Timeline for Infinite residue field extensions and algebraic closure of residue fields
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2013 at 22:54 | comment | added | Will Chen | In particular, using the same argument you can show that there is some integer $n$ for which $X^n - x_1$ has no root in $K(x_1,...,x_n)$. Now you just have to convince yourself that if $X^n - x_1$ had a root in $L(x_1,...,x_n)$, then it would have to have a root in $K(x_1,...,x_n)$. This should be pretty clear from our setup. | |
| Sep 5, 2013 at 22:47 | comment | added | Will Chen | extending the base from $K$ to any extension field $L$ will result in the residue field of that point becoming $L(x_1,..,x_n)$, where the $x_i$'s again satisfy the same relation. The residue field embedding then is just defined by sending $x_i\mapsto x_i$, and sending $K\hookrightarrow L$. This residue field extension can't possibly factor through the algebraic closure of $K(x_1,...,x_n)$. Why? Well, first note that $K(x_1,...,x_n)$ cannot be algebraically closed (this follows from the fact that it's the fraction field of a FINITELY generated $K$-algebra). | |
| Sep 5, 2013 at 22:44 | comment | added | Will Chen | If you want something more formal, we could argue as follows. A scheme of finite type over $K$ is given locally by a finitely generated $K$-algebra. Ie, something like $K[x_1,...,x_n]/(f_1,...,f_m)$. Consider the generic point of some irreducible component. This point has residue field equal to the function field of the irreducible component, which is just the fraction field of some integral finitely generated $K$-algebra. These fraction fields basically look like $K(x_1,...,x_n)$, where the $x_i$'s may satisfy some equation with coefficients in $K$. | |
| Sep 5, 2013 at 22:33 | comment | added | Will Chen | It's not (except maybe in some trivial situations like when your scheme is Spec $K$ or something). Basically, you should ask yourself, where can the $t$ go? The natural embedding $K(t) = k(p(x))\hookrightarrow k(x) L(t)$ has to send the coordinate $t$ to, essentially, $t$. Any such embedding $k(p(x)) \hookrightarrow k(x)$ coming from a change of base morphism $p$ will essentially only be realized on the coefficient field. The variables (in our case $t$) will still be mapped to $t$, and is necessarily algebraically independent from the elements of the coefficient field. | |
| Sep 5, 2013 at 20:27 | comment | added | Georg S. | You are of course right about the containment! I probably want the embedding $k(p(x)) \hookrightarrow k(x)$ to factor through an algebraic closure of $k(p(x))$. Is this possible? | |
| Sep 5, 2013 at 20:20 | history | answered | Will Chen | CC BY-SA 3.0 |