Sub-module of a Tensor Product of ring and module.

free-modulesmodulesring-theorytensor-productstensors

Suppose $R$ is ring with unit and $M$ be unital $R$-module and $N$ be sub-module of $M$. Also let $S$ be a ring containing R as subring and $1_S=1_R$. I have two questions.

$1.$ Can we claim that $S\otimes_R N$ is sub-module of $S\otimes_R M$ ?

I think this should not be the case as the same pair $(s,m)$ can give different cosets in two tensor products despite $N$ being sub-module of $M$. But, I have not been able to find examples. Please help.

$2.$ Can we put some condition on $R$ so that $S\otimes_R N$ will always be sub-module of $S\otimes_R M$ if $N$ is submodule of $M$?

Best Answer

Let $S=k[x,y]/(xy)$ and $R=k[x]/(xy)\subseteq S$. Now, $R$ is an integral domain, so we can set $M$ to be its field of fractions, and $N=R\subseteq M$. Then $S\otimes_R N=S$, but $$(y,1)=\left(y,\frac{x}{x}\right)=\left(xy,\frac{1}{x}\right)=\left(0,\frac{1}{x}\right)=(0,0)$$ in $S\otimes_R M$, and so the inclusion is not an injection.

For your second question, it is necessary and sufficient for $S$ to be flat as an $R$-module since then tensoring preserves injections.

Edit: For necessity, if $S$ is not flat then there must be some injection $\varphi:L\to M$ which is no longer an injection after tensoring with $S$. If we set $N=\text{Im}(\varphi)\subseteq M$ then we can view this as the inclusion of $N$ into $M$, and so the inclusion of $S\otimes_R N$ into $S\otimes_R M$ is not injective.

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