I'm asked to prove that the ideal $I=(x,y)$ in $R=k[x,y]$ is not a flat R-module.

My approach was to use the exact sequence $$0\rightarrow I \to R \to R/I \to 0$$ to induce a non injective map $$I\otimes I\to R\otimes I$$ because the element $x\otimes y-y\otimes x$ is sent to $0$ by the inclusion.

But I need to show that such element is not zero in the domain. My intuitive approach is that such element is nonzero because in $I\otimes I$ I'm not allowed to move $x$ or $y$ "inside" the tensor because $1\notin I$. But I cannot formalise it properly. So how to prove that such tensor is not zero?


It might be easier to approach this problem the following way: We have the following free resolution of $I$:

$$0 \to R \to R \oplus R \to I \to 0,$$ where the first map is given by $1 \mapsto (-y,x)$ and the second map is given by $(1,0) \mapsto x, (0,1) \mapsto y$.

After tensoring with $R/I$ we get the following exact sequence:

$$0 \to Tor_1(I,R/I) \to R/I \to R/I \oplus R/I \to I/I^2 \to 0$$

The map $R/I \to R/I \oplus R/I$ is given by $1 \mapsto (-y,x)=0 \in R/I \oplus R/I$, hence this is the zero map. So we deduce $Tor_1(I,R/I) \cong R/I$, in particular $I$ is not flat.

Furthermore we deduce that the kernel of your map (since this kernel is precisely $Tor_1(I,R/I)$) is generated by an element, which is annihilated by $I$. This is just your element $x \otimes y - y \otimes x$.

  • $\begingroup$ Dear MooS, why the sequence with Tor is left exact? I.e. Why is $\text{Tor}^R_1(R/I,R/I)$ trivial? $\endgroup$ – Luigi M Apr 23 '15 at 11:32
  • $\begingroup$ Never mind, figured out! The resolution of $R/I$ has length 1, because it's a field $\endgroup$ – Luigi M Apr 23 '15 at 11:39
  • $\begingroup$ The next term on the left would be $Tor_1(R \oplus R,R/I)$ which is zero, since $R \oplus R$ is free, hence flat. $\endgroup$ – MooS Apr 23 '15 at 13:19
  • $\begingroup$ By the way: Tensoring the exact sequence $$0 \to I \to R \to R/I \to 0$$ with $R/I$, we get an exact sequence $$0 \to Tor_1(R/I,R/I) \to I/I^2 \to R/I$$, hence $Tor_1(R/I,R/I)$ is the kernel of the map $I/I^2 \to R/I$, which is the zero map. We deduce $Tor_1(R/I,R/I) = I/I^2 \neq 0$. But as I said above, this Tor-group does not even matter in this context. $\endgroup$ – MooS Apr 23 '15 at 13:32

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