- #1
mathmari
Gold Member
MHB
- 4,984
- 7
Hey! 
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)
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)