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The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation (at least when the characteristic is bigger than $n$). [[User:Anton|Anton]] 21:06, 17 November 2011 (PST) | The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation (at least when the characteristic is bigger than $n$). [[User:Anton|Anton]] 21:06, 17 November 2011 (PST) | ||
=Chat with Matt [[User:Anton|Anton]] 17:58, 26 November 2011 (PST)= | |||
[[Category:Note]] | [[Category:Note]] | ||
Revision as of 17:58, 26 November 2011
<protect>Anton</protect>
Hilbert component
Suppose $E$ is an algebra over $B$, free of rank $n$ as a module. Then we have the Grothendieck-Deligne norm map $\sigma:TS^nE\to B$. This map is determined by two properties:
- it is a $B$-algebra homomorphism,
- $\sigma(x\otimes x\otimes\cdots\otimes x)=N_{E/B}(x)$ for any $x\in E$.
The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation (at least when the characteristic is bigger than $n$). Anton 21:06, 17 November 2011 (PST)