<protect>Anton,Matt</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)

Question: show that various properties about Galois closure follow from this description

Chat with Matt Anton 17:58, 26 November 2011 (PST)

Grothendieck-Deligne map can be though of as follows: if $E$ over $B$ is an $n$-dimensional algebra, then the Hilbert scheme of $n$ points in $E$ is trivial, but $Sym^n(E)$ is not. The Chow map from Hilb to Sym is a closed immersion which corresponds to the GD map.
Question: is the GD map stable under arbitrary base change? it should be
Note that for any smoothable family $E$ over $B$, there is a nilpotent extension of the family which witnesses smoothability. As soon as the discriminant is non-zero at every point of $B$, $E$ must be smoothable.

Idea: Suppose $E$ over $B$ smoothable. Perhaps there is a universal nilpotent thickening of the family which witnesses smoothability, and maybe this is given by the GD map. For example, consider $k[x]/x^2$ over $k$, then the GD map is from $TS^2 k[\epsilon]$ (different letter used for clarity in what follows). Then the family $k[x,\epsilon]/(x^2-\epsilon\otimes \epsilon)$ witnesses that the original family is smoothable.

There are some obvious candidates for other cases, but they don't work so well. For example, consider the family $k[x,y](x^2,xy,y^2)$ over $k$. It feels natural to use relations like $X^2-sym(x\otimes x\otimes 1)$ (where $sym$ means $1/6$ of the symmetrization) or equivalently $X^2-sym(x\otimes 1\otimes 1)^2$. However, it looks like the discriminants you get this way turn out to be zero.