MHB Does \(B\) Span Algebraically \(E\) Over \(F\)?

  • Thread starter Thread starter mathmari
  • Start date Start date
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

Let $E/F$ be an extension, $S=\{a_1,\ldots,a_n\}\subseteq E$ algebraically independent over $F$ and $S\subseteq T$, $T$ a subset of $E$, that spans $E$ algebraically over $F$.

I want to show that there exists a set $B$ between $S$ and $T$, that is a trancendental basis of $E/F$, as follows:

Let $T\setminus S=\{\beta_1,\ldots ,\beta_m\}$.

If $T=\varnothing$, then $B=S$ is the trancendental basis.

Otherwise, we define $S_0=S$ and for $i=1,\ldots ,m$

$S_i=\left\{\begin{matrix} S_{i-1} & \text{ if } \beta_i \text{ is algebraic } /F(S_{i-1})\\ S_{i-1}\cup \{\beta_i\} & \text{ if } \beta_i \text{ is not algebraic } /F(S_{i-1}) \end{matrix}\right.$

I want to show that that $B=S_m$ is the trancendental basis of $E/F$.

I have shown that $S_m$ is $F$-algebraically independent.

So, it is left to show that $S_m$ spans algebraically $E$ over $F$. Could you give me some hints how we could show that? (Wondering)
 
Physics news on Phys.org
To show that $S_m$ spans algebraically $E$ over $F$, we have to show that $E/F(S_m)$ is algebraic, right?

We have that the extension $E/F(T)$ is algebraic.

It holds that $S_m\subseteq T$, or not?

Do we know if $F(S_m)\leq F(T)$ or $F(T)\leq F(S_m)$ ? (Wondering)
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
Back
Top