Let $R=\mathbb{F}_5[X,Y]/(Y^2-X^3-2X)$. I have to determine the structure of the $R$-module $M=R/XR\times R/YR$ using the structure theorem for finitely generated modules over Dedekind domain. Since in this case we have a torsion module, $M$ is isomorphic to $R/I_1\times\cdots\times R/I_n$ for some ideals $I_i$ where $I_1\subset\dots\subset I_n$.

I was trying to find such ideals by finding the prime decomposition of $X$ and $Y$ in $R$ and then applying the Chinese remainder theorem. In order to find the prime decomposition, I was looking for polynomials $f$ and $g$ such that $fg=X+(Y^2-X^3-2X)h$ for some polynomial $h$. However, I did not succeed. Does this approach seem fruitful, if yes what are those polynomials, or should I try something differently?


$R/XR$ is more than just an $R$-module; it is actually a ring as well.

A common way to understand rings is to do arithmetic on their presentations. For example,

$$\begin{align} R/(X) &= \left( \mathbf{F}_5[X,Y]/(Y^2 - X^3 - 2X)\right) / (X) \\&\cong \left( \mathbf{F}_5[X,Y]/(X)\right) / (Y^2 - X^3 - 2X) \\&\cong \mathbf{F}_5[Y] / (Y^2) \end{align}$$

  • $\begingroup$ I tried to do the same for $R/(Y)$ and got $\mathbb{F}_5[X]/(X^3-2X)=\mathbb{F}_5\times\mathbb{F}_5[X]/(X^2-2)$ by Chinese remainder. I tried to translate these rings back into $R/I$ so that I get the desired form. But I got stuck on $\mathbb{F}_5[X]/(X^2-2)$. Any hints for that? $\endgroup$ – MightyGuy Apr 27 '18 at 17:16

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