# $\mu$-regular lines.

Let $V$ be linear vector space of dimension $n$ over an algebraically closed field $k$. Fix $X$ an affine algebraic variety $X$ and a finite morphism of algbebraic varieties: $\mu: X \rightarrow V$.

A line $l \subset V$ is $\mu$ regular if the composition morphism $pr\circ \mu: X \rightarrow V/l$ is finite.

Let $S$ denote the set of $\mu$ regular lines. Then I want to show that $S$ contains a dense open subset in $\mathbb{P}(V)$, the projective space.

Any references or answers or hints are welcomed.

edit: I forgot to mention that I suppose that $X$ has $\dim X <n$.

Write $k[x_1,...,x_n]$ for the coordinate ring of $V$ and $X=Spec(R)$. The map $\mu^*: k[x_1,...,x_n] \rightarrow R$ has kernel $I$. It will be enough to show this in the case $X=Spec(k[x_1,...,x_n]/I)$ and $\mu^*$ is the quotient map.
Suppose the line $l$ is of the form $(a_1 t, a_2 t, ..., a_n t)$ as $t$ varies in $k$. The elements $y_i = a_nx_i - a_ix_n$ for $1 \leq i < n$ are in the image of the map $$k[V/l] \rightarrow k[V].$$
Why do the images of $y_i$ in $k[x_1,...,x_n]/I$ generate the algebra $k[x_1,...,x_n]/I$ for generic choices of $a_1,...,a_n$?
• I understand that they generate this algebra, because $y_i$ are combinations of $x_n,x_i$ for $1\leq i <n$, and $y_i$ never vanish (I mean I can pick coeffecients $a_i$ such that they never get vanish). Can you please elaborate how construct this open dense set? Thanks in advance. Feb 6, 2014 at 10:24
• Try this example. If $n=2$ and $I = (x_1-x_2)$ what are the conditions I'd need to require on $a_1$ and $a_2$? Feb 6, 2014 at 18:00
• $a_1=a_2=1$, I think. Feb 6, 2014 at 19:51
• $a_1=-1, a_2 =1$ works. Remember you want it to be an open condition, not a closed one. What other choices of $a_1$ and $a_2$ will result in $a_2x_1-a_1x_2$ generating the $k$-algebra $k[x_1,x_2]/(x_1-x_2)$? (Caveat: $char(k)$ not 2 here) Feb 6, 2014 at 19:56
• $a_1 = -a_2$. Can you please elaborate how does this relate to the question I asked, i.e that $S$ has an open dense set in $\mathbb{P}(V)$; Thanks. Feb 6, 2014 at 20:19