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Let $R$ be an integral domain, and let $K$ be its field of fractions. Let $C$ be an $R$-module. Then is $C \cong C \otimes_R R$ injectively contained in $C \otimes_R K$?
Define $\phi: C \otimes_R R \to C \otimes_R K$ by $c \otimes r \mapsto c \otimes r$ and extend it additively. I'm having trouble showing that this map is well-defined and injective.
Hint: $R=\mathbb{Z}$, $C=\mathbb{Z}/2\mathbb{Z}$.
There is no problem in $\phi$ being well defined; there's some problem in showing it's injective. ;-)
The map is well defined, because such is the map $\tilde\phi\colon C\to C\otimes_RK$ defined by $$ x\mapsto x\otimes1 $$ and composing with the natural isomorphism $C\otimes_RR\to C$ gives your $\phi$.
In $K \otimes_R C$ all tensors are elementary, i.e., they all have the form $x \otimes c$. That comes from using a common denominator for any finite list of elements in $K$ (as ratios of elements of $R$). The kernel of the $R$-linear mapping $C \rightarrow C \otimes_R K$ given by $c \mapsto 1 \otimes c$ is precisely the torsion submodule $C_{\rm tor}$, so this mapping of $C$ to $C \otimes_R K$ is injective precisely when $C$ is torsion-free. See Corollary 4.27 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf
The example given by egreg uses for $C$ a torsion module, which is why the mapping is not injective in that case. As long as $C$ has nonzero torsion the mapping won't be injective, and (harder) if $C$ is torsion-free then it is injective.
Hint on well-definedness.
define $\Phi: C \times R \to C \otimes_R K$ by $c \times r \mapsto c \otimes r$ , then by the universal property of tensor $\Phi$ induces a $R$-hom $C \otimes_{R} R \to C \otimes_R K$ with $c \otimes r \mapsto c \otimes r$, of course, it is well-defined.
