Difference between revisions of "Ring is integral extension of fixed-point subring under finite automorphism group"

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(New page: ==Statement== Let <math>A</math> be a commutative unital ring and <math>G</math> be a finite group acting as automorphisms on <math>A</math> (in other words, <math>G</math> is...)
 
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Let <math>A</math> be a [[commutative unital ring]] and <math>G</math> be a [[finite group]] acting as [[automorphism]]s on <math>A</math> (in other words, <math>G</math> is a finite subgroup of the automorphism group of <math>A</math>). Let <math>B = A^G</math> be the subring of <math>B</math> comprising those elements fixed by ''every'' element of <math>G</math>. Then, <math>B</math> is an [[integral extension]] of <math>A</math>. In fact, every element of <math>B</math> satisfies a monic polynomial over <math>A</math> of degree equal to the order of <math>G</math>.
 
Let <math>A</math> be a [[commutative unital ring]] and <math>G</math> be a [[finite group]] acting as [[automorphism]]s on <math>A</math> (in other words, <math>G</math> is a finite subgroup of the automorphism group of <math>A</math>). Let <math>B = A^G</math> be the subring of <math>B</math> comprising those elements fixed by ''every'' element of <math>G</math>. Then, <math>B</math> is an [[integral extension]] of <math>A</math>. In fact, every element of <math>B</math> satisfies a monic polynomial over <math>A</math> of degree equal to the order of <math>G</math>.
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==Related facts==
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* [[Automorphism group acts transitively on fibers of spectrum over fixed-point subring]]
  
 
==Proof==
 
==Proof==
  
 
Let <math>x \in B</math>. Consider the elements <math>g \cdot x</math> for <math>g \in G</math>. Then, all the elementary symmetric polynomials in these elements take values in <math>A</math>. Hence, we can construct a monic polynomial of degree equal to the order of <math>G</math>, with all coefficients in <math>A</math>, and whose roots are precisely the elements <math>g \cdot x</math>.
 
Let <math>x \in B</math>. Consider the elements <math>g \cdot x</math> for <math>g \in G</math>. Then, all the elementary symmetric polynomials in these elements take values in <math>A</math>. Hence, we can construct a monic polynomial of degree equal to the order of <math>G</math>, with all coefficients in <math>A</math>, and whose roots are precisely the elements <math>g \cdot x</math>.

Revision as of 00:24, 9 March 2008

Statement

Let A be a commutative unital ring and G be a finite group acting as automorphisms on A (in other words, G is a finite subgroup of the automorphism group of A). Let B = A^G be the subring of B comprising those elements fixed by every element of G. Then, B is an integral extension of A. In fact, every element of B satisfies a monic polynomial over A of degree equal to the order of G.

Related facts

Proof

Let x \in B. Consider the elements g \cdot x for g \in G. Then, all the elementary symmetric polynomials in these elements take values in A. Hence, we can construct a monic polynomial of degree equal to the order of G, with all coefficients in A, and whose roots are precisely the elements g \cdot x.