# Idempotents in $\Bbb Z_2[x]/(x^7+1)$

I am trying to find idempotent elements in $$R:=\Bbb Z_2[x]/(x^7+1)$$. Of course $$0,1$$ are idempotents.

My attempt: For $$f \in \Bbb Z_2[x]$$, let $$\bar{f}$$ denote its residue class. We may assume that $$\deg (f)<7$$. Suppose $$\bar{f}$$ is an idempotent. Then $$\bar{f}^2-\bar{f}=0$$ in $$R$$, so $$f^2-f$$ is a multiple of $$x^7+1=x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$$. But I can't find how to proceed. Any hints?

• While there are some special case optimizations that apply in this case (as in Jyrki's answer), I strongly recommend that you also learn the more general ideas in the other answers if you wish to understand the essence of the matter, – Gone Apr 9 at 16:50

We can write any element of $$R$$ using the lowest degree polynomial in its coset. So if $$I=\langle x^7+1\rangle$$, then the general element $$f$$ looks like $$f=\sum_{i=0}^6b_ix^i+I$$ with $$b_i\in\Bbb{Z}_2$$, $$i=0,1,\ldots,6$$.

The hints (prove these if you don't know them already):

• By Freshman's Dream $$f^2=\sum_{i=0}^6b_ix^{2i}+I.$$
• Because $$x^7+I=1+I$$ we have $$x^a+I=x^b+I$$ whenever $$a\equiv b\pmod7$$.
• $$f$$ is an idempotent if and only if $$b_i=b_{2i}$$ for all $$i$$. The subscript $$2i$$ is calculated modulo $$7$$.

When the fog has cleared up you should see a total of eight idempotents in this ring (you can freely choose a carefully picked subset of the $$b_i$$s but the other coefficients are constrained).

• I assumed that $\Bbb{Z}_2$ stands for the field of two elements. If the ring of $2$-adic integers is intended, the conclusion actually still holds, but the details will be more complicated. – Jyrki Lahtonen Apr 9 at 5:44
• The last bulleted coefficient trick was the thing I was having trouble bringing to mind from my days in coding theory :) I know you are fond of the subject as well, so I'm glad you had it in mind! There were a lot of astute observations that made these calculations easier. – rschwieb Apr 9 at 12:06
• Indeed, @rschwieb. This standard algorithm was worked out to find the idempotents of cyclic binary codes of an odd length. An extra method for finding the idempotents generating minimal ideals is also known, but not needed to answer this particular question. – Jyrki Lahtonen Apr 9 at 16:13

In this answer, I assume $$\mathbb{Z}_2=\mathbb{Z}/2\mathbb{Z}=\mathbb{F}_2$$. Note that the splitting field of $$x^8-x$$ over $$\mathbb{F}_2$$ is $$\mathbb{F}_{2^3}$$. Therefore, $$x^8-x=x(x-1)\,q(x)\,r(x)$$ for some irreducible polynomials $$q(x),r(x)\in\mathbb{F}_2[x]$$ of degree $$3$$. From here, we can easily see that, without loss of generality, $$q(x):=x^3+x+1\text{ and }r(x):=x^3+x^2+1\,.$$ Thus, if $$f(x)+\langle x^7+1\rangle \in\mathbb{F}_2[x]/\langle x^7+1\rangle$$ is idempotent, then $$f(x)\,\big(f(x)-1\big)=\big(f(x)\big)^2-f(x)$$ is divisible by $$x^7+1=p(x)\,q(x)\,r(x)\,,$$ where $$p(x):=x-1=x+1$$. Thus, there are $$2^3=8$$ possible polynomials $$f(x)$$ modulo $$x^7+1$$ that works, depending on the subset $$S\subseteq \big\{p(x),q(x),r(x)\big\}$$ which contains the factors that divides $$f(x)$$ such that $$f(x)-1$$ is divisible by the factors that are in $$\big\{p(x),q(x),r(x)\big\}\setminus S$$. Here is the list of all of them (found using the Chinese Remainder Theorem):

• for $$S=\emptyset$$, $$f(x)=1$$;
• for $$S=\big\{p(x)\big\}$$, $$f(x)=x^6+x^5+x^4+x^3+x^2+x$$;
• for $$S=\big\{q(x)\big\}$$, $$f(x)=x^4+x^2+x$$;
• for $$S=\big\{r(x)\big\}$$, $$f(x)=x^6+x^5+x^3$$;
• for $$S=\big\{p(x),q(x)\big\}$$, $$f(x)=x^6+x^5+x^3+1$$;
• for $$S=\big\{r(x),p(x)\big\}$$, $$f(x)=x^4+x^2+x+1$$;
• for $$S=\big\{q(x),r(x)\big\}$$, $$f(x)=x^6+x^5+x^4+x^3+x^2+x+1$$;
• for $$S=\big\{p(x),q(x),r(x)\big\}$$, $$f(x)=0$$.

(Observe that the four polynomials in the bottom of the list are obtained from the four polynomials in the top of the list by adding $$1$$, so you need to determine only four of the polynomials.)

• Remark that the general idea behind this is explained in my answer (and its links). But, alas, at least two readers seem to dislike highlighting general ideas. – Gone Apr 14 at 16:25
• @Gone It looks like you got some kind of revenge-downvoting. Anyway, I appreciate your answer here. – Batominovski Apr 14 at 16:29
• Yes, possibly also related to 3 serial downvotes today. I too appreciate your fine answer(s). It's truly unfortunate that many votes have little to do with mathematics. – Gone Apr 14 at 16:33

Key Idea  For $$f\in {\rm UFD}\, R,\,$$ idempotents $$\,e\in R/f\,$$ correspond to coprime splittings of $$\,f\,$$ since

$$e^2=e\ \,{\rm in}\,\ R/f\iff f\mid e(1-e)\iff f = gh,\, g\mid e,\,h\mid 1-e\qquad$$

In OP $$\,f = (x\!+\!1)(x^3\!+\!x\!+\!1)(x^3\!+\!x^2\!+\!1)$$ is a product of $$\,\color{#c00}3\,$$ primes yielding $$2^{\large \color{#c00}3}$$ such splittings (whose associated idempotents are easily computable by CRT as here, e.g. by solving the system $$\,e\equiv 0\pmod{\!x\!+\!1},\ e\equiv 1\pmod{\!f/(x\!+\!1)}\,$$ etc.

• This does work, but for rings of the particular type $\Bbb{Z}_2[x]/\langle x^n+1\rangle$, $n$ odd, the method in my answer gives the answer rather more quickly. – Jyrki Lahtonen Apr 9 at 16:12
• Since this is just a repackaging of Batominovski's answer, maybe it would be better as a comment? (It fits, I checked.) – rschwieb Apr 9 at 17:37
• @rschwieb As usual we'll have to disagree on that. – Gone Apr 9 at 17:41
• I'm not sure you got to the bottom of it. When I was still actively working with these rings, it was sometimes more fruitful to see them as group rings of cyclic groups (when the central idempotents give the Wedderburn-Artin components). Admittedly my point of view here is that of a coding theorist first and algebraist second - comes with the territory actually. – Jyrki Lahtonen Apr 9 at 19:34
• @JyrkiLahtonen Of course there are often many generalizations ("essences"). But I don't think any of them will be clear to beginners from what is written in your answer. Why not elaborate? – Gone Apr 9 at 20:37