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I would like to ask if there is a simple way to prove the non-flatness of the above morphism of rings using just the definition of a flat module.

agleaner
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1 Answers1

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Consider the injection $(s,t) \rightarrow \mathbb{C}[s,t]$. Tensor with $\mathbb{C}[x,y]$ and notice that the result is not injective. Indeed, the natural map $(s,t) \otimes_{\mathbb{C}[s,t]}\mathbb{C}[x,y]\rightarrow \mathbb{C}[x,y]$ sends $s \otimes y$ and $t \otimes 1$ to the same place. It is easy to check that these are not the same element of the tensor product.

In general, if something is not flat then there is some finitely generated ideal $I \subset R$ that will witness this failure.

Dylan Wilson
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  • Dylan, many thanks for your answer! – agleaner Nov 29 '12 at 14:13
  • In fact, I have one more question. How have you found out this proof, i.e. why did you chose $I=(s,t)$? What is the intuition? – agleaner Nov 29 '12 at 14:31
  • I knew there had to be some finitely-generated ideal that worked... and it sort of felt right that the equation $sy = t$ would be important for reasons that I don't know how to put in words... so I tried $(s,t)$, and it worked. – Dylan Wilson Nov 29 '12 at 17:17
  • I'm failing to see why $s \otimes y$ and $t \otimes 1$ are distinct in the tensor product. They, of course, LOOK different, but can you give an explanation as to why they are different? Like, why there's no way to rewrite $s \otimes y$ as $t \otimes 1$ with some tricky maneuvering? – gravitybeatle Dec 22 '21 at 19:00