Let $k$ be a field and $R=k[x]$. Let $M=\dfrac{k[x, y]}{(x)\cap(x, y)^2}$. How to show that $M$ is not a flat $R$-module?
I am sorry but I am really blank on this one.
Asked
Active
Viewed 655 times
1
user26857
- 53,190
caffeinemachine
- 19,646
-
Your ring only has an $x$, the module has an $x$ and a $y$ and a $t$ in the denominator. Your title had a typo. – Mariano Suárez-Álvarez Sep 20 '15 at 18:10
-
@MarianoSuárez-Alvarez Sorry about that. Fixed it. – caffeinemachine Sep 20 '15 at 19:20
-
Can you give generators for the ideal inthe denominator? – Mariano Suárez-Álvarez Sep 20 '15 at 19:22
-
We have $(x^2, xy)\subseteq (x)\cap (x, y)^2$. I am thinking this must be an equality but am not sure at this point. – caffeinemachine Sep 20 '15 at 19:34
-
Well, a good start would be to make sure! – Mariano Suárez-Álvarez Sep 20 '15 at 19:46
-
@MarianoSuárez-Alvarez We have $I:=(x)\cap (x, y)^2=(x)\cap ((x^2)+(xy)+(y^2))$, giving $I=(x)\cap (x^2)+(x)\cap (xy)+(x)\cap (y^2)$, and thus $I= (x^2)+(xy)=(x,^2 xy)$. – caffeinemachine Sep 21 '15 at 06:43
-
Ok. What criteria do you have to check if a module is flat? – Mariano Suárez-Álvarez Sep 21 '15 at 06:47
-
The definition I know is that $M$ is a flat $R$-module if tensoring with $M$ preserves exactness of short exact sequences. – caffeinemachine Sep 21 '15 at 06:50
-
Are you sure $R$ is supposed to be $k[x]$? – Mariano Suárez-Álvarez Sep 21 '15 at 07:04
2 Answers
1
If $M$ is flat, then it must be torsion-free (see here), but this is not the case: $xy=0$ and $y\ne 0$ in $M$.
0
Let $M=k[x]/(x^2)$, $N=k[x]/(x)$ and $f:M\to N$ the map induced by the identity of $k[x]$. Let $K$ be the kernel of $f$, so that we have an exact sequence $$0\to K\to M\to N\to 0$$ What happens if you tensor this with $k[x,y]/(x^2,xy)$?
Mariano Suárez-Álvarez
- 139,300