(New page: Show that no finite field is algebraically closed. ---- There are just so many ways this can be proven. Here's one: Consider a finite field <math>\scriptstyle F</math> with elements <mat...) |
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There are just so many ways this can be proven. Here's one: | There are just so many ways this can be proven. Here's one: | ||
| − | Consider a finite field <math>\scriptstyle F</math> with elements <math>\scriptstyle a_1,a_2,\ldots,a_n</math>. Because the polynomial <math>\scriptstyle(x-a_1)(x-a_2)\cdots(x-a_n)+1</math> is | + | Consider a finite field <math>\scriptstyle F</math> with elements <math>\scriptstyle a_1,a_2,\ldots,a_n</math>. Because the polynomial <math>\scriptstyle(x-a_1)(x-a_2)\cdots(x-a_n)+1</math> is irreducible in <math>\scriptstyle F</math>, there exists a splitting field <math>\scriptstyle E</math> for the polynomial over <math>\scriptstyle F</math> by Theorem 20.2. That is, <math>\scriptstyle E</math> is a proper algebraic extension of <math>\scriptstyle F</math>, so <math>\scriptstyle F</math> cannot be algebraically closed. |
:--[[User:Narupley|Nick Rupley]] 00:40, 30 April 2009 (UTC) | :--[[User:Narupley|Nick Rupley]] 00:40, 30 April 2009 (UTC) | ||
Latest revision as of 20:42, 29 April 2009
Show that no finite field is algebraically closed.
There are just so many ways this can be proven. Here's one:
Consider a finite field $ \scriptstyle F $ with elements $ \scriptstyle a_1,a_2,\ldots,a_n $. Because the polynomial $ \scriptstyle(x-a_1)(x-a_2)\cdots(x-a_n)+1 $ is irreducible in $ \scriptstyle F $, there exists a splitting field $ \scriptstyle E $ for the polynomial over $ \scriptstyle F $ by Theorem 20.2. That is, $ \scriptstyle E $ is a proper algebraic extension of $ \scriptstyle F $, so $ \scriptstyle F $ cannot be algebraically closed.
- --Nick Rupley 00:40, 30 April 2009 (UTC)
