Timeline for algebraic closure of a subgroup of GL
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 5, 2011 at 21:28 | answer | added | shamovic | timeline score: 2 | |
| Mar 5, 2011 at 19:31 | comment | added | Johannes Ebert | But only by a second. | |
| Mar 5, 2011 at 18:33 | comment | added | Steven Landsburg | Looks like Johannes beat me to this. | |
| Mar 5, 2011 at 18:33 | comment | added | Steven Landsburg | If Mariano has guessed the definitions correctly, then the "linear span" of $\Gamma$ is closed and contains $\Gamma$, hence contains the Zariski closure of $\Gamma$. The fact that $\Gamma$ is a group has nothing to do with it. | |
| Mar 5, 2011 at 18:28 | comment | added | Johannes Ebert | Let $V$ be a finitedimensional $C$-vector space and $S \subset V$ a set. Let $x$ be in the Zariski closure of $S$. By definition, each polynomial that vanishes on $S$ has to vanish at $x$. Take a linear form $f$ that vanishes on $S$. Since $f$ is a polynomial, it follows $f(x)=0$. Thus any linear form which is trivial on $S$ is trivial on the Zariski closure. Conclusion: the Zariski closure is a subset of the linear span. So the answer is yes, and has nothing to do with groups. | |
| Mar 5, 2011 at 18:11 | comment | added | Mariano Suárez-Álvarez | By linear span you mean the subspace spanned by $\Gamma$ in the vector space of matrices? By algebraic closure you mean the Zariski closure? | |
| Mar 5, 2011 at 18:08 | history | asked | user9552 | CC BY-SA 2.5 |