Let $R=\mathbb{Z}/2\mathbb{Z}[x]$. Find all non-isomorphic $R$-modules that each have four elements.
The answer given was $$R/(x^2);\quad R/(x^2+x);\\R/(x^2+x+1);\quad R/(x^2+1);\\ R/(x+1) \oplus R/(x+1);\quad R/(x+1) \oplus R/(x)$$
Now I am having quite some difficulty determining why these are not isomorphic. Prior to this question, I believed for each degree $2$ polynomial $f(x)$ in $R$, $$R/(f(x))\cong\{a+bx\mid a,b \in \mathbb{Z}/2\mathbb{Z}\}\cong R \oplus R$$
Now I think this would only be true if $f(x)$ is irreducible and I know the only irreducible polynomial with degree $2$ here is $x^2+x+1$. I also realize that the isomorphism with $R \oplus R$ is for the underlying abelian groups. Would the isomorphism with $R \oplus R$ still hold for each quotient by the ideal generated by a monic polynomial with degree $2$, just the operation of multiplication on $R \oplus R$ is different for each? So the rings are nonisomorphic? I also cannot see why $R/(x+1) \oplus R/(x+1)$ and $ R/(x+1) \oplus R/(x)$ are not isomorphic. Similarly, I cannot see why $$R/(x^2);\quad R/(x^2+x);\quad R(x^2+1)$$ are nonisomorphic. I get that $R/(x^2+x) \cong R/(x+1) \oplus R/(x)$ so the decomposition into cyclics is different for this module. However, now I cannot see why this module is not isomorphic to the two direct sums of modules above. Other than that I am lost with this. How do you determine this?
Finally, I get that I should use the classification theorem of finitely generated modules over a PID to help me answer this question. Should I examine the invariant factors to do this and can these factors tell me that these modules are not isomorphic?
Best Answer
The annihilators of the $R$-modules you listed are, respectively,
$(x^2)$, $(x^2+x)$, $(x^2+x+1)$, $(x^2+1)$, $(x+1)$, and $\{0\}$.
Isomorphic $R$-modules share the same annihilator in $R$.
I suppose that is essentially what I'm suggesting above. The invariant factors would certainly be helpful for justifying the original candidates, too. The thing is that there are only finitely many polynomials of degree $1$ and $2$ to work with, and that gives you an exhaustive list of cyclic $R$ modules to build a four-element module.