# Tensor product of two left $R$-modules $A$ and $B$

In the case $R$ is not a commutative ring.

I don't know what happens if I define the tensor product of two left $R$-modules $A$ and $B$ as follows:

Step 1: Let $F(A,B,R)$ be the free $R$-modules with basis $A\times B$

step 2: Consider the submodule $N$ of $F(A,B,R)$ which is generated by elements with the following forms:

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

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

$(ra,b)-r(a,b)$

$(a,rb)-r(a,b)$

for $a\in A$, $b\in B$ and $r\in R$.

Then tensor product of the two modules $A$ and $B$ is $F/N$.

Nobody stops you from making such a definition. However, what you get is not really useful. Let's see why.

Let $a\mathbin{\tau}b$ denote the element $(a,b)+N\in F/N$. Suppose $r,s\in R$; then, by definition, $$rs(a\mathbin{\tau}b)=r(sa\mathbin{\tau}b)=(rsa)\mathbin{\tau}b$$ but also $$r(sa\mathbin{\tau}b)=(sa)\mathbin{\tau}(rb)= s(a\mathbin{\tau}(rb))= s(r(a\mathbin{\tau}b))=sr(a\mathbin{\tau}b)$$ Thus $$rs(a\mathbin{\tau}b)=sr(a\mathbin{\tau}b)$$ Hence $F/N$ is a module with the following property:

for each $x\in F/N$ and $r,s\in R$, $rsx=srx$.

When the ring is not commutative, this limits very much a possible usefulness of the concept.

For instance, suppose $A=B=R$. Then it is easily seen that $F/N$ is generated by $1\mathbin{\tau}1$, so there is a surjective module homomorphism $f\colon R\to F/N$ with $f(1)=1\mathbin{\tau}1$. The kernel is a left ideal $I$ such that $rs-sr\in I$, for every $r,s\in R$.

Now take the easiest noncommutative ring, the quaternions $\mathbb{H}$. The left ideal above is not $\{0\}$, so it is the whole ring and $F/N$ is the zero module. Similarly for the ring of $2\times2$ matrices over a field.

• Sorry, but I really don't know why it will limit the concept... – PHU CUONG LE VAN Jan 24 '17 at 18:11
• Because we started from a module, then get a quotient. And an element will get the property: rsx=srx. – PHU CUONG LE VAN Jan 24 '17 at 18:12
• @PHUCUONGLEVAN I added some info. – egreg Jan 24 '17 at 18:57