Field Extension over the Field of Rational Functions is Finite












1












$begingroup$


Let $K$ be a field and $K(X)$ the field of rational functions with one variable over $K$. If $T(X) in K(X)$ show that the field extension $K(X)/K(T)$ is finite.




The definition of the field of rational function was:



$$K(X):={frac{f}{g} | f,g in K[X], g neq 0}$$



The field operations for $K(X)$ are:



$$frac{f_1}{g_1} + frac{f_1}{g_1} = frac{f_1 g_2 + f_2 g_1}{g_1 g_2}$$ $$frac{f_1}{g_1} + frac{f_1}{g_1} = frac{f_1 f_2}{g_1 g_2}$$




I have the feeling it has something to do with separability. There is a subfield of $K(T)$ which is isomorph to $mathbb{Q}$, therefore the charateristic of $K(T)$ is $0$, which means it is a perfect field. Then $K(X)/K(T)$ is seperable.



But I have no idea how to connect this to $[K(X):K(T)]<infty$. Maybe I have a completely wrong approach. Any help would be appreciated and I'm sorry if this is a duplicate.






abstract-algebra field-theory extension-field rational-functions function-fields







share|cite|improve this question











$endgroup$












  • $begingroup$
    I meant there is a subfield of $K(T)$ such that it is isomorphic to $mathbb{Q}$. I will correct it.
    $endgroup$
    – matt
    Nov 29 '18 at 23:05






  • 3




    $begingroup$
    Just show that $X$ is algebraic over $K(T)$
    $endgroup$
    – user8268
    Nov 29 '18 at 23:59






  • 1




    $begingroup$
    Is $T(X)$ the same thing as $T$?
    $endgroup$
    – Servaes
    Nov 30 '18 at 0:30






  • 1




    $begingroup$
    $T(X) =frac{sum_{j=0}^J a_j X^j}{sum_{l=0}^L b_l X^l}$ then $X$ is a root of $T(X)sum_{l=0}^L b_l z^l-sum_{j=0}^J a_j z^jin K(T(X))[z]$. If $gcd(sum_{j=0}^J a_j X^j,sum_{l=0}^L b_l X^l) = 1$ then it is its minimal polynomial.
    $endgroup$
    – reuns
    Nov 30 '18 at 1:13
















1












$begingroup$


Let $K$ be a field and $K(X)$ the field of rational functions with one variable over $K$. If $T(X) in K(X)$ show that the field extension $K(X)/K(T)$ is finite.




The definition of the field of rational function was:



$$K(X):={frac{f}{g} | f,g in K[X], g neq 0}$$



The field operations for $K(X)$ are:



$$frac{f_1}{g_1} + frac{f_1}{g_1} = frac{f_1 g_2 + f_2 g_1}{g_1 g_2}$$ $$frac{f_1}{g_1} + frac{f_1}{g_1} = frac{f_1 f_2}{g_1 g_2}$$




I have the feeling it has something to do with separability. There is a subfield of $K(T)$ which is isomorph to $mathbb{Q}$, therefore the charateristic of $K(T)$ is $0$, which means it is a perfect field. Then $K(X)/K(T)$ is seperable.



But I have no idea how to connect this to $[K(X):K(T)]<infty$. Maybe I have a completely wrong approach. Any help would be appreciated and I'm sorry if this is a duplicate.






abstract-algebra field-theory extension-field rational-functions function-fields







share|cite|improve this question











$endgroup$












  • $begingroup$
    I meant there is a subfield of $K(T)$ such that it is isomorphic to $mathbb{Q}$. I will correct it.
    $endgroup$
    – matt
    Nov 29 '18 at 23:05






  • 3




    $begingroup$
    Just show that $X$ is algebraic over $K(T)$
    $endgroup$
    – user8268
    Nov 29 '18 at 23:59






  • 1




    $begingroup$
    Is $T(X)$ the same thing as $T$?
    $endgroup$
    – Servaes
    Nov 30 '18 at 0:30






  • 1




    $begingroup$
    $T(X) =frac{sum_{j=0}^J a_j X^j}{sum_{l=0}^L b_l X^l}$ then $X$ is a root of $T(X)sum_{l=0}^L b_l z^l-sum_{j=0}^J a_j z^jin K(T(X))[z]$. If $gcd(sum_{j=0}^J a_j X^j,sum_{l=0}^L b_l X^l) = 1$ then it is its minimal polynomial.
    $endgroup$
    – reuns
    Nov 30 '18 at 1:13














1












1








1





$begingroup$


Let $K$ be a field and $K(X)$ the field of rational functions with one variable over $K$. If $T(X) in K(X)$ show that the field extension $K(X)/K(T)$ is finite.




The definition of the field of rational function was:



$$K(X):={frac{f}{g} | f,g in K[X], g neq 0}$$



The field operations for $K(X)$ are:



$$frac{f_1}{g_1} + frac{f_1}{g_1} = frac{f_1 g_2 + f_2 g_1}{g_1 g_2}$$ $$frac{f_1}{g_1} + frac{f_1}{g_1} = frac{f_1 f_2}{g_1 g_2}$$




I have the feeling it has something to do with separability. There is a subfield of $K(T)$ which is isomorph to $mathbb{Q}$, therefore the charateristic of $K(T)$ is $0$, which means it is a perfect field. Then $K(X)/K(T)$ is seperable.



But I have no idea how to connect this to $[K(X):K(T)]<infty$. Maybe I have a completely wrong approach. Any help would be appreciated and I'm sorry if this is a duplicate.






abstract-algebra field-theory extension-field rational-functions function-fields







share|cite|improve this question











$endgroup$




Let $K$ be a field and $K(X)$ the field of rational functions with one variable over $K$. If $T(X) in K(X)$ show that the field extension $K(X)/K(T)$ is finite.




The definition of the field of rational function was:



$$K(X):={frac{f}{g} | f,g in K[X], g neq 0}$$



The field operations for $K(X)$ are:



$$frac{f_1}{g_1} + frac{f_1}{g_1} = frac{f_1 g_2 + f_2 g_1}{g_1 g_2}$$ $$frac{f_1}{g_1} + frac{f_1}{g_1} = frac{f_1 f_2}{g_1 g_2}$$




I have the feeling it has something to do with separability. There is a subfield of $K(T)$ which is isomorph to $mathbb{Q}$, therefore the charateristic of $K(T)$ is $0$, which means it is a perfect field. Then $K(X)/K(T)$ is seperable.



But I have no idea how to connect this to $[K(X):K(T)]<infty$. Maybe I have a completely wrong approach. Any help would be appreciated and I'm sorry if this is a duplicate.






abstract-algebra field-theory extension-field rational-functions function-fields




abstract-algebra field-theory extension-field rational-functions function-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 13:12









Servaes

23.9k33893




23.9k33893










asked Nov 29 '18 at 22:56









mattmatt

857




857












  • $begingroup$
    I meant there is a subfield of $K(T)$ such that it is isomorphic to $mathbb{Q}$. I will correct it.
    $endgroup$
    – matt
    Nov 29 '18 at 23:05






  • 3




    $begingroup$
    Just show that $X$ is algebraic over $K(T)$
    $endgroup$
    – user8268
    Nov 29 '18 at 23:59






  • 1




    $begingroup$
    Is $T(X)$ the same thing as $T$?
    $endgroup$
    – Servaes
    Nov 30 '18 at 0:30






  • 1




    $begingroup$
    $T(X) =frac{sum_{j=0}^J a_j X^j}{sum_{l=0}^L b_l X^l}$ then $X$ is a root of $T(X)sum_{l=0}^L b_l z^l-sum_{j=0}^J a_j z^jin K(T(X))[z]$. If $gcd(sum_{j=0}^J a_j X^j,sum_{l=0}^L b_l X^l) = 1$ then it is its minimal polynomial.
    $endgroup$
    – reuns
    Nov 30 '18 at 1:13


















  • $begingroup$
    I meant there is a subfield of $K(T)$ such that it is isomorphic to $mathbb{Q}$. I will correct it.
    $endgroup$
    – matt
    Nov 29 '18 at 23:05






  • 3




    $begingroup$
    Just show that $X$ is algebraic over $K(T)$
    $endgroup$
    – user8268
    Nov 29 '18 at 23:59






  • 1




    $begingroup$
    Is $T(X)$ the same thing as $T$?
    $endgroup$
    – Servaes
    Nov 30 '18 at 0:30






  • 1




    $begingroup$
    $T(X) =frac{sum_{j=0}^J a_j X^j}{sum_{l=0}^L b_l X^l}$ then $X$ is a root of $T(X)sum_{l=0}^L b_l z^l-sum_{j=0}^J a_j z^jin K(T(X))[z]$. If $gcd(sum_{j=0}^J a_j X^j,sum_{l=0}^L b_l X^l) = 1$ then it is its minimal polynomial.
    $endgroup$
    – reuns
    Nov 30 '18 at 1:13
















$begingroup$
I meant there is a subfield of $K(T)$ such that it is isomorphic to $mathbb{Q}$. I will correct it.
$endgroup$
– matt
Nov 29 '18 at 23:05




$begingroup$
I meant there is a subfield of $K(T)$ such that it is isomorphic to $mathbb{Q}$. I will correct it.
$endgroup$
– matt
Nov 29 '18 at 23:05




3




3




$begingroup$
Just show that $X$ is algebraic over $K(T)$
$endgroup$
– user8268
Nov 29 '18 at 23:59




$begingroup$
Just show that $X$ is algebraic over $K(T)$
$endgroup$
– user8268
Nov 29 '18 at 23:59




1




1




$begingroup$
Is $T(X)$ the same thing as $T$?
$endgroup$
– Servaes
Nov 30 '18 at 0:30




$begingroup$
Is $T(X)$ the same thing as $T$?
$endgroup$
– Servaes
Nov 30 '18 at 0:30




1




1




$begingroup$
$T(X) =frac{sum_{j=0}^J a_j X^j}{sum_{l=0}^L b_l X^l}$ then $X$ is a root of $T(X)sum_{l=0}^L b_l z^l-sum_{j=0}^J a_j z^jin K(T(X))[z]$. If $gcd(sum_{j=0}^J a_j X^j,sum_{l=0}^L b_l X^l) = 1$ then it is its minimal polynomial.
$endgroup$
– reuns
Nov 30 '18 at 1:13




$begingroup$
$T(X) =frac{sum_{j=0}^J a_j X^j}{sum_{l=0}^L b_l X^l}$ then $X$ is a root of $T(X)sum_{l=0}^L b_l z^l-sum_{j=0}^J a_j z^jin K(T(X))[z]$. If $gcd(sum_{j=0}^J a_j X^j,sum_{l=0}^L b_l X^l) = 1$ then it is its minimal polynomial.
$endgroup$
– reuns
Nov 30 '18 at 1:13










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$begingroup$

Let $f,gin K[X]$ be such that $T=frac{f}{g}$, and consider the polynomial
$$h:=Tg(Y)-f(Y)in (K(T))[Y].$$
Note that $h=0$ if and only if $f=g$, in which case $Tin K$ is constant and the extension $K(X)/K(T)$ is in fact not finite.



If $hneq0$ then the fact that $h(X)=0$ implies that $X$ is algebraic over $K(T)$, and hence that $K(X)$ is a finite extension of $K(T)$.






share|cite|improve this answer











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    $begingroup$

    Let $f,gin K[X]$ be such that $T=frac{f}{g}$, and consider the polynomial
    $$h:=Tg(Y)-f(Y)in (K(T))[Y].$$
    Note that $h=0$ if and only if $f=g$, in which case $Tin K$ is constant and the extension $K(X)/K(T)$ is in fact not finite.



    If $hneq0$ then the fact that $h(X)=0$ implies that $X$ is algebraic over $K(T)$, and hence that $K(X)$ is a finite extension of $K(T)$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Let $f,gin K[X]$ be such that $T=frac{f}{g}$, and consider the polynomial
      $$h:=Tg(Y)-f(Y)in (K(T))[Y].$$
      Note that $h=0$ if and only if $f=g$, in which case $Tin K$ is constant and the extension $K(X)/K(T)$ is in fact not finite.



      If $hneq0$ then the fact that $h(X)=0$ implies that $X$ is algebraic over $K(T)$, and hence that $K(X)$ is a finite extension of $K(T)$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $f,gin K[X]$ be such that $T=frac{f}{g}$, and consider the polynomial
        $$h:=Tg(Y)-f(Y)in (K(T))[Y].$$
        Note that $h=0$ if and only if $f=g$, in which case $Tin K$ is constant and the extension $K(X)/K(T)$ is in fact not finite.



        If $hneq0$ then the fact that $h(X)=0$ implies that $X$ is algebraic over $K(T)$, and hence that $K(X)$ is a finite extension of $K(T)$.






        share|cite|improve this answer











        $endgroup$



        Let $f,gin K[X]$ be such that $T=frac{f}{g}$, and consider the polynomial
        $$h:=Tg(Y)-f(Y)in (K(T))[Y].$$
        Note that $h=0$ if and only if $f=g$, in which case $Tin K$ is constant and the extension $K(X)/K(T)$ is in fact not finite.



        If $hneq0$ then the fact that $h(X)=0$ implies that $X$ is algebraic over $K(T)$, and hence that $K(X)$ is a finite extension of $K(T)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 13:17

























        answered Nov 30 '18 at 0:43









        ServaesServaes

        23.9k33893




        23.9k33893






























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