Prove that elementary tensor is non-zero in proof of non-flatness

Question

This is a follow-up question to this question (and the answer there by René Schipperus) about proving that $k[t]$ is non-flat as $k[t^2,t^3]$-module.

I have reduced this to showing that $t\otimes t$ is non-zero in $k[t]/(t^2)\otimes_{k[t^2,t^3]} k[t]$. To show that an elementary tensor is non-zero, I have to find a $k[t^2,t^3]$-bilinear map from $k[t]/(t^2) \times k[t]$ to an abelian group such that the image of $(t,t)$ is non-zero.

Does someone have a hint?

Let $a: k[T]\to k$ mapping $f$ to $f'(0)$, it is $k[T^2, T^3]=R$-linear, where you view $k$ as a $R$-module where $T^2$ and $T^3$ act trivially, you can consider $k[T]/(T^2)\times k[T] \to k[T]/(T^2)\times k\to k[T]/(T^2)$, where the map on the left is $1\otimes a$ and the second is the multiplication, it is clearly $k$-linear and in fact $R$-linear as $t^2, t^3$ both act trivially. — Ahr, Nov 02 '20 at 14:45
But then $1x=x$ axiom implies then $1=0$ right ? But this is not true in $k$ ? — Dr. Heinz Doofenshmirtz, Nov 02 '20 at 15:50
It looks like you might have meant to link to another question? — Badam Baplan, Nov 02 '20 at 15:50
I mean that $T^2$ and $T^3$ acts by multiplication by $0$, not all $R$. — Ahr, Nov 02 '20 at 15:52
If you prefer you view $k$ as an $R$-module via $R\times k\to k$ given by $(f,\alpha)\mapsto f(0).\alpha$, you clearly have $T^2.\alpha=T^3.\alpha=0$ — Ahr, Nov 02 '20 at 15:58

score 2 · Accepted Answer · answered Nov 02 '20 at 16:50

2

Let $A=k[t]/(t^2)$ and $u=[t]\in A$ satisfying $u^2=0$.
The action is given by $f\cdot(a+bu)=f(u)\cdot(a+bu)$ for $f\in k[t]$, in particular $f=t^2$ annihilates all of $A$.

Then consider $\varphi:A\times k[t]\to A$ mapping $$(a+bu,\,f)\,\mapsto\,b\cdot f(u)$$ We get $\varphi((a+bu)c,\,f)=\varphi(a+bu,\,cf)$ for any $c\in k$, and the actions of $t^2,t^3$ make both sides $0$, so it's $k[t^2,t^3]$-bilinear, and it maps $$\varphi([t],\,t)=\varphi(u,t)=1\cdot u=u\,\ne 0\,.$$

answered Nov 02 '20 at 16:50

Berci

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Note that $b=(a+bt)'(0)$, so it's basically the same answer as in the comments. – Berci Nov 02 '20 at 17:01
Thank you Berci ! – Dr. Heinz Doofenshmirtz Nov 02 '20 at 18:45

Prove that elementary tensor is non-zero in proof of non-flatness

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