MHB Show That $F(S)$ Is the Smallest Subfield of $K$

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

Let $F$ be a subfield of the field $K$ and $S$ a non-empty subset (not necessarily, subfield) of $K$, finite or infinite. Let $F(S)$ be the subset of $K$ that is defined as follows:
An element $u\in K$ is in $F(S)$ iff there are finitely many elements of $S$, say $s_1, \dots , s_n$, and polynomials with $n$ variables $f,g\in F[x_1, \dots , x_n]$, with $g(s_1, \dots , s_n)\neq 0$, so that $u=f(s_1, \dots , s_n)/g(s_1, \dots , s_n)$.

I want to show that $F(S)$ is a subfield of $K$ and even the smallest subfield of $K$, that contains $F$ and $S$. So, if $E$ is the subfield of $K$ and $F\cup S\subseteq E$, then $F(S)\subseteq E$. I have done the following:

$F(S)$ is a subring of $K$ :

We have that $a,b\in F(S)$ then $a=\frac{f_1(s_1, \dots , s_n)}{g_1(s_1, \dots , s_n)}$ and $b=\frac{f_2(s_1, \dots , s_n)}{g_2(s_1, \dots , s_n)}$.
Then $a\cdot b=\frac{f_1(s_1, \dots , s_n)f_2(s_1, \dots , s_n)}{g_1(s_1, \dots , s_n)g_2(s_1, \dots , s_n)}\in F(S)$ and $a-b=\frac{f_1(s_1, \dots , s_n)g_2(s_1, \dots , s_n)-f_2(s_1, \dots , s_n)g_1(s_1, \dots , s_n)}{g_1(s_1, \dots , s_n)g_2(s_1, \dots , s_n)}\in F(S)$.

Since $K$ is a field, it is also an integral domain and since $F(S)$ is a subring of $K$, $F(S)$ is also an integral domain.

We have that $F(S)$ is a subfield of $K$, since it is an integral domain and for each non-zero element $u\in F(S)$, $u=\frac{f(s_1, \dots , s_n)}{g(s_1, \dots , s_n)}$ there is its inverse, $u^{-1}=\frac{g(s_1, \dots , s_n)}{f(s_1, \dots , s_n)}$.
Let $E$ a subfield of $K$ and $F\cup S\subseteq E$.
The elements of $F(S)$ are of the form $\frac{f(s_1, \dots , s_n)}{g(s_1, \dots , s_n)}$. Since $F\cup S\subseteq E$ we have that $f(s_1, \dots , s_n)\in E$ and $g(s_1, \dots , s_n)\in E$. Since $E$ is a field we have that $\frac{f(s_1, \dots , s_n)}{g(s_1, \dots , s_n)}\in E$. Therefore, $F(S)\subseteq E$. Is everything correct? Could I improve something? (Wondering)
 
Physics news on Phys.org
mathmari said:
Hey! :o

Let $F$ be a subfield of the field $K$ and $S$ a non-empty subset (not necessarily, subfield) of $K$, finite or infinite. Let $F(S)$ be the subset of $K$ that is defined as follows:
An element $u\in K$ is in $F(S)$ iff there are finitely many elements of $S$, say $s_1, \dots , s_n$, and polynomials with $n$ variables $f,g\in F[x_1, \dots , x_n]$, with $g(s_1, \dots , s_n)\neq 0$, so that $u=f(s_1, \dots , s_n)/g(s_1, \dots , s_n)$.

I want to show that $F(S)$ is a subfield of $K$ and even the smallest subfield of $K$, that contains $F$ and $S$. So, if $E$ is the subfield of $K$ and $F\cup S\subseteq E$, then $F(S)\subseteq E$. I have done the following:

$F(S)$ is a subring of $K$ :

We have that $a,b\in F(S)$ then $a=\frac{f_1(s_1, \dots , s_n)}{g_1(s_1, \dots , s_n)}$ and $b=\frac{f_2(s_1, \dots , s_n)}{g_2(s_1, \dots , s_n)}$.
Then $a\cdot b=\frac{f_1(s_1, \dots , s_n)f_2(s_1, \dots , s_n)}{g_1(s_1, \dots , s_n)g_2(s_1, \dots , s_n)}\in F(S)$ and $a-b=\frac{f_1(s_1, \dots , s_n)g_2(s_1, \dots , s_n)-f_2(s_1, \dots , s_n)g_1(s_1, \dots , s_n)}{g_1(s_1, \dots , s_n)g_2(s_1, \dots , s_n)}\in F(S)$.

Since $K$ is a field, it is also an integral domain and since $F(S)$ is a subring of $K$, $F(S)$ is also an integral domain.

We have that $F(S)$ is a subfield of $K$, since it is an integral domain and for each non-zero element $u\in F(S)$, $u=\frac{f(s_1, \dots , s_n)}{g(s_1, \dots , s_n)}$ there is its inverse, $u^{-1}=\frac{g(s_1, \dots , s_n)}{f(s_1, \dots , s_n)}$.
Let $E$ a subfield of $K$ and $F\cup S\subseteq E$.
The elements of $F(S)$ are of the form $\frac{f(s_1, \dots , s_n)}{g(s_1, \dots , s_n)}$. Since $F\cup S\subseteq E$ we have that $f(s_1, \dots , s_n)\in E$ and $g(s_1, \dots , s_n)\in E$. Since $E$ is a field we have that $\frac{f(s_1, \dots , s_n)}{g(s_1, \dots , s_n)}\in E$. Therefore, $F(S)\subseteq E$. Is everything correct? Could I improve something? (Wondering)
This seems fine to me.
 
mathmari said:
Let $E$ a subfield of $K$ and $F\cup S\subseteq E$.
The elements of $F(S)$ are of the form $\frac{f(s_1, \dots , s_n)}{g(s_1, \dots , s_n)}$. Since $F\cup S\subseteq E$ we have that $f(s_1, \dots , s_n)\in E$ and $g(s_1, \dots , s_n)\in E$. Since $E$ is a field we have that $\frac{f(s_1, \dots , s_n)}{g(s_1, \dots , s_n)}\in E$. Therefore, $F(S)\subseteq E$.

I thought about it again and I got stuck right now...
Why do we have that $f(s_1, \dots , s_n), g(s_1, \dots , s_n) \in F\cup S$ ?
$f(s_1, \dots , s_n), g(s_1, \dots , s_n)$ are polynomials with coefficients in $F$. Are the variables then elements of $S$ ? (Wondering)
 
mathmari said:
Why do we have that $f(s_1, \dots , s_n), g(s_1, \dots , s_n) \in F\cup S$ ?

Or doesn't it stand? (Wondering)
 
caffeinemachine said:
This seems fine to me.

Thank you! (Yes)
 
Back
Top