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Show that $4x^2+6x+3$ is a unit in $\mathbb{Z}_8[x]$.

Once you have found the inverse like here, the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = $1$ imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?

user26857
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ZSMJ
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    You don't really need to find an inverse. $2$ is nilpotent in $\mathbb{Z}/8$, so $2x$ is nilpotent in $(\mathbb{Z}/8)[x]$ and hence $p(2x)$ is a unit in $(\mathbb{Z}/8)[x]$ iff its constant term is a unit in $\mathbb{Z}/8$. – user10354138 May 14 '19 at 11:36
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    $\frac{1}{3+6x+4x^2}=\frac{1}{3}\frac{1}{1+(6x+4x^2)/3}=\frac{1}{3}\left(1-(6x+4x^2)/3+((6x+4x^2)/3)^2+((6x+4x^2)/3)^3+...\right)$. Compute that $1/3=3$ and those powers, which are not as many as they look. – logarithm May 14 '19 at 11:44
  • @user10354138 What theorem about nilpotent elements is being applied here? I'd need to understand its proof first. – ZSMJ May 14 '19 at 13:39
  • @Rhaldryn That result is proven directly by the formula that I wrote. The invertibility of the constant term ensures that you can do the first step. The fact that the other terms are nilpotent ensures that eventually the powers in the infinite series all vanish. – logarithm May 14 '19 at 13:46
  • @logarithm I was having trouble proving that the sum of two nilpotent elements is also nilpotent but I see it now using the binomial th. Also shouldn't the $+$ and $-$ signs in the series alternate? – ZSMJ May 14 '19 at 14:51
  • @Rhaldryn Yes, they should alternate. It is just the geometric series $\frac{1}{1-A}=1+A+A^2+...$ and $A$ happens to be nilpotent. – logarithm May 14 '19 at 14:52

6 Answers6

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Write $$ 4x^2+6x+3 = 3(4x^2+2x+1) = 3((2x)^2+(2x)+1) = 3 \frac{(2x)^3-1}{2x-1} = \frac{-3}{2x-1} $$ Therefore, $$ \frac{1}{4x^2+6x+3} = \frac{2x-1}{-3} = 3(1-2x) = 3-6x = 2x+3 $$

lhf
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If $R$ is a commutative ring: the units in $R[x]$ are the polynomials whose constant term is a unit, and whose higher order coefficients are nilpotent. You can apply this directly to your example.

Ehsaan
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Hint: As in the hinted paper, a possible ansatz would be

$(4x^2+6x+3) (ax+b) = 4ax^3+(4b+6a)x^2+ (6b+3a)x+3b=1$.

This requires $4a\equiv 0\mod 8$ (so $a$ must be even), $4b+6a\equiv 0\mod 8$, and $6b+3a\equiv 0\mod 8$ and $3b\equiv 1\mod 8$ (so $b=3$).

The cases left are $a$ even with $b=3$.

Wuestenfux
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To find an inverse polynomial for that holds $p(x)(4x^2+6x+3)=1$ so it has to be $3y=1\mod 8$ [edit: For more context on $y$, see the comments below]. So $y=3$ and the polynomial might look like this:

$p(x)=(ax+3)$

Then $(4x^2+6x+3)(ax+3)=4ax^3+(6a+12)x^2+(3a+18)x+9$. Now it has to be $4a\equiv 0\mod 8$ and $6a+12\equiv 0\mod 8$ and $3a+18\equiv 0\mod 8$.

Is there such an $a$?. Yes indeed. For $a=2$ we have $8\equiv 0\mod 8$

$24\equiv 0\mod 8$ and $24\equiv 0\mod 8$.

If we would fail to find this $a$ in this step, we would have to try with $p(x)=(ax^2+bx+3)$ and proceed as above, which gets more and more complicated.

So it is $(4x^2+6x+3)(2x+3)\equiv 1\mod 8$

Cornman
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    What is $y$ ? the constant term of $p(x)$? – J. W. Tanner May 14 '19 at 12:23
  • @J.W.Tanner Yes, I meant that when we have a linear polynomial ax+y, then y has to be 3. – Cornman May 14 '19 at 12:42
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    Perhaps you should edit to make that obvious – J. W. Tanner May 14 '19 at 12:48
  • @J.W.Tanner I refered to the comments now. I knew, that it is a little dubiuos, but I thought one might still understand it, espacially with how I proceed in the calculation. I should have wrote (ax+b) first and then explain why $b\equiv 3$, I guess. – Cornman May 14 '19 at 13:01
  • @Cornman Thanks. What values can the degree of the inverse polynomial take? – ZSMJ May 14 '19 at 14:07
  • @Rhaldryn I do not think, that you can say something about the degree of such a polynomial in general. It is 'random' that the $p$ here happens to have degree 1. – Cornman May 14 '19 at 15:41
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Using $ $ simpler multiples, $ $ we can invert $\,\ a \:\!-\:\! f,\,$ for invertible $\,a,\,$ say $\,\color{#0a0}{ab = 1},\,$ & nilpotent $\,f,\,$ say $\color{#c00}{f^n = 0},\,$ by inverting a simpler multiple $\, a^n-\color{#c00}{f^n} = \color{#0a0}{a^n},\,$ which has obvious inverse $\,\color{#0a0}{b^n},\,$ i.e.

$\ \ \ \ \ \ \ \ \ \color{#0a0}{ab=1},\, \color{#c00}{f^{\large n} = 0}\ \Rightarrow\ \overbrace{\dfrac{1}{a-f} = \dfrac{a^{\large n-1}\!+\!\cdots\! +\! f^{\large n-1}}{\!\!\!\!\!\color{#0a0}{a^{\large n}}-\color{#c00}{f^{\large n}}}}^{\large \text{check via cross multiply}} =\, \color{#0a0}{b^{\large n}}(a^{\large n-1}+\cdots + f^{\large n-1})$ $\!\begin{align}{\rm so}\ \ &\color{#0a0}{3(3)=1},\, \color{#c00}{f^{\large 3} = 0}\ \Rightarrow\ \dfrac{1}{3-f} = \dfrac{3^{\large 2}+\,3f\,+\, f^{\large 2^{\phantom{|^{|^|}}\!\!\!\!\!}}}{\color{#0a0}{3^{\large 3}-\color{#c00}{f^{\large 3}}}}\ \ \, =\, \ \ \color{#0a0}{3^{\large 3}}(3^{\large 2}\! +3f + f^{\large 2}) = \bbox[5px,border:1px solid #c00]{2x+3}\\[.1em] &{\rm because}\ \ \ \color{#c00}{2^{\large 3}\mid f^{\large 3^{\phantom{ |^|}}}} {\rm by}\ \ 2\mid f = -6x-4x^2\,\ \ [\,{\rm to\ invert}\ \ 3\!-\!f = 3\!+\!6x\!+\!4x^2 \in \smash[b]{\Bbb Z_{\large \color{#c00}{2^{\Large 3}} }]}\end{align}$

Remark $ $ We may view this as a constructive proof that divisors of units are units, i.e. if $\,d\mid u\,$ unit, then $\,dd' = u\,$ so $\,\dfrac{1}d = \dfrac{d'}u.\,$ Above we chose $\,u = a^n-f^n,\,$ a unit multiple $\,a-f,\,$ which is "simpler" since it has $ $ obvious $ $ inverse, because $\,u= a^n\,$ for unit $\,a,\,$ by $\,f^n = 0$.

The same idea shows how to invert all cyclotomics $\Phi_{k}(f)$ for $\,k\ge n,\,$ since they have simpler multiple $\,1-f^k = 1.\,$ There are simple algorithms for recognizing polynomials of cyclotomic form (i.e. divisors of $\,1-x^k)$ and products of such, so the above method generalizes widely.

This idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously in the method of rationalizing denominators and in Gauss's algorithm for computing modular inverses. Analogous methods may be employed for for computing remainders via mod arithmetic.

Generally we can use standard (nonconstructive) methods of commutative algebra to give quicker proofs, e.g see here, which mentions how it is related to Euclid's proof that there are infinitely many primes (and generalizations). Compare also the methods used here to prove that a polynomial is a unit iff its constant term is a unit and all other coefficients are nilpotent (the method of proof there can be made constructive - analogous to above)

Bill Dubuque
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    The entire first paragraph is just obfuscation of what is really important, which is the last paragraph. Many colors, but quite bad exposition. Key ideas: Invertible constant term, other terms nilpotent, and geometric series. Period. – logarithm May 14 '19 at 16:00
  • @logarithm Quite bizarre claims. The sought constructive proof in this special case certainly does $\rm\color{#c00}{highlight}$ the $\rm\color{#0a0}{key}$ ideas. As for colors, many readers have remarked that they find them very helpful. – Bill Dubuque May 14 '19 at 16:42
  • It is simple, the idea in that link is neither crucial there nor here. 'Simpler denominator'? That's $1$, which you find computing the inverse, which you constructively compute using the three crucial ideas that I said. Just like the multiplying by $x-1$ in the link, and what you did here, which hides the key idea of the geometric series, can go in the trash without affecting the solution of the problem. As I said, your first paragraph is just obfuscation of what is really important. – logarithm May 14 '19 at 17:09
  • @logarithm In fact I purposely chose to simplify by using polynomials vs power series, i.e. to use $\dfrac{1}{a-f} = \dfrac{b^n (a^n!-!f^n)}{a-f} = b^n(a^{n-1}+\cdots + f^{n-1}),$ vs. the equivalent power series form $\dfrac{1}{a-f} \ =\ \dfrac{b}{1-bf}\ =\ b(1+bf + b^2f^2 + \cdots),$ since students may not yet have the background to work with (formal) power series. Nothing is "hidden" in my approach. Further it helps to highlight the emphasized analogies with the other methods mentioned. – Bill Dubuque May 14 '19 at 18:18
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    I have no problem with that choice, which only now you are adding explicitly. That is just geometric series and nilponency of $f$ merged together. As you change the answer the obfuscation might disappear. There is nothing different or new in 'your approach', only presentation. That is why presentation is what I criticized. A little now, and even more in previous versions of that presentation, there was no idea made explicit but 'making the denominator simpler', which says nothing because inverting is making the denominator as simple as it can be. – logarithm May 14 '19 at 18:47
  • @logarithm Au contraire, I didn't change the original proof at all - I just added the remark about the generalization from $,3\to n,,$ which I thought would be obvious to anyone who grasped the idea that I emphasize. I think you may have missed that since for some reason you don't want to view it in terms of the more general viewpoint that I emphasize (which makes the generalization obvious). – Bill Dubuque May 14 '19 at 18:54
  • The original proof is only computation. The link that the OP had, in which the inverse is presented and verified informs as much as this one without the remarks and the last paragrah. The substance is what you have been adding in that remark and what has been written in the last paragraph. Again, 'make denominator simpler' is not an idea, it is a goal. Finding inverse is a particular case of 'making denominators simpler', but that doesn't say anything about how or why it can be done. This is why I said that what was written kept the actual ideas hidden. – logarithm May 14 '19 at 19:15
  • @logarithm The idea proves quite useful. I've taught it for decades and it works quite well for my students. Follow the links to learn more if you so desire. Best of luck in your studies. – Bill Dubuque May 14 '19 at 19:45
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    Arrogant. I am probably older than you with as much if no more experience than you teaching and doing research. Hiding the key ideas that make a proof work doesn't make anything work. Better luck to your students. Using colored chalk is not what makes teaching better. – logarithm May 14 '19 at 19:55
  • @logarithm Sorry to hear that ideas and colors irk you. Best of luck with that too. – Bill Dubuque May 14 '19 at 20:22
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    Not telling the students the actual ideas is what I don't like. Hiding the ideas is done either when you don't know them or to present a proof in a way that looks like magic. The latter is a valid reason in some cases, and is only successful if the resulting argument actually looks more elegant. Not the case in the computation that you did either. – logarithm May 14 '19 at 20:35
  • @logarithm Being new here, you probably are not aware that many of my answers are purposely designed to be hints - in the hope that they will spark the reader to discover the key ideas on their own (which usually results in a far better learning experience). I use colors to help guide the reader in making such inferences, e.g. highlighting the use of $\rm \color{#c00}{nilpotency}$ and $\color{#0a0}{\rm invertibly\ simpler}$ above. Of course a single hint can't work for all readers, so when someone asks for clarification I am happy to oblige - as above. – Bill Dubuque May 14 '19 at 20:56
  • For the record, the formal power series approach is here. but beware that many folks (including some professional mathematicians) don't have a good grasp of formal (vs. functional) power series. For a typical example see the confusion here. (if you can't see joriki's deleted answer then read his comments). This is one reason why I avoided using formal power series. – Bill Dubuque May 15 '19 at 00:38
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Lets show that $\frac{1}{3+6x+4x^2}=3+2x$ in $\Bbb Z_8[x]$ by our division method that we learned in primary school (I couldn't make this shape tough: |____ ): $$ \begin{array}{c|l} 1&\hline 3+6x+4x^2\\ -(9+18x+12x^2)&\hline\color{red}{3+2x}\\ 6x+4x^2&\hline\\ -(6x+4x^2)&\hline\\ 0&\hline \end{array} $$

Bob Dobbs
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