Proving/Disproving $M$ has the structure of an $R$-module

Question

Given an abelian group $M$ and a ring $R$, how can one prove or disprove that $M$ has the structure of an $R$-module? When proving $M$ is an $R$-module, if it is not obvious how to define an action $R\times M\to M$ that satisfies the module axioms, what theorems/techniques can be used to test if $M$ is an $R$-module?

What techniques can be applied to show $M$ is not an $R$-module?

For instance, how would I test if $\mathbb{Z}/5\mathbb{Z}$ is a $\mathbb{Z}[i]$-module?

All examples, nonexamples, links, and theorems (that can be ingested by someone familiar with Dummit/Foote) are welcome!

Answer 1

A structure of (unitary left) $R$-module over an abelian group $G$ (written additively) is determined by a (unitary) ring homomorphism $R\to\operatorname{End}(G)$, where $\operatorname{End}(G)$ consists of the endomorphisms of $G$ under the standard sum operation and map composition.

To see why, suppose $G$ is an $R$-module. For $r\in R$, define $\lambda_r\colon G\to G$ by $\lambda_r(x)=rx$. It is easy to verify that $\lambda_r\in\operatorname{End}(G)$ and that the map $\lambda\colon R\to\operatorname{End}(G)$ is indeed a ring homomorphism.

Conversely, give a ring homomorphism $\varphi\colon R\to\operatorname{End}(G)$, define, for $r\in R$ and $x\in G$, $$ rx=\varphi(r)(x) $$ (the endomorphism $\varphi(r)$ evaluated at $x$). Again, it is easy to verify this gives $G$ a structure of $R$-modules.

Also easy is to see that doing the two steps in sequence takes back to the starting point: an $R$-module structure defines a ring homomorphism which in turn defines the same $R$-module structure and conversely.

In order to see whether $\mathbb{Z}/5\mathbb{Z}$ can be given a structure of $\mathbb{Z}[i]$-module, we need to find a ring homomorphism as above. The task is made easier by the observation that $\operatorname{End}(\mathbb{Z}/5\mathbb{Z}$ is isomorphic to the ring $\mathbb{Z}/5\mathbb{Z}$. If $f\in\operatorname{End}(\mathbb{Z}/5\mathbb{Z}$, we see that $f(x)=f(1)x$ (multiplication in $\mathbb{Z}/5\mathbb{Z}$).

A ring homomorphism $\mathbb{Z}[i]\to\mathbb{Z}/5\mathbb{Z}$ is determined uniquely by a ring homomorphism $\psi\colon\mathbb{Z}[X]\to\mathbb{Z}/5\mathbb{Z}$ such that $\ker\psi\supseteq\langle X^2+1\rangle$, by definition of $\mathbb{Z}[i]$.

Such a $\psi$ is determined once we fix $t=\psi(X)$. Such an element must satisfy $t^2+1=0$, which means $t=2$ or $t=3$.

Thus there are two module structures on $\mathbb{Z}/5\mathbb{Z}$. One is $$ (a+bi)[x]=[ax+2bx] $$ and the other one is $$ (a+bi)[x]=[ax+3bx] $$ as it turns out once we explicit the computations.

Can we find a structure of $\mathbb{Z}[i]$-module on $\mathbb{Q}$? No, because $\operatorname{End}(\mathbb{Q})\cong\mathbb{Q}$ and there's no rational number whose square is $-1$.

Neither we can find one on $\mathbb{Z}/3\mathbb{Z}$, for a similar reason.

The general problem is much harder: here we could use knowledge of how $\mathbb{Z}[i]$ can be obtained as a quotient ring of $\mathbb{Z}[X]$ and also of the endomorphism ring of the abelian group. However, endomorphism rings of abelian group are quite difficult to study and so are ring homomorphisms.

Answer 2

In the particular example you gave, in fact you can make $\mathbb{Z} / 5 \mathbb{Z}$ into a $\mathbb{Z}[i]$-algebra by "mapping $i$ to 2" -- the specific formula for module multiplication would be: $$ (a+bi) \cdot c := (a+2b)c $$ for $a+bi \in \mathbb{Z}[i]$, $c \in \mathbb{Z} / 5 \mathbb{Z}$. (Here, the essential property of 2 which will make $\alpha \cdot (\beta \cdot x) = (\alpha \beta) \cdot x$ work is that $2^2 \equiv -1 \pmod{5}$.)

For a possible technique for a negative proof, we'll prove $\mathbb{Z} / 7 \mathbb{Z}$ cannot be made into a $\mathbb{Z}[i]$-module in a way compatible with its usual abelian group structure. So, suppose it could be. Then consider the induced action of the multiplicative group $\{ \pm 1, \pm i \}$ on the module. Each orbit is of size either 1, 2, or 4. However, the only possible orbit of size 1 (or fixed point) is $\{ 0 \}$ since if $x$ is a fixed point, then $x = -x$ so $x = 0$. On the other hand, an orbit size of exactly 2 is impossible: that would imply for a generator $x$ that $-x = i^2 x = x$ which again implies $x = 0$ so it's actually a fixed point. Therefore, we would get a decomposition of $\mathbb{Z} / 7 \mathbb{Z}$ into a single orbit of size 1 and then orbits of size 4, which is clearly impossible.