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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.

This leads to the question:

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.

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  • $\begingroup$ The proof of the first claim seems have nothing to do with integral domain, $\endgroup$
    – Bowei Tang
    Commented Jun 10 at 18:41
  • $\begingroup$ @Bowei Tang Thank you. Do you have a counterexample? $\endgroup$
    – Liang Chen
    Commented Jun 10 at 18:50
  • $\begingroup$ 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 $f:I^{+}\to R^{+}$? $\endgroup$
    – JAG131
    Commented Jun 10 at 18:51

1 Answer 1

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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.

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  • $\begingroup$ 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 $K$? $\endgroup$
    – Liang Chen
    Commented Jun 10 at 20:30
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    $\begingroup$ Here is a short proof of this fact, using only that $K$ is torsion-free and divisible $\endgroup$
    – C. icosahedra
    Commented Jun 10 at 20:43
  • $\begingroup$ 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. $\endgroup$
    – Liang Chen
    Commented Jun 10 at 21:01
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    $\begingroup$ 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 $\endgroup$
    – C. icosahedra
    Commented Jun 10 at 21:01

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