Possible Duplicate:
Equality of two notions of tensor products over a commutative ring
Let $A$ be a commutative ring. There are two common definitions of tensor product of two $A$-modules $M$ and $N$ in terms of universal property. One definition defines it as a universal object of a map $\tau:M\times N\rightarrow G,$ where $G$ is an abelian group, and $\tau$ is $A$-balanced, i.e.,
- $\tau(m+m',n)=\tau(m,n)+\tau(m',n)$,
- $\tau(m,n+n')=\tau(m,n)+\tau(m,n')$,
- $\tau(a\cdot m,n)=\tau(m,a\cdot n)$.
Another definition defines it as a universal object of a map $\sigma:M\times N\rightarrow L,$ where $L$ is an $A$-module, and $\sigma$ is $A$-bilinear, i.e.,
- $\sigma(m+m',n)=\sigma(m,n)+\sigma(m',n)$,
- $\sigma(m,n+n')=\sigma(m,n)+\sigma(m,n')$,
- $\sigma(a\cdot m,n)=a\cdot\sigma(m,n)$,
- $\sigma(m,a\cdot n)=a\cdot\sigma(m,n)$.
I always thought they were equivalent definitions, but now I made a couple of observations:
- Let $\tau:M\times N\rightarrow G $ be a universal object in the first sense. If $G$ happens to be an $A$-module and $\tau$ happens to be $A$-bilinear, then $\tau$ is a universal object in the second sense.
- Let $\sigma:M\times N\rightarrow L$ be a universal object in the second sense. It doesn't seem to follow that $\sigma$ is a universal object in the first sense.
So intuitively, it seems that "the set of the first universal objects contains the set of the second universal objects". Am I correct? If so, then it seems necessary to distinguish the two definitions of tensor product.