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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?

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    $\begingroup$ 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 $\mathbb{Z}/8$. $\endgroup$ – user10354138 May 14 at 11:36
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    $\begingroup$ $\frac{1}{3+6x+4x^2}=\frac{1}{3}\frac{1}{1+(6x+4x^2)/3}=\frac{1}{3}\left(1-(6x+4x^2)/3+((6x+4x^2)/3)^2+((6x+4x^2)/3)^3+...\right)$. Compute that $1/3=3$ and those powers, which are not as many as they look. $\endgroup$ – logarithm May 14 at 11:44
  • $\begingroup$ @user10354138 What theorem about nilpotent elements is being applied here? I'd need to understand its proof first. $\endgroup$ – Akash Gaur May 14 at 13:39
  • $\begingroup$ @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. $\endgroup$ – logarithm May 14 at 13:46
  • $\begingroup$ @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? $\endgroup$ – Akash Gaur May 14 at 14:51
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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.

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

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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 $b=3$.

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  • $\begingroup$ Thanks. What values can the degree of the inverse polynomial take? $\endgroup$ – Akash Gaur May 14 at 14:06
  • $\begingroup$ Arbitrarily large as long as the coefficients cancel. $\endgroup$ – Wuestenfux May 14 at 14:09
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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 $(4x^2+6x+3)(2x+3)\equiv 1\mod 8$

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  • $\begingroup$ What is $y$ ? the constant term of $p(x)$? $\endgroup$ – J. W. Tanner May 14 at 12:23
  • $\begingroup$ @J.W.Tanner Yes, I meant that when we have a linear polynomial ax+y, then y has to be 3. $\endgroup$ – Cornman May 14 at 12:42
  • $\begingroup$ Perhaps you should edit to make that obvious $\endgroup$ – J. W. Tanner May 14 at 12:48
  • $\begingroup$ @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. $\endgroup$ – Cornman May 14 at 13:01
  • $\begingroup$ @Cornman Thanks. What values can the degree of the inverse polynomial take? $\endgroup$ – Akash Gaur May 14 at 14:07
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By the idea here we find a simpler multiple $\,\color{#0a0}{3^{\large 3}}$ of the denominator $3+f\ $ (here $f = 4x+6x^2)$

$\ \ \ \underbrace{{2\mid f}}_{\large \color{#c00}{ 2^{\LARGE 3}\ \mid \ f^{\LARGE 3}}}\!\Rightarrow\, \bmod\,\color{#c00}{2^{\large 3}}\!:\,\ \dfrac{1}{3+f} = \dfrac{3^{\large 2}\!-3f+f^{\large 2}}{\color{#0a0}{3^{\large 3}}+\color{#c00}{f^{\large 3}}} \equiv \dfrac{1-3f+f^{\large 2}} {\color{#0a0}3\,+\,\color{#c00}0}\equiv 3-f+3f^{\large 2} \equiv\, 2x+3$

Remark $ $ This idea generalizes: to invert $\, a - f\,$ where $a$ is invertible, say $\,ab = 1\,$ and $f$ is nilpotent $\,f^n = 0\,$ it suffices to invert its simpler multiple $\, a^n-f^n = a^n,\,$ which has obvious inverse $\,b^n,\,$ i.e.

$$\color{#0a0}{ab=1},\, \color{#c00}{f^{\large n} = 0}\ \Rightarrow\ \dfrac{1}{a-f} = \dfrac{a^{\large n-1}+\cdots + f^{\large n-1}}{\color{#0a0}{a^{\large n}}-\color{#c00}{f^{\large n}}} = \color{#0a0}{b^{\large n}}(a^{\large n-1}+\cdots + f^{\large n-1})\qquad \qquad\qquad\ $$

Generally it is easy 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 - similar to above).

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 modular arithmetic.

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  • $\begingroup$ 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. $\endgroup$ – logarithm May 14 at 16:00
  • $\begingroup$ @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. $\endgroup$ – Bill Dubuque May 14 at 16:42
  • $\begingroup$ 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. $\endgroup$ – logarithm May 14 at 17:09
  • $\begingroup$ @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. $\endgroup$ – Bill Dubuque May 14 at 18:18
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    $\begingroup$ 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. $\endgroup$ – logarithm May 14 at 18:47

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