<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.

Chat with Matt Anton 16:52, 1 December 2011 (PST)

Not clear if vanishing locus of the discriminent actually contains the complement of the principal component, but since the hilbert scheme is finite type, there does exist some power of the discriminent which contains the complement of the principal component. So to show that a given 0-scheme is smoothable, it is enough to show it is the "closed fiber" of a family where the discriminent has sufficiently large order of nilpotence.