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So I am being asked how many $\mathbb{Z}[x]$ module structures does the abelian group $\mathbb{Z}/5\mathbb{Z}$ have. Thinking about it, we really only need to define the action of $x$ on $[1]$ because if we do so, then we know what the action of $\sum_{i=0}^n a_ix^i$ would be on $[m]$ by:$$(\sum_{i=0}^na_ix^i)\cdot [m]=(\sum_{i=0}^na_ix^i)\cdot[1+...+1]=\sum_{i=0}^na_i(x^i\cdot [1+...+1])$$and if we have that $x\cdot [1]=[t]$, then we would have that $x^2\cdot [1]=x\cdot[t]=x\cdot[1+...+1]=x\cdot[1]+...+x\cdot[1]=[t]+...+[t]=[t^2]$ and in general, $x^i\cdot [1]=[t^i]$, note that we also have that the action of $\mathbb{Z}$ on $\mathbb{Z}/5\mathbb{Z}$ is fixed by $a\cdot[m]=[am]$. Hence, if we know $x\cdot [1]$ we would know the entire action of our ring on our module. Looking at it, I think that we can let $x\cdot [1]$ be equal to anything and I dont see what problems can be caused, so my answer would be that there are $5$ possible structures.

I see the similarity of having $f(x)\in \mathbb{Z}[x]$ and then taking $k\in \mathbb{Z}$ and doing $[f(k)]$ modulo $5$, but I dont know if there is any meaning behind it or what.

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up vote 3 down vote accepted

("Ring" in the following should always mean "ring with unity" and thus ring homomorphisms are understood to map $1\mapsto 1$)

Indeed, the polynomial ring $A[x]$ is (sometimes used as its definition) universal in the following sense: For every ring homomorphism $f\colon A\to B$ and element $y\in B$ there exists a unique ring homomorphsim $h\colon A[x]\to B$ such that $h|_A=f$ and $h(x)=y$.

Since a module structure is essentially a ring homomorphism into the ring of endomorpphisms of an abelian group, your argument is valid as there is a unique ring homomorphism $\mathbb Z\to \operatorname{End}(\mathbb Z/5\mathbb Z)$ (which is in fact also a case of universality: For any ring $A$ there is a unique $\mathbb Z\to A$) and - as you notice - an element $y\in \operatorname{End}(\mathbb Z/5\mathbb Z)$ can be chosen arbitrariyly as image of $x$.

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