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(Link to Vector product for the special common case.)
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''crossed product, of a [[group]] $G$ and a [[ring]] $K$''
 
''crossed product, of a [[group]] $G$ and a [[ring]] $K$''
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''of a vector $a$ by a vector $b$ in $\mathbb{R}^3$, see [[Vector product]].''
 
''of a vector $a$ by a vector $b$ in $\mathbb{R}^3$, see [[Vector product]].''
  
An associative ring defined as follows. Suppose one is given a mapping $\sigma$ of a group $G$ into the isomorphism group of an associative ring $K$ with identity, and a family
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An associative ring defined as follows. Suppose one is given a [[mapping]] $\sigma$ of a group $G$ into the [[isomorphism]] group of an associative ring $K$ with identity, and a family
  
 
$$ \rho = \{ \rho_{g,h} | g,h \in G\} $$
 
$$ \rho = \{ \rho_{g,h} | g,h \in G\} $$
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This ring is denoted by $K(G, \rho, \sigma)$; the elements $t_g$ form a $K$-basis for the ring.
 
This ring is denoted by $K(G, \rho, \sigma)$; the elements $t_g$ form a $K$-basis for the ring.
  
If $\sigma$ maps $G$ onto the identity automorphism of $K$, then $K(G, \rho)$ is called a twisted or crossed group ring, and if, in addition, $\rho_{g,h}=1$ for all $g,h\in G$, then $K(G,\rho,\sigma)$ is the group ring of $G$ over $K$ (see [[Group algebra|Group algebra]]).
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If $\sigma$ maps $G$ onto the identity [[automorphism]] of $K$, then $K(G, \rho)$ is called a twisted or crossed group ring, and if, in addition, $\rho_{g,h}=1$ for all $g,h\in G$, then $K(G,\rho,\sigma)$ is the group ring of $G$ over $K$ (see [[Group algebra]]).
  
Let $K$ be a field and $\sigma$ a monomorphism. Then $K(G,\rho,\sigma)$ is a simple ring, being the cross product of the field with its Galois group.
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Let $K$ be a field and $\sigma$ a [[monomorphism]]. Then $K(G,\rho,\sigma)$ is a simple ring, being the cross product of the field with its [[Galois group]].
  
====References====
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  S.K. Sehgal,  "Topics in group rings" , M. Dekker  (1978)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  A.A. Bovdi,  "Cross products of semi-groups and rings"  ''Sibirsk. Mat. Zh.'' , '''4'''  (1963)  pp. 481–499  (In Russian)</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  A.E. Zalesskii,  A.V. Mikhalev,  "Group rings"  ''J. Soviet Math.'' , '''4'''  (1975)  pp. 1–74  ''Itogi Nauk. i Tekhn. Sovrem. Probl. Mat.'' , '''2'''  (1973)  pp. 5–118</TD></TR><TR><TD valign="top">[4]</TD> <TD valign="top">  D.S. Passman,  "The algebraic structure of group rings" , Wiley  (1977)</TD></TR></table>
 
  
  
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====Comments====
 +
 +
In the defining relations for a factor system above $\rho^{\sigma(g_1)}_{g_2,g_3}$, e.g., of course stands for the result of applying the automorphism $\sigma(g_1)$ to the element $\rho_{g_2,g_3}$. If $\rho_{g,h}=1$ for all $g,h\in G$, then one obtains the skew group ring $K(G,1,\sigma)$. Cross products arise naturally when dealing with extensions. Indeed, let $N$ be a normal subgroup of $G$. Choose a set of representatives $\{\bar{g}\}$ of $G/N$ in $G$. Then every $\alpha\in KG$, the [[group algebra]] of $G$, can be written as a unique sum $\alpha=\sum\alpha_{\bar{g}}\bar{g}$, $\alpha_{\bar{g}}\in KN$. Now write
 +
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$$ \bar{g}\bar{h}=\rho_{\bar{g},\bar{h}}\bar{g}\bar{h}, \quad \bar{g}\alpha=\sigma_{\bar{g}}(\alpha)\bar{g} . $$
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 +
Then the $\rho_{\bar{g},\bar{h}}$ define a factor system (for the group $G/N$ and the ring $KN$ relative to the set of automorphisms $\sigma_{\bar{g}}$) and
 +
 +
$$ KG=KN(G/N,\rho,\sigma) . $$
  
====Comments====
+
Up to Brauer equivalence every [[central simple algebra]] is a cross product, but not every [[division algebra]] is isomorphic to a cross product. Two algebras $A$, $B$ over $K$ are Brauer equivalent if $A\otimes M_{n_1}(K)$ is isomorphic to $B\otimes M_{n_2}(K)$ for suitable $n_1$ and $n_2$. Here $M_n(K)$ is the algebra of $n\times n$ matrices over $K$. Cf. also [[Brauer group]].
In the defining relations for a factor system above <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712037.png" />, e.g., of course stands for the result of applying the automorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712038.png" /> to the element <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712039.png" />. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712040.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712041.png" />, then one obtains the skew group ring <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712042.png" />. Cross products arise naturally when dealing with extensions. Indeed, let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712043.png" /> be a normal subgroup of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712044.png" />. Choose a set of representatives <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712045.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712046.png" /> in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712047.png" />. Then every <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712048.png" />, the group algebra of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712049.png" />, can be written as a unique sum <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712050.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712051.png" />. Now write
 
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712052.png" /></td> </tr></table>
 
  
Then the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712053.png" /> define a factor system (for the group <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712054.png" /> and the ring <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712055.png" /> relative to the set of automorphisms <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712056.png" />) and
 
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712057.png" /></td> </tr></table>
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====References====
  
Up to Brauer equivalence every [[Central simple algebra|central simple algebra]] is a cross product, but not every [[Division algebra|division algebra]] is isomorphic to a cross product. Two algebras <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712058.png" /> over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712059.png" /> are Brauer equivalent if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712060.png" /> is isomorphic to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712061.png" /> for suitable <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712062.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712063.png" />. Here <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712064.png" /> is the algebra of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712065.png" /> matrices over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c027/c027120/c02712066.png" />. Cf. also [[Brauer group|Brauer group]].
+
<table>
 +
<TR><TD valign="top">[1]</TD> <TD valign="top">  S.K. Sehgal,  "Topics in group rings" , M. Dekker  (1978)</TD></TR>
 +
<TR><TD valign="top">[2]</TD> <TD valign="top">  A.A. Bovdi,  "Cross products of semi-groups and rings" ''Sibirsk. Mat. Zh.'' , '''4'''  (1963)  pp. 481–499  (In Russian)</TD></TR>
 +
<TR><TD valign="top">[3]</TD> <TD valign="top">  A.E. Zalesskii,  A.V. Mikhalev,  "Group rings" ''J. Soviet Math.'' , '''4'''  (1975)  pp. 1–74  ''Itogi Nauk. i Tekhn. Sovrem. Probl. Mat.'' , '''2'''  (1973)  pp. 5–118</TD></TR>
 +
<TR><TD valign="top">[4]</TD> <TD valign="top">  D.S. Passman,  "The algebraic structure of group rings" , Wiley  (1977)</TD></TR>
 +
</table>

Latest revision as of 08:53, 2 June 2016


crossed product, of a group $G$ and a ring $K$

of a vector $a$ by a vector $b$ in $\mathbb{R}^3$, see Vector product.

An associative ring defined as follows. Suppose one is given a mapping $\sigma$ of a group $G$ into the isomorphism group of an associative ring $K$ with identity, and a family

$$ \rho = \{ \rho_{g,h} | g,h \in G\} $$

of invertible elements of $K$, satisfying the conditions

$$ \rho_{g_1,g_2}\rho_{g_1g_2,g_3} = \rho^{\sigma(g_1)}_{g_2,g_3}\rho_{g_1,g_2g_3} $$ $$ \alpha^{\sigma(g_2)\sigma(g_1)} = \rho_{g_1,g_2}\alpha^{\sigma(g_1g_2)}\rho^{-1}_{g_1,g_2} $$

for all $\alpha\in K$ and $g_1,g_2,g_3\in G$. The family $\rho$ is called a factor system. Then the cross product of $G$ and $K$ with respect to the factor system $\rho$ and the mapping $\sigma$ is the set of all formal finite sums of the form

$$ \sum_{g\in G} \alpha_g t_g $$

where $\alpha_g \in K$ and the $t_g$ are symbols uniquely assigned to every element $g\in G$, with binary operations defined by

$$ \sum_{g\in G} \alpha_g t_g + \sum_{g\in G} \beta_g t_g = \sum_{g\in G} (\alpha_g+\beta_g)t_g,$$ $$ \left(\sum_{g\in G}\alpha_gt_g\right) \left(\sum_{g\in G}\beta_gt_g\right) = \sum_{g\in G} \left(\sum_{h_1h_2=g}\alpha_{h_1}\beta^{\sigma(h_1)}_{h_2}\rho_{h_1,h_2}\right) t_g $$

This ring is denoted by $K(G, \rho, \sigma)$; the elements $t_g$ form a $K$-basis for the ring.

If $\sigma$ maps $G$ onto the identity automorphism of $K$, then $K(G, \rho)$ is called a twisted or crossed group ring, and if, in addition, $\rho_{g,h}=1$ for all $g,h\in G$, then $K(G,\rho,\sigma)$ is the group ring of $G$ over $K$ (see Group algebra).

Let $K$ be a field and $\sigma$ a monomorphism. Then $K(G,\rho,\sigma)$ is a simple ring, being the cross product of the field with its Galois group.


Comments

In the defining relations for a factor system above $\rho^{\sigma(g_1)}_{g_2,g_3}$, e.g., of course stands for the result of applying the automorphism $\sigma(g_1)$ to the element $\rho_{g_2,g_3}$. If $\rho_{g,h}=1$ for all $g,h\in G$, then one obtains the skew group ring $K(G,1,\sigma)$. Cross products arise naturally when dealing with extensions. Indeed, let $N$ be a normal subgroup of $G$. Choose a set of representatives $\{\bar{g}\}$ of $G/N$ in $G$. Then every $\alpha\in KG$, the group algebra of $G$, can be written as a unique sum $\alpha=\sum\alpha_{\bar{g}}\bar{g}$, $\alpha_{\bar{g}}\in KN$. Now write

$$ \bar{g}\bar{h}=\rho_{\bar{g},\bar{h}}\bar{g}\bar{h}, \quad \bar{g}\alpha=\sigma_{\bar{g}}(\alpha)\bar{g} . $$

Then the $\rho_{\bar{g},\bar{h}}$ define a factor system (for the group $G/N$ and the ring $KN$ relative to the set of automorphisms $\sigma_{\bar{g}}$) and

$$ KG=KN(G/N,\rho,\sigma) . $$

Up to Brauer equivalence every central simple algebra is a cross product, but not every division algebra is isomorphic to a cross product. Two algebras $A$, $B$ over $K$ are Brauer equivalent if $A\otimes M_{n_1}(K)$ is isomorphic to $B\otimes M_{n_2}(K)$ for suitable $n_1$ and $n_2$. Here $M_n(K)$ is the algebra of $n\times n$ matrices over $K$. Cf. also Brauer group.


References

[1] S.K. Sehgal, "Topics in group rings" , M. Dekker (1978)
[2] A.A. Bovdi, "Cross products of semi-groups and rings" Sibirsk. Mat. Zh. , 4 (1963) pp. 481–499 (In Russian)
[3] A.E. Zalesskii, A.V. Mikhalev, "Group rings" J. Soviet Math. , 4 (1975) pp. 1–74 Itogi Nauk. i Tekhn. Sovrem. Probl. Mat. , 2 (1973) pp. 5–118
[4] D.S. Passman, "The algebraic structure of group rings" , Wiley (1977)
How to Cite This Entry:
Cross product. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Cross_product&oldid=38905
This article was adapted from an original article by A.A. Bovdi (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article