Csdrhrt

Consider $R=k[x,y]$, where $k$ is a field. Consider the $R$-module $M=langle x,yrangle$.

I would like to see that $x otimes y neq y otimes x$ in $M otimes_R M$.

My try to prove it:

Let $F$ be the free abelian group with basis $M times M$; that is, $F$ is free on all ordered pairs $(a,b)$ with $a,b in M$. Define $S$ to be the subgroup of $F$ generated by all elements of the following types:

1. $(a,b+b')-(a,b)-(a,b')$


2. $(a+a',b)-(a,b)-(a',b)$


3. $(ar,b)-(a,rb)$

Since $M otimes_R M = F/S$, $x otimes y = y otimes x$ would imply that $(x,y)-(y,x)$ lies in $S$. This would imply that $(x,y)-(y,x)$ has one of the forms mentioned above.

Let's say

$$(x,y)-(y,x) = (ar,b)-(a,rb)$$ for some $a,b in M$ and some $r in R$.

Consider $M=langle x,y rangle = {cx+dy hspace{0,2cm}| c,d in k[x,y]}$.

The equality above would imply $ar=x$ and for that we would need $r=1$ and $a=x$ since $ain M$ can not be $a=1$. But the equation avobe says that $a=y$ and we know that $x neq y$.

So $(x,y)-(y,x) notin S$, hence $x otimes y neq y otimes x$.

Is that correct? Anyone can give me any feed back?

Thank you.

From the Universal Property of the tensor product, we can develop a more concrete approach which amounts to constructing any $R$-bilinear map that witnesses the inequality of the simple tensors $x otimes y$ and $y otimes x$.

Let $a otimes b, c otimes d in M_1 otimes_R M_2$. The following our equivalent:

(1) $a otimes b = c otimes d$ in $M_1 otimes_R M_2$

(2) No $R$-bilinear map on $M_1 times M_2$ can distinguish $(a,b)$ from $(c,d)$.

(3) No $R$-linear map on $M_1 otimes M_2$ can distinguish $a otimes b$ from $c otimes d$.

Proof: (1) -> (2). Let $phi$ be an $R$-bilinear map. By the universal property, $phi$ factors as $tilde{phi}f$ where $f$ is the canonical balanced product $f(m,n) = m otimes n$. Thus $a otimes b = c otimes d implies f(a,b) = f(c,d) implies tilde{phi}{f}(a,b) = tilde{phi}{f}(c,d) implies phi(a,b) = phi(c,d)$.

(2) -> (1) Argue by contrapositive. If $a otimes b not= c otimes d = 0$ then the canonical balanced product is an $R$-bilinear map on $M_1 times M_2$ which distinguishes $(a,b)$ from $(c,d)$.

(3) -> (2) Let $phi$ be an $R$-bilinear map. Again, by the universal property, factor it as $tilde{phi}f = phi$. By assumption $tilde{phi}$ cannot distinguish between $a otimes b$ and $c otimes d$, so $phi(c,d) = tilde{phi}(aotimes b) = tilde{phi}(c otimes d) = phi(c,d)$.

(2) -> (3) Argue the contrapositive, noting that the composition of a bilinear map (in this case the canonical $R$-balanced product) and a linear map is again a bilinear map.

$square$

This turns the task of showing that two simple tensors are not equal into the task of fashioning a bilinear map which can distinguish them.
Thus, to show $x otimes y not= y otimes x$, our task is to construct a bilinear map $phi: M times M rightarrow N$ for any $R$-module $N$ such that $phi(x,y) not= phi(y,x)$.

I encourage you to work through it from here, because this sort of construction turns up often, but I'll leave a hint in the spoiler below.

How about setting $phi(ax + by, cx + dy) = ad - bc$?