# $\Bbb F_2[X]$ modules with 8 elements

Problem 4.
Let $\Bbb F_2$ be the field with 2 elements and let $R=\Bbb F_2[X]$. List, up to isomorphism, all $R$-modules with 8 elements.

Solution.
We use the classification theorem of modules over a PID. Since $R$ is a finite module, it is in particular an $\Bbb F_2$ vector space. We can write $$M\cong R/n_1R\oplus R/n_2 R\oplus\cdots\oplus R/n_r R$$ for polynmials $n_1\mid n_2\mid\cdots\mid n_r$. In our case, we have $\sum_{i=1}^r\deg n_i=3$, so we have three options: $r=1,\deg n_1=3$, $r=2,\deg n_2=2$, and $r=3,\deg n_3=1$. The first case yields 8 options. For the second case, we need $n_2$ to be reducible, so we have $X(X+1),X^2,(X+1)^2$ as choices. The first choice yields 2 decompositions, and the latter choices yield 1 decomposition each, for a total of 4. For the linear case, we need the same linear term repeated thrice, which is 2 choices. Therefore there are 14 in all, listed by invariant factors below: \begin{align} &\{X^3+(0/1)X^2+(0/1)X+(0/1)\}\\ &\{X^2,X\},\{X^2+X,X+(0/1)\},\{X^2+1,X+1\}\\ &\{X,X,X\},\{X+1,X+1,X+1\}. \end{align}

My question is the problem above. I can understand the solution, but how can I see that the solutions given are from distinct isomorphism classes? For instance, is the below true? $$R/(X)\oplus R/(X)\oplus R/(X)\cong R/(X+1)\oplus R/(X+1)\oplus R/(X+1)\\ \cong F_2\oplus F_2\oplus F_2$$

Thanks for any help.

• For your question, the last two are not isomorphic as $R$-module. Consider how $X$ acts on both. Commented Oct 1, 2016 at 3:44
• What i707107 said! $R/(X)$ and $R/(X+1)$ are isomorphic as rings (both isomorphic to $\Bbb{F}_2$ as you observed), but they are not isomorphic as $R$-modules. Commented Oct 1, 2016 at 5:32
• Aha... I think I got it. Thanks so much. Commented Oct 1, 2016 at 5:37

Let $$L$$ be the comprehensive list of nonisomorphic $$\Bbb{F}_2[X]$$-modules of order $$8$$. Then by the counting argument you cite above $$L$$ must be contained in the following list:

1. $$\Bbb{F}_2[X]/(X^3)$$
2. $$\Bbb{F}_2[X]/(X^3+X^2)=\Bbb{F}_2[X]/(X^2(X+1))\\\cong \Bbb{F}_2[X]/(X^2)\oplus \Bbb{F}_2[X]/(X+1)$$
3. $$\Bbb{F}_2[X]/(X^3+X)=\Bbb{F}_2[X]/(X(X+1)^2)\\\cong \Bbb{F}_2[X]/(X)\oplus \Bbb{F}_2[X]/((X+1)^2)$$
4. $$\Bbb{F}_2[X]/(X^3+1)=\Bbb{F}_2[X]/((X+1)(X^2+X+1))\cong {\Bbb{F}_2[X]/(X+1)\oplus \Bbb{F}_2[X]/(X^2+X+1)}$$
5. $$\Bbb{F}_2[X]/(X^3+X^2+X)=\Bbb{F}_2[X]/(X(X^2+X+1))\\\cong \Bbb{F}_2[X]/(X)\oplus \Bbb{F}_2[X]/(X^2+X+1)$$
6. $$\Bbb{F}_2[X]/(X^3+X^2+1)$$
7. $$\Bbb{F}_2[X]/(X^3+X+1)$$
8. $$\Bbb{F}_2[X]/(X^3+X^2+X^1+1)= \Bbb{F}_2[X]/((X+1)^3)$$
9. $$\Bbb{F}_2[X]/(X)\oplus \Bbb{F}_2[X]/(X^2)$$
10. $$\Bbb{F}_2[X]/(X)\oplus \Bbb{F}_2[X]/(X(X+1))\\\cong \Bbb{F}_2[X]/(X)\oplus \Bbb{F}_2[X]/(X)\oplus\Bbb{F}_2[X]/(X+1)$$
11. $$\Bbb{F}_2[X]/(X+1)\oplus \Bbb{F}_2[X]/(X(X+1))\\\cong \Bbb{F}_2[X]/(X)\oplus \Bbb{F}_2[X]/(X+1)\oplus\Bbb{F}_2[X]/(X+1)$$
12. $$\Bbb{F}_2[X]/(X+1)\oplus \Bbb{F}_2[X]/((X+1)^2)$$
13. $$\Bbb{F}_2[X]/(X)\oplus \Bbb{F}_2[X]/(X)\oplus\Bbb{F}_2[X]/(X)$$
14. $$\Bbb{F}_2[X]/(X+1)\oplus \Bbb{F}_2[X]/(X+1)\oplus\Bbb{F}_2[X]/(X+1)$$,

where the $$\Bbb{F}_2[X]$$-module isomorphisms $$\cong$$ are due to the structure theorem (and the coprimeness of pertaining polynomials). Note that $$X$$ and $$X+1$$ are irreducible in $$\Bbb{F}_2[X]$$ as they are of degree $$1$$. Likewise $$X^2+X+1,X^3+X^2+1$$ and $$X^3+X+1$$ are irreducible in $$\Bbb{F}_2[X]$$ since they have no roots in $$\Bbb{F}_2$$.

Again applying the structure theorem to the final forms of each of these (isomorphism classes of) modules we can conclude that $$L$$ is precisely the list provided.

Also note that 1-8 in $$L$$ is the comprehensive list of nonisomorphic cyclic $$\Bbb{F}_2[X]$$-modules of order $$8$$.

We can write $$L$$ in a more compact way using the so-called module-type notation. Define $$p_{1,1}(X):=X, p_{1,2}(X):=X+1, p_{2,1}(X):=X^2+X+1,p_{3,1}(X):=X^3+X^2+1$$,$$p_{3,2}(X):=X^3+X+1\in\Bbb{F}_2[X]$$. Then $$L$$ is

1. $$(p_{1,1}^3)$$
2. $$(p_{1,1}^2,p_{1,2}^1)$$
3. $$(p_{1,1}^1,p_{1,2}^2)$$
4. $$(p_{1,1}^1,p_{2,1})$$
5. $$(p_{1,2}^1,p_{2,1})$$
6. $$(p_{3,1}^1)$$
7. $$(p_{3,2}^1)$$
8. $$(p_{1,2}^3)$$
9. $$(p_{1,1}^1,p_{1,1}^2)$$
10. $$(p_{1,1}^1,p_{1,1}^1,p_{1,2}^1)$$
11. $$(p_{1,1}^1,p_{1,2}^1,p_{1,2}^1)$$
12. $$(p_{1,2}^1,p_{1,2}^2)$$
13. $$(p_{1,1}^1,p_{1,1}^1,p_{1,1}^1)$$
14. $$(p_{1,2}^1,p_{1,2}^1,p_{1,2}^1)$$

Remark: The initial counting argument does not provide a priori that distint polynomials provide distint isomorphism classes. We prove distinctness by factoring them into irreducibles and referring to the structure theorem. It should also be noted that the same strategy is not that practical in general, for instance consider the question of classifying all $$\Bbb{F}_3[X]$$-modules of order $$3^4$$. Then there are $$5$$ options:

$$4; 1+3; 2+2; 1+1+2; 1+1+1+1,$$

and just for the cyclic case one has to worry about $$3^3$$ polynomials to begin with, wlog taking them to be monic.

Regarding the last part of your question, I'd like to add a comment. I believe you build those (ring) isomorphisms by using the evaluations at $$0$$ and $$1$$, and then you say e.g. $$\Bbb{F}_2[0]\cong\Bbb{F}_2(0)=\Bbb{F}_2$$. But for this to make sense as an $$\Bbb{F}_2[X]$$-module homomorphism you need to have a map $$\operatorname{sca}:\Bbb{F}_2[X]\times\Bbb{F}_2\to\Bbb{F}_2$$ that defines an $$\Bbb{F}_2[X]$$-module structure on $$\Bbb{F}_2$$ to begin with, which does not make much sense.

• Thanks for the details! But I'm confused why 1-8 is cyclic while the rest are not? Commented Dec 17, 2021 at 4:40
• I believe I used the adjective "cyclic" in the same sense as the first sentence of encyclopediaofmath.org/wiki/Cyclic_module. Commented Dec 17, 2021 at 8:15
• For example, your 14 is a direct sum of cyclic modules (being $R$ quotient some ideals), so this is not a cyclic module (plus an alternative reason that it's finitely-generated but has more than one generator), right? But 2 also factors into two summands, and $x^3+x^2$ is reducible. So neither the decomposition, nor the reducibility impacts being cyclic? Indeed, only 6 and 7 are irreducible over $\mathbb{F}_2$. Perhaps I'm getting disconnected on some basic notions.. Commented Dec 17, 2021 at 18:58
• I think you are correct. One could think of the analogous question for abelian groups (= $\mathbb{Z}$-modules) of order $12$. $\mathbb{Z}/(12)\cong \mathbb{Z}/(4) \oplus \mathbb{Z}/(3)$ is cyclic, but $\mathbb{Z}/(6) \oplus \mathbb{Z}/(2)$ and $\mathbb{Z}/(2) \oplus \mathbb{Z}/(2) \oplus \mathbb{Z}/(3)$ are not. Commented Dec 17, 2021 at 19:54
• I see, thank you! Commented Dec 18, 2021 at 20:21