# coefficient extension for fraction field $K(T) \otimes_K L$

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.

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 $$\sum T^i\otimes (f_i+(f))$$)
• 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 $deg(f)$ – Tim Grosskreutz Apr 3 at 17:42
• 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 – Tim Grosskreutz Apr 3 at 18:07