# Tensor Product of Quotient.

Let $$K$$ be a field and $$L/K$$ a field extension. Suppose $$A$$ is a $$K$$-algebra and $$I$$ is an ideal. I want to show that $$(A/I\otimes_K L) \to (A\otimes_K L)/(I\otimes_K L)$$

So i define a map $$f: A\otimes_K L \to (A/I)\otimes_K L$$ by $$f(a\otimes \lambda) =(a+I) \otimes \lambda$$ So I wish to show that the kernel of $$f$$ is $$(I\otimes_K L)$$. It is clear to me that $$(I\otimes_K L) \subseteq \operatorname{Ker}(f)$$.

However, I'm having difficulties showing the reverse inclusion. If $$\{b_i \}_i$$ is a $$K$$-basis for $$L$$, then if $$a\otimes \lambda \in \operatorname{Ker}(f)$$, then we write $$\lambda = \sum_i \alpha_ib_i$$ for $$\alpha_i\in K$$. Then $$0=f(a\otimes \lambda) = \sum_i f(\alpha_ia\otimes b_i) = \sum_i (\alpha_ia+I)\otimes b_i.$$ Now, I want to conclude that since the sum is zero, we must have that each term is zero and hence $$\alpha_i a+I=0$$. But I'm not sure I can conclude this. Any help would be greatly appreciated.

• Do you know about flat modules? – Bernard Feb 2 at 21:56

Your statement can proved very briefly by applying exactness properties of tensor functor as follows. First note the exact sequence of $$K$$-modules $$\{0\}\to I\to A\to A/I\to\{0\}$$ Since $$L$$ is a free, hence flat, $$K$$-module, tensoring with $$L$$ gives the exact sequence of $$L$$-modules $$\{0\}\to I\otimes_KL\to A\otimes_KL\to(A/I)\otimes_KL\to\{0\}$$ from which we get an isomorphisms of $$L$$-module $$(A/I)\otimes_KL\cong(A\otimes_KL)/(I\otimes_KL)$$