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I am not sure whether there exists a specific procedure for the computation of certain tensor products. For instance, suppose we are required to compute $ k[t]/(t-\lambda) \otimes_{k[t]} k[x,y,t]/(xy-\lambda), $ where $ k $ is an algebraically closed field and $ \lambda \in k. $

Unless I am mistaken,

$ k[t]/(t-\lambda) \otimes_{k[t]} k[x,y,t]/(xy-t) \cong k[x,y,t]/(t-\lambda,xy-t) = k[x,y]/(xy-\lambda). $

I have only been able to state acknowledge the above isomorphism(if it is indeed correct) by my having seen similar statements several times before. But since this is no way to learn, I'm wondering what the proper way to explain such results is.

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    $\begingroup$ This is an instance of $R/I\otimes_R M\cong M/IM$. $\endgroup$ – Lord Shark the Unknown Jan 25 at 6:26
  • $\begingroup$ I see. Will this be in Atiyah-Macdonald? $\endgroup$ – Confused Student Jan 25 at 6:29
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    $\begingroup$ @OverwhelmedAGApprentice probably it's in Atiyah-Macdonald-- but in this case you can prove it with your bare hands -- the map is induced by $r \otimes m \mapsto rm$ with inverse map induced by $m \mapsto 1 \otimes m$. You have to check that the forward map is well-defined on the tensor product, i.e. bilinear (it is) and that the inverse map is well defined mod $IM$ (it is). Then these two maps are clearly inverse to each other so you win. $\endgroup$ – hunter Jan 25 at 7:07

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