Timeline for Proof of the result that the Galois group of a specialization is a subgroup of the original group?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2016 at 14:44 | comment | added | 352506 | @KConrad, is the condition that $f(t_0,x)$ is separable enough to imply that $B/\mathfrak P$ is the splitting field of $f(t_0,x)$? I can only prove this if I exclude finitely many additional values of $t_0$. | |
| May 19, 2016 at 11:34 | comment | added | 352506 | Why is $B/\mathfrak P$ the splitting field of $f(t_0,x)$? I can only see that it contains a splitting field. I can show it is a splitting field only if I exclude finitely many values $t_0$: choose a primitive element $z$ for $E/F$ that is a $\mathbb Q[t]$-linear combination of the $x$-roots of $f(t,x)$, and let $P$ be its minimal polynomial over $\mathbb Q(t)$. If the discriminant of $P(t_0,x)$ is nonzero, then I can show that $B/\mathfrak P$ is contained in the splitting field of $f(t_0,x)$. Is there a way to prove this without excluding any value of $t_0$? | |
| Aug 5, 2010 at 14:50 | comment | added | Emerton | Dear Keith, I agree completely. | |
| Aug 5, 2010 at 14:47 | vote | accept | Adam | ||
| Aug 5, 2010 at 7:33 | history | edited | KConrad | CC BY-SA 2.5 |
added 206 characters in body
|
| Aug 5, 2010 at 7:30 | comment | added | KConrad | Matt: That was an accidental omission. I meant to assume that $f(x,t)$ is smooth so that $A'$ is int. closed. But since the question being asked is local on the choice of $t_0$, such a global hypothesis is not really necessary and the soln. I wrote doesn't even require knowing $A'$ concretely: let $F'$ be $F$ adjoined with one $x$-root of $f(x,t)$ and then let $A'$ be int. closure of $A$ in $F'$. The soln. barely uses $A'$ or $F'$ (just to know unram. prime stays unram. in Galois closure) and it'd be better to have defined $A$ after picking $t_0$ as the localn. of $\mathbf Q[t]$ at $t-t_0$. | |
| Aug 5, 2010 at 5:16 | comment | added | Emerton | Dear Keith, Are you sure that $A'$ is integrally closed in general? What if $f(x,t)$ were $x^3 - t^2$? The rest is okay though, since, as Brian notes in this comment, $A'$ will be integrally closed in a n.h. of a separable specialization $t_0$, since it will be etale over $A$ there. | |
| Aug 5, 2010 at 2:19 | comment | added | BCnrd | The SGA1 version: For infinite field $L$ of any char. (e.g., $\mathbf{Q}$) & $f \in L[t,x]$ with $x$-deg $n > 0$ & $f$ sep'ble over $L(t)$, let $c, d \in L[t]$ resp. be lead $x$-coeff. and $x$-discr, $A := L[t][1/cd]$. Let $K/L(t)$ be finite Galois ext'n, Gal. gp $G$, splitting $f$, so splits $L(t)[x]/(f)$. Since $A[x]/(f)$ is finite etale $A$-alg, int. closure $B$ in $K$ is finite etale $G$-torsor over $A$ splitting $f$. Specialize at $L$-pt $t_0$ of $A$ (!) to get finite etale $G$-torsor $B_0$ over $L$ splitting $f(t_0,x)$. A factor field of $B_0$ is Galois over $L$ with Gal. gp in $G$. QED | |
| Aug 4, 2010 at 23:54 | history | edited | KConrad | CC BY-SA 2.5 |
added 160 characters in body
|
| Aug 4, 2010 at 22:04 | comment | added | KConrad | Pete: I did leave it to Adam to check $B/\mathfrak P$ is the Galois closure of $g(x)$, so I will not be posting this elsewhere with all details. | |
| Aug 4, 2010 at 21:31 | comment | added | Pete L. Clark | Very nice answer. Blurbworthy, perhaps? | |
| Aug 4, 2010 at 21:13 | history | answered | KConrad | CC BY-SA 2.5 |