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.

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

What you are looking for is a direct proof of the right exactness of the tensor product functor. This is usually proved by applying left exactness of hom functor. However, you can find a direct proof here.

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 $$(A/I)\otimes_KL\cong(A\otimes_KL)/(I\otimes_KL)$$

  • $\begingroup$ I like this proof but i'm looking to see why this isomorphism holds, so i'm want to get "my hands dirty" so to speak and complete the above proof. $\endgroup$ – Rdrr Feb 2 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.