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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.

Bernard
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Rdrr
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1 Answers1

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

Fabio Lucchini
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  • 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. – Rdrr Feb 02 '19 at 22:21