# Finding units of a quotient ring.

I've been given a two part question to first find the number of elements and then number of units is in $$R/I$$ where $$R = \mathbb{F}_7[x]$$, $$f(x) = x^3 + 4$$ and $$I = f(x)$$

I found the number of elements through my own notes, where we have $$p^k$$ where $$p$$ is the size of the ring, in this case $$7$$ and $$k$$ is the degree of the polynomial element, in this case $$3$$. Therefore the number of elements is $$7^3 = 343$$.

Now i know that a unit is an element $$u \in R$$ such that for any element $$v \in R$$, $$uv = 1$$, therefore a unit is an element in $$R$$ with an inverse also in $$R$$. However I'm unsure how to use this information to help find how many units there are out of the 343 elements as it would be impractical to start testing each element one by one.

Hint $$\,\ x^3 + 4\,$$ is irreducible, since it has no roots: $$\bmod 7\!:\ a^{\large 3}\equiv -4\,\overset{(\,\ \ )^{\Large 2}}\Longrightarrow\,1\equiv a^{\large 6}\equiv 16\Rightarrow\!\Leftarrow$$

Thus $$\,(x^3+4)$$ is maximal so $$\, \Bbb F_7[x]/(x^3+4)\,$$ is a field, so every nonzero element is a unit.

• Note  Though it may appear that there is some duplication between this answer and rschwieb's, that was not the case when I posted this (his answer was later edited because he apparently misread the polynomial so posted an answer assuming that the polynomial was reducible). – Bill Dubuque Apr 30 at 17:31
• Looks to me like the only things that overlap are the final answers, which one would hope would overlap. The means to the solutions seem totally different. I don't think anyone is going to accuse you of duplication, so don't worry. Thanks for the feedbackless downvote that drew my attention to my mistake. it was silly indeed and was glad to have the opportunity to fix it. – rschwieb Apr 30 at 19:23
• @rschwieb Imo, it is not honorable to publicly guess / accuse another user of downvoting. And speaking of honor, I would have deleted that wrong answer rather than morph it into one that is close to a dupe of a prior answer. But this is not the place to discuss meta matters. Btw, correct is 'gcd', not 'lcd'. – Bill Dubuque Apr 30 at 19:44
• What a very strange way to say "you're welcome"! The thanks wasn't intended to be facetious. Would you really delete one of your answers if it was on the whole true with useful content but suffered a tragicomical mistake that was easily rectified? If so, I would encourage you not to, as it seems like a classic baby-with-bathwater scenario, and there's no sense in pride getting in the way of useful content. Thanks again, (for the typo thing too.) – rschwieb May 1 at 15:21
• @rschwieb Yes, I have deleted my answer in such cases (as have many others). The intent of my note was solely to help readers avoid any possible confusion around duplication. Could you please not ping me any further about these meta topics. Further note to readers: the above comments refer to a much shorter prior version of rschwieb's answer. – Bill Dubuque May 1 at 15:35

In general you could do something like this:

Since $$F_7[x]$$ is a principal ideal domain, an element $$u$$ of $$F_7[x]$$ is going to have an inverse in $$F_7[x]/(f(x))$$ precisely when $$lcd(u, f(x))=1$$. Otherwise they would both lie in the proper ideal generated by their least common divisor.

If you can count the number of polynomials of degree less than that of $$f$$ that are coprime with $$f$$, you have the number of units.

So, a good first step would be to factor $$f(x)$$ over $$F_7$$ completely. In your case it's not too hard to verify all the residues of $$x^3$$ are in $$\{-1,0,1\}$$ mod $$7$$, so $$x^3+4$$ has no roots, and is irreducible. So it turns out all elements are coprime to $$x^3+4$$, and are invertible.

Note: apologies for the last version of the solution. I had muddled $$x^3+4$$ for $$x^4+3$$ at the end from a problem I had previously seen. I had thought the problem would be more challenging than just a field.

Since I started out analyzing $$f(x)=x^4+3=(x-2)(x+2)(x^2+2)$$ anyways, I may as well finish. It is a diverting combinatorial task to count the units. One can count the monic polynomials that aren't coprime with $$f(x)$$, and then multiply by $$6$$ (to recover the nonmonic ones.) Just do one degree at a time.

Firstly, there are (obviously) two monic polynomials of degree $$1$$ which have a common factor with $$f(x)$$.

For degree $$2$$, there are $$14$$ such polynomials: $$x^2+2$$ itself, the product $$(x+2)(x-2)$$, and $$12$$ more products of one of the two with some other $$x-a$$.

For degree $$3$$: for $$x-2$$, there are $$49$$ monic degree $$3$$ polynomials which it divides. One of these it shares with $$x^2+2$$, and $$7$$ are shared with $$x+2$$. Symmetrically the same thing can be said for $$x+2$$. Finally, $$x^2+2$$ can be paired with seven linear factors, and one case overlaps each of the linear factors $$x+2$$ and $$x-2$$. So if you draw up a little Venn diagram of the situation, you can see its total population is $$96$$.

So we count $$112$$ monic polynomials sharing a factor, accounting for $$672$$ nonzero nonunits. $$0$$ is also a nonunit that we haven't counted so far, so the total is $$673$$ nonunits. Out of the entirity of $$7^4=2401$$ polynomials, you have $$1728$$ units.

If you know the Chinese Remainder theorem, this isn't hard to double-check. We'd have that $$F_7[x]/(f(x))\cong F_7[x]/(x-2)\times F_7[x]/(x+2)\times F_7[x]/(x^2+2)\cong F_7\times F_7\times F_{49}$$. THe units of the product ring are the product of the unit groups, so we'd expect $$6\cdot 6\cdot 48$$ units, which you see matches: $$1728$$.

Since that still only involved fields, I'll include one last example: suppose our $$f(x)$$ had been $$(x-2)(x+2)^3$$. The analysis along similar lines above says:

1. there are 2 relevant monic degree $$1$$ polynomials
2. $$7$$ degree $$2$$ polynomials for $$x-2$$, one of which is shared with $$x+2$$, and symmetrically $$x-2$$ has $$7$$ as well, accounting for $$13$$ distinct monic polynomials
3. There are $$49$$ polynomials divisible by $$x-2$$, $$7$$ of which are shared with $$x+2$$ (you need to choose the last linear factor.) The same can be said for $$x+2$$, accounting for $$91$$ distinct monic polynomials.

The totals is $$106$$ monic polynomials, or $$636$$ polynomials, therefore $$1764$$ units.

Double checking with the Chinese remainder theorem, we have that $$F_7[x]/(f(x))\cong F_7[x]/(x-2)\times F_7[x]/(x+2)^3$$, where the first factor is $$F_7$$ again, but the second factor is no longer a field. It's a local ring with maximal ideal $$(x+2+(x+2)^3)$$. As a local ring, its units are precisely the things not inside the maximal ideal: in this case, that ideal contains $$7^2$$ elements, leaving $$7^3-7^2=294$$ units behind. Then in the product, you guessed it, that accounts for $$6\cdot 294=1764$$ units.

• That seems to make sense to me, thank you for the link I'll check it out – L G Apr 30 at 14:55