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This is a follow up from HERE.

Suppose $K$ is a field and consider $K$ as a $K[x,y]-$module where the scalar product is defined by $f(x,y)\cdot k = f(0,0)\cdot k$.

Is $K$ injective or flat as $k[x,y]-$module?

COMMENT: It is proved by @GeorgesElencwajg that $K$ is not projective.

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Hint: $K=K[x,y]/(x,y)$, and $R/I$ is almost never flat over $R$. –  Martin Brandenburg Jan 5 at 15:41

1 Answer 1

up vote 6 down vote accepted

Non injectivity
An injective module $M$ over a domain $D$ is divisible: this means that for every non-zero $d\in D$ and every $m\in M$ there exists $n\in N$ with $m=d\cdot n$ .
But here, choosing $m=1\in K=M$, it is impossible to write $1=x\cdot n\in K$ since for every $n\in K$ one actually has $x\cdot n=0$ .
So $K$ is not injective.

Non flatness
1) The $K[x,y]$-module $K$ is of finite presentation. Hence it is flat if and only if it is projective.
Since I already proved that $K$ is not projective, it is not flat either.

2) Here is another proof, in the vein of Martin's comment.
If $I\subset R$ is an ideal in a ring, for the quotient $R/I$ to be flat over $R$ it is necessary that $I=I^2$.
Since $(x,y)\neq(x,y)^2$, $K=K[x,y]/(x,y)$ is not flat over $K[x,y]$.

3) A third proof is obtained by remembering that a flat module over a domain is torsionless.
Since $x\cdot 1=0\in K$ even though $x\neq 0\in K[x,y]$ and $1\neq 0\in K$, the module $K$ has torsion and is thus not flat.

Edit
As a generalization of the above, it might be worth remembering that if $0\subsetneq I\subsetneq R$ is a non-trivial ideal of a domain, the $R$-module $R/I$ is never projective, nor injective, nor flat.

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Dear @rschwieb, you are perfectly right of course: I have corrected my blunder. Thank you very much for your attentive reading [and I found your "n injective" quite amusing ...] . –  Georges Elencwajg Jan 5 at 19:07
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Dear Georges : No problem :) I always enjoy your answers... –  rschwieb Jan 5 at 23:29

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