# If $R$ is an integral domain, $I$ is an ideal of $R$, and $0\neq f: I \to R$ is an $R$-module homomorphism, can we conclude that $f$ is injective?

If $$R = \mathbb{Z}$$, $$0 \neq I \unlhd \mathbb{Z}$$, and $$0 \neq f: I \to \mathbb{Z}$$ is an arbitrary $$\mathbb{Z}$$-module homomorphism, then $$f$$ must be injective.

If $$R$$ is an integral domain, $$0 \neq I$$ is an ideal of $$R$$, and $$0 \neq f: I \to R$$ is an arbitrary $$R$$-module homomorphism, can we conclude that $$f$$ is injective? Here, the module structures of $$I$$ and $$R$$ over $$R$$ are assumed to be $$r \cdot x := r * x$$ and $$r \cdot y := r * y$$, where $$r \in R$$, $$x \in I$$, and $$y \in R$$, with $$*$$ representing the multiplication operation in the ring $$R$$.

If $$R$$ is not an integral domain, then there is a counterexample. Let $$R = \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$, $$I = 2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$.

$$f: 2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \longrightarrow R = \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z},$$

$$(x, y) \longmapsto (0, y),$$

then we can see that $$f$$ is not injective.

• The proof of the first claim seems have nothing to do with integral domain, Commented Jun 10 at 18:41
• @Bowei Tang Thank you. Do you have a counterexample? Commented Jun 10 at 18:50
• Since we're discussing $R$-module homomorphisms, maybe an alternative way to view this question: Let abelian group $R^{+}$ denote the underlying additive group of an integral domain $R$. Denote $I^+\triangleleft_{grp} R^{+}$ a normal subgroup which is the underlying additive group of an ideal in $R$. If we equip $R^{+},I^{+}$ with $R$-module structure, then is it necessary that $\text{ker}(f)=0$ will hold true for any given $R$-module homomorphism $f:I^{+}\to R^{+}$? Commented Jun 10 at 18:51

I believe so. The inclusion $$R\rightarrow K$$ of $$R$$ into its field of fractions is $$R$$-linear, and by composition we obtain an $$R$$-linear map $$I\rightarrow K$$. Since $$K$$ is an injective $$R$$-module this map extends to a map $$R\rightarrow K$$, which must be of the form $$r\mapsto ar$$ for some $$a\ne 0$$ in $$K$$. So all maps $$I\rightarrow R$$ are multiplication by an element of $$K$$, and when this element is nonzero the map is injective.
• Since $K$ is an injective $R$-module this map extends to a map $R\rightarrow K$, which must be of the form $r\mapsto ar$ for some $a\ne 0$ in $K$? Commented Jun 10 at 20:30
• Here is a short proof of this fact, using only that $K$ is torsion-free and divisible Commented Jun 10 at 20:43
• I know $K$ is an injective module over $R$. Why can any nonzero homomorphism $f: I \to R$ be extended to a map $f': R \to K$? We do not know if $f$ is injective. Therefore, we cannot use the lifting property of injective modules so far. Commented Jun 10 at 21:01
• We are extending the map $I\rightarrow K$ to a map $R\rightarrow K$ with respect to the inclusion $I\rightarrow R$, which is injective, rather than $f$. The link is essentially a direct proof of all this anyway, if you aren't convinced Commented Jun 10 at 21:01