Difference between revisions of "Algebra (structure)"

(New page: Let <math>R</math> be a commutative ring. We say that a set <math>E</math> is an '''<math>R</math>-algebra''' if <math>E</math> is an <math>R</math>-module and we have a <math>A</...)
 
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Let <math>R</math> be a [[commutative ring]].  We say that a set <math>E</math> is an '''<math>R</math>-algebra'''
 
Let <math>R</math> be a [[commutative ring]].  We say that a set <math>E</math> is an '''<math>R</math>-algebra'''
if <math>E</math> is an <math>R</math>-[[module]] and we have a <math>A</math>-[[bilinear mapping]] of <math>E\times E</math> into
+
if <math>E</math> is an <math>R</math>-[[module]] and we have a <math>R</math>-[[bilinear mapping]] of <math>E\times E</math> into
 
<math>E</math>, denoted multiplicatively.  That is, we have a multiplication between elements of <math>E</math>,
 
<math>E</math>, denoted multiplicatively.  That is, we have a multiplication between elements of <math>E</math>,
and between elements of <math>A</math> and elements of <math>E</math> such that for any <math>r \in A</math>, <math>x,y \in E</math>,
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and between elements of <math>R</math> and elements of <math>E</math> such that for any <math>r \in R</math>, <math>x,y \in E</math>,
 
<cmath> r(xy) = (rx)y = x(ry) , </cmath>
 
<cmath> r(xy) = (rx)y = x(ry) , </cmath>
 
and
 
and
 
<cmath> r(x+y) = rx + ry. </cmath>
 
<cmath> r(x+y) = rx + ry. </cmath>
We identify elements <math>r</math> of <math>A</math> with the corresponding elements <math>r1</math> of <math>E</math>.
+
We identify elements <math>r</math> of <math>R</math> with the corresponding elements <math>r1</math> of <math>E</math>.
  
 
Note that multiplication in <math>E</math> need not be [[associative]] or [[commutative]]; however,
 
Note that multiplication in <math>E</math> need not be [[associative]] or [[commutative]]; however,
the elements of <math>A</math> must commute and associate with all elements of <math>E</math>.  We can thus think
+
the elements of <math>R</math> must commute and associate with all elements of <math>E</math>.  We can thus think
of <math>E</math> as an <math>A</math>-module endowed with a certain kind of multiplication.
+
of <math>E</math> as an <math>R</math>-module endowed with a certain kind of multiplication.
  
 
Equivalently, we can say that <math>E</math>
 
Equivalently, we can say that <math>E</math>
is an <math>A</math>-algebra if it is a not-necessarily-associative ring that contains <math>A</math> as a sub-ring.
+
is an <math>R</math>-algebra if it is a not-necessarily-associative ring that contains <math>R</math> as a sub-ring.
  
 
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Latest revision as of 16:19, 8 September 2009

Let $R$ be a commutative ring. We say that a set $E$ is an $R$-algebra if $E$ is an $R$-module and we have a $R$-bilinear mapping of $E\times E$ into $E$, denoted multiplicatively. That is, we have a multiplication between elements of $E$, and between elements of $R$ and elements of $E$ such that for any $r \in R$, $x,y \in E$, \[r(xy) = (rx)y = x(ry) ,\] and \[r(x+y) = rx + ry.\] We identify elements $r$ of $R$ with the corresponding elements $r1$ of $E$.

Note that multiplication in $E$ need not be associative or commutative; however, the elements of $R$ must commute and associate with all elements of $E$. We can thus think of $E$ as an $R$-module endowed with a certain kind of multiplication.

Equivalently, we can say that $E$ is an $R$-algebra if it is a not-necessarily-associative ring that contains $R$ as a sub-ring.

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See also

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