residue field$R/P$ has a finite length projective resolution. Now localize--this is exact, so we get a finite length projective$R_P$-projective resolution overof $(R/P)_P$, which is the localization.residue field of $R_P$)

This stuff is in Chapter 19 of Eisenbud's Commutative Algebra.

It's not clear to me how one would try to prove the first theorem from the definitions of regular.

Edit: fixed some mistakes

This is a bit late, but here is one example I like:

Theorem. A localization of a regular local ring is still regular.

One way to prove is to deduce it from

Theorem. Let $R$ be a local ring. Then the following are equivalent:

  1. $R$ is regular

  2. Every $R$-module has a finite length projective resolution

  3. The residue field has a finite length projective resolution.

(To use it, note that if $R$ is regular, then its residue field has a finite length projective resolution. Now localize--this is exact, so we get a finite length projective resolution over the localization.)

This stuff is in Chapter 19 of Eisenbud's Commutative Algebra.

It's not clear to me how one would try to prove the first theorem from the definitions of regular.

This is a bit late, but here is one example I like:

Theorem. A localization of a regular local ring at a prime ideal is still regular.

One way to prove this is to deduce it from

Theorem. Let $R$ be a local ring. Then the following are equivalent:

  1. $R$ is regular

  2. Every $R$-module has a finite length projective resolution

  3. The residue field has a finite length projective resolution.

(To use it, let $P$ be the prime ideal. Since $R$ is regular, $R/P$ has a finite length projective resolution. Now localize--this is exact, so we get a finite length $R_P$-projective resolution of $(R/P)_P$, which is the residue field of $R_P$)

This stuff is in Chapter 19 of Eisenbud's Commutative Algebra.

It's not clear to me how one would try to prove the first theorem from the definitions of regular.

Edit: fixed some mistakes

Source Link
Steven Sam
  • 10.7k
  • 3
  • 47
  • 80

This is a bit late, but here is one example I like:

Theorem. A localization of a regular local ring is still regular.

One way to prove is to deduce it from

Theorem. Let $R$ be a local ring. Then the following are equivalent:

  1. $R$ is regular

  2. Every $R$-module has a finite length projective resolution

  3. The residue field has a finite length projective resolution.

(To use it, note that if $R$ is regular, then its residue field has a finite length projective resolution. Now localize--this is exact, so we get a finite length projective resolution over the localization.)

This stuff is in Chapter 19 of Eisenbud's Commutative Algebra.

It's not clear to me how one would try to prove the first theorem from the definitions of regular.