Showing that a certain map is not flat by explicit counterexample

Question

I wish to show that the injection $k[y^2, y^3] \rightarrow k[y]$ is not flat.

I know of geometric ways to see this, but I wish to see explicitly $k[y^2, y^3]$-modules (or localizations thereof) $0 \rightarrow M' \rightarrow M$ which does not remain injective upon tensoring with $k[y]$. Is such a search futile?

score 6 · Answer 1 · edited Apr 13 '17 at 12:21

I'd like to add a short algebraic proof (although this is not exactly what you asked for) which shows again the usefulness of Exercise 7.2 in Matsumura, Commutative Ring Theory:

If $A\subset B$ are integral domains with the same field of fractions and $B$ is faithfully flat over $A$, then $A=B$.

In your case $A=k[y^2,y^3]$ and $B=k[y]$. If the extension is flat, then it is faithfully flat (since it is integral) and therefore $A=B$, a contradiction.

Martin Brandenburg · Accepted Answer · 2013-12-02T23:52:41.813

5

The "geometric proof" automatically produces an algebraic proof. The singularity on the cusp corresponds to the maximal ideal $(y^2,y^3) \subseteq k[y^2,y^3]$.

The inclusion $(y^2,y^3) \subseteq k[y^2,y^3]$ doesn't stay injective when tensored with $k[y^2,y^3] \to k[y]$. In fact, $y^2 \otimes y$ and $y^3 \otimes 1$ have the same image under $(y^2,y^3) \otimes_{k[y^2,y^3]} k[y] \to k[y]$, but are not equal.

edited Dec 02 '13 at 23:52

answered Dec 01 '13 at 11:22

Martin Brandenburg

163,620

Hi @Martin could you elaborate further? What is the injection of $k[y^2, y^3]$-modules that you are hinting it? – Elden Elmanto Dec 01 '13 at 22:31
I hope it's clearer now. – Martin Brandenburg Dec 02 '13 at 23:54
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Could you give a hint for $y^2 \otimes y\ne y^3 \otimes 1$? – Dec 03 '13 at 08:14
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Mod out $(y^2,y^3)^2$. – Martin Brandenburg Dec 03 '13 at 14:19
This works. Thanks! – Elden Elmanto Dec 03 '13 at 19:04
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What have to mod out by the square of maximal ideal and how? Are you suggesting somehow to tensor $(y^2,y^3) \otimes_{k[y^2,y^3]} k[y]$ by $k$? – Dec 04 '13 at 08:37
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Yes. This will give you the map on cotangent spaces, which is $k^2 \to k$. – Martin Brandenburg Dec 04 '13 at 08:45
@MartinBrandenburg Sorry to rekindle this question after many years... But I'm really not clear how to mod out $(y^2, y^3)^2$. Could you explain this? This is also stated in https://math.stackexchange.com/questions/669818/direct-proof-of-non-flatness but I don't see why it holds... Thanks in advance. – smalldog Dec 13 '17 at 13:56

Showing that a certain map is not flat by explicit counterexample

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