# Show that $(M \otimes_R N) \otimes_R P \simeq M \otimes_R (N \otimes_R P)$.

Let $M,N$ and $P$ be $R$-modules. Then

$$(M \otimes_R N) \otimes_ R P \simeq M \otimes_R (N \otimes_R P).$$

The proof of the above as given in the lecture note given by our instructor is as follows $:$

The map $(M \otimes_R N) \times P \longrightarrow M \otimes_R (N \otimes_R P)$ given by

$$(x \otimes y , z) \mapsto x \otimes (y \otimes z)$$

is a well defined $R$-bilinear map on $(M \otimes_R N) \times P$, which can be extended to an $R$-bilinear map $(M \otimes_R N) \otimes_R P \longrightarrow M \otimes_R (N \otimes_R P)$ given by

$$(x \otimes y) \otimes z \mapsto x \otimes (y \otimes z).$$

This is an isomorphism with inverse given by $x \otimes (y \otimes z) \mapsto (x \otimes y) \otimes z$ defined similarly.

Now I am trying to prove the well-definedness of the first map i.e. the map $(M \otimes_R N) \times P \longrightarrow M \otimes_R (N \otimes_R P)$ defined by

$$(x \otimes y , z) \mapsto x \otimes (y \otimes z)$$

For this I take two elements $(x_1 \otimes y_1, z_1), (x_2 \otimes y_2, z_2) \in (M \otimes_R N) \times P$ with $(x_1 \otimes y_1, z_1) = (x_2 \otimes y_2, z_2)$. Then to show the well-definedness we must show that $x_1 \otimes (y_1 \otimes z_1) = x_2 \otimes (y_2 \otimes z_2)$. But I have failed to show that.

You need $R$ commutative here.
To show your map is well-defined, fix $z\in P$. Then $$(x,y)\mapsto x\otimes (y\otimes z)$$ is bilinear from $M\times N$ to $M\otimes(N\otimes P)$. Thus it induces a linear map from $M\otimes N$ to $M\otimes(N\otimes P)$ with $x\otimes y\mapsto x\otimes(y\otimes z)$. So $(x\otimes y,z) \mapsto x\otimes(y\otimes z)$ is well-defined from $(M\otimes N)\times P$ to $M\otimes(N\otimes P)$.
Of course the "right" way to prove this associativity is to prove that both sides are universal objects representing trilinear maps from $M\times N\times P$.
• How can I show the well-definedness of the linear map you mentioned? I first take two elements $x_1 \otimes y_1$ and $x_2 \otimes y_2$ from the generating set of $M \otimes_R N$ with $x_1 \otimes y_1 = x_2 \otimes y_2$ then how do we show that $x_1 \otimes (y_1 \otimes z) = x_2 \otimes (y_2 \otimes z)$? – Arnab Chatterjee. Apr 27 '18 at 6:52
• The map $M ×N \longrightarrow M \otimes N$ is not always injective. Right? – Arnab Chatterjee. Apr 27 '18 at 6:55
• So we may have $(m_1,n_1),(m_2,n_2) \in M \times N$ with $(m_1,n_1) \neq (m_2,n_2)$ but $m_1 \otimes n_1 = m_2 \otimes n_2$. – Arnab Chatterjee. Apr 27 '18 at 6:59
• @ArnabChatterjee. Linear maps from $M\otimes N$ correspond exactly to bilinear maps from $M\times N$. This is the universal property of tensor products. – Lord Shark the Unknown Apr 27 '18 at 8:00