equality of two simple tensors in $R=k[x,y]$

Question

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

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 $a\in M$ 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.

Answer 1

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 $\phi(c,d) = \tilde{\phi}(a\otimes b) = \tilde{\phi}(c \otimes d) = \phi(c,d)$.

(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 $\phi(ax + by, cx + dy) = ad - bc$?