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let $L/K$ be an algebraic field extension. denote by $K(T)= Frac(K[T])$ the transcendental field extension of $K$. I would like to find out how to show that the equation $$K(T) \otimes_K L = L(T)$$

holds and especially where the requirement that $L$ algebraic flows in. That seems to be essential since for transcendent $L:= K(T)$ the formula above fails since $K(T) \otimes_K K(T)$ is not a field.

Thanks in advance!

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by induction you must show that $K(T)\otimes K[\alpha]=K[\alpha](T)$ write $K[\alpha]=\frac{K[T]}{(f)}$ and prove the morphism $g\otimes (h+(f))\to g\pi(h)$ ($\pi$ is the natural morphism from $K[T]/(f)$ to L)is isomorphism.

(construct the converse by sending $\sum f_i(\alpha)T^i$ to $\sum T^i\otimes (f_i+(f))$)

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  • $\begingroup$ induction on degree of $f$? how to realize the induction step? we have to reduce the problem to a irreducible poynomial of lower degree then $deg(f)$ $\endgroup$ – Tim Grosskreutz Apr 3 at 17:42
  • $\begingroup$ no I mean L is algebraic over K so in general you need more than one algebraic elements to generate it over K(although if the base field is prefect you need only one element) so induction on the number of generators of L over K $\endgroup$ – ali Apr 3 at 17:46
  • $\begingroup$ not every algebraic extension is finite. or can this argument be passed to direct limits? $\endgroup$ – Tim Grosskreutz Apr 3 at 17:54
  • $\begingroup$ yes you must the direct limit $\endgroup$ – ali Apr 3 at 17:55
  • $\begingroup$ let me summarize it: by inductive limit argument we reduce the problem to finite algebraic extension $L =K(a_1,..., a_n)= L'(a_n)$ with $L'= K(a_1,..., a_{n-1})$. then $L \otimes_K K(T)= L'(a_n)\otimes_{L'} L' \otimes_K K(T)= L'(a_n)\otimes_{L'}L'(T)$ by ih. so wlog L=K(a)$. and for this you give explicitely an iso $\endgroup$ – Tim Grosskreutz Apr 3 at 18:07

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