Why $M \otimes M$ does not have a ring structure?

Question

I am reading some section about tensor algebras, and I don't have clear the idea on why $M \otimes M$ dont have a ring structure, where $M$ is an $R$-module. R is commutative and $1 \in R$. So far my knowledge goes, i think it is because the product $m_1 \otimes m_2 \notin M$. can someone explain better to me why it is?

In general I don't understand quite well the idea of tensor products at all.

Thanks

Answer 1

An $R$-algebra structure on $M\otimes M$ would looks like a map $(M\otimes M) \otimes (M \otimes M) \to M\otimes M$. The natural structure map you are equipped with is merely a bilinear map $M \times M \to M\otimes M$, which does not have the right domain to tell you how to define how to multiply two basic tensors $m_1 \otimes m_2$ and $m_3\otimes m_4$.

Note that there is a reasonable map of this form if you already have a multiplication $M\otimes M \to M$, in which case you can define $(m_1 \otimes m_2)\cdot (m_3 \otimes m_4) = (m_1m_3, m_2m_4)$.