Rabinowitch's trick: Difference between revisions

Revision as of 01:06, 8 January 2008

Statement

Let $R$ be a commutative unital ring. The following are equivalent:

$R$ is a Jacobson ring
If $P$ is a prime ideal of $R$ and if $S : = R / P$ contains an element $b \neq 0$ such that $S [b^{- 1}]$ is a field, then $S$ is a field

Proof

1 implies 2

As in the hypotheses, let $P$ be a prime ideal in $R$ . Then $S : = R / P$ is an integral domain, and hence 0 is a prime ideal in this. Further, $S$ is also a Jacobson ring since Jacobson is quotient-closed, and thus the zero ideal in $S$ is an intersection of maximal ideals.

Now, the primes of $S [b^{- 1}]$ correspond to the primes of $S$ that do not contain $b$ . Since $S [b - 1]$ is a field, $b$ is contained in every nonzero prime ideal of $S$ . Thus if $S$ were not a field, zero would not be a maximal ideal, and hence $b$ would be contained in every maximal ideal, contradicting the fact that the intersection of all maximal ideals is zero. Thus $S [b^{- 1}]$ is a field.

2 implies 1

Fill this in later

@@ Line 5: / Line 5: @@
 # <math>R</math> is a [[Jacobson ring]]
 # If <math>P</math> is a [[prime ideal]] of <math>R</math> and if <math>S := R/P</math> contains an element <math>b \ne 0</math> such that <math>S[b^{-1}]</math> is a [[field]], then <math>S</math> is a field
+==Proof==
+===1 implies 2===
+As in the hypotheses, let <math>P</math> be a [[prime ideal]] in <math>R</math>. Then <math>S := R/P</math> is an [[integral domain]], and hence 0 is a prime ideal in this. Further, <math>S</math> is also a [[Jacobson ring]] since [[Jacobson is quotient-closed]], and thus the zero ideal in <math>S</math> is an [[intersection of maximal ideals]].
+Now, the primes of <math>S[b^{-1}]</math> correspond to the primes of <math>S</math> that do not contain <math>b</math>. Since <math>S[b{-1}]</math> is a [[field]], <math>b</math> is contained in ''every'' nonzero prime ideal of <math>S</math>. Thus if <math>S</math> were ''not'' a [[field]], zero would not be a maximal ideal, and hence <math>b</math> would be contained in every [[maximal ideal]], contradicting the fact that the intersection of all maximal ideals is zero. Thus <math>S[b^{-1}]</math> is a field.
+===2 implies 1===
+{{fillin}}